Algebra Examples

$f (x) = 3 x^{3} - 2 x^{2} - 4 x - 3$

Step 1

Find the first derivative of the function.

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By the Sum Rule, the derivative of $3 x^{3} - 2 x^{2} - 4 x - 3$ with respect to $x$ is $\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [- 2 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [- 2 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 3]$

Evaluate $\frac{d}{d x} [3 x^{3}]$ .

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Since $3$ is constant with respect to $x$ , the derivative of $3 x^{3}$ with respect to $x$ is $3 \frac{d}{d x} [x^{3}]$ .

$3 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 2 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 3]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$3 (3 x^{2}) + \frac{d}{d x} [- 2 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 3]$

Multiply $3$ by $3$ .

$9 x^{2} + \frac{d}{d x} [- 2 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 3]$

Evaluate $\frac{d}{d x} [- 2 x^{2}]$ .

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Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x^{2}$ with respect to $x$ is $- 2 \frac{d}{d x} [x^{2}]$ .

$9 x^{2} - 2 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 3]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$9 x^{2} - 2 (2 x) + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 3]$

Multiply $2$ by $- 2$ .

$9 x^{2} - 4 x + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 3]$

Evaluate $\frac{d}{d x} [- 4 x]$ .

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Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$9 x^{2} - 4 x - 4 \frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$9 x^{2} - 4 x - 4 \cdot 1 + \frac{d}{d x} [- 3]$

Multiply $- 4$ by $1$ .

$9 x^{2} - 4 x - 4 + \frac{d}{d x} [- 3]$

Differentiate using the Constant Rule.

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Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$9 x^{2} - 4 x - 4 + 0$

Add $9 x^{2} - 4 x - 4$ and $0$ .

$9 x^{2} - 4 x - 4$

Step 2

Find the second derivative of the function.

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By the Sum Rule, the derivative of $9 x^{2} - 4 x - 4$ with respect to $x$ is $\frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 4]$ .

$f'' (x) = \frac{d}{d x} (9 x^{2}) + \frac{d}{d x} (- 4 x) + \frac{d}{d x} (- 4)$

Evaluate $\frac{d}{d x} [9 x^{2}]$ .

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Since $9$ is constant with respect to $x$ , the derivative of $9 x^{2}$ with respect to $x$ is $9 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 9 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 4 x) + \frac{d}{d x} (- 4)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = 9 (2 x) + \frac{d}{d x} (- 4 x) + \frac{d}{d x} (- 4)$

Multiply $2$ by $9$ .

$f'' (x) = 18 x + \frac{d}{d x} (- 4 x) + \frac{d}{d x} (- 4)$

Evaluate $\frac{d}{d x} [- 4 x]$ .

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Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$f'' (x) = 18 x - 4 \frac{d}{d x} x + \frac{d}{d x} (- 4)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = 18 x - 4 \cdot 1 + \frac{d}{d x} (- 4)$

Multiply $- 4$ by $1$ .

$f'' (x) = 18 x - 4 + \frac{d}{d x} (- 4)$

Differentiate using the Constant Rule.

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Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$f'' (x) = 18 x - 4 + 0$

Add $18 x - 4$ and $0$ .

$f'' (x) = 18 x - 4$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$9 x^{2} - 4 x - 4 = 0$

Step 4

Find the first derivative.

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Find the first derivative.

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By the Sum Rule, the derivative of $3 x^{3} - 2 x^{2} - 4 x - 3$ with respect to $x$ is $\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [- 2 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [- 3]$ .

$f' (x) = \frac{d}{d x} (3 x^{3}) + \frac{d}{d x} (- 2 x^{2}) + \frac{d}{d x} (- 4 x) + \frac{d}{d x} (- 3)$

Evaluate $\frac{d}{d x} [3 x^{3}]$ .

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Since $3$ is constant with respect to $x$ , the derivative of $3 x^{3}$ with respect to $x$ is $3 \frac{d}{d x} [x^{3}]$ .

$f' (x) = 3 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 2 x^{2}) + \frac{d}{d x} (- 4 x) + \frac{d}{d x} (- 3)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f' (x) = 3 (3 x^{2}) + \frac{d}{d x} (- 2 x^{2}) + \frac{d}{d x} (- 4 x) + \frac{d}{d x} (- 3)$

Multiply $3$ by $3$ .

$f' (x) = 9 x^{2} + \frac{d}{d x} (- 2 x^{2}) + \frac{d}{d x} (- 4 x) + \frac{d}{d x} (- 3)$

Evaluate $\frac{d}{d x} [- 2 x^{2}]$ .

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Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x^{2}$ with respect to $x$ is $- 2 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 9 x^{2} - 2 \frac{d}{d x} x^{2} + \frac{d}{d x} (- 4 x) + \frac{d}{d x} (- 3)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f' (x) = 9 x^{2} - 2 (2 x) + \frac{d}{d x} (- 4 x) + \frac{d}{d x} (- 3)$

Multiply $2$ by $- 2$ .

$f' (x) = 9 x^{2} - 4 x + \frac{d}{d x} (- 4 x) + \frac{d}{d x} (- 3)$

Evaluate $\frac{d}{d x} [- 4 x]$ .

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Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$f' (x) = 9 x^{2} - 4 x - 4 \frac{d}{d x} x + \frac{d}{d x} (- 3)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f' (x) = 9 x^{2} - 4 x - 4 \cdot 1 + \frac{d}{d x} (- 3)$

Multiply $- 4$ by $1$ .

$f' (x) = 9 x^{2} - 4 x - 4 + \frac{d}{d x} (- 3)$

Differentiate using the Constant Rule.

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Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$f' (x) = 9 x^{2} - 4 x - 4 + 0$

Add $9 x^{2} - 4 x - 4$ and $0$ .

$f' (x) = 9 x^{2} - 4 x - 4$

The first derivative of $f (x)$ with respect to $x$ is $9 x^{2} - 4 x - 4$ .

$9 x^{2} - 4 x - 4$

Step 5

Set the first derivative equal to $0$ then solve the equation $9 x^{2} - 4 x - 4 = 0$ .

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Set the first derivative equal to $0$ .

$9 x^{2} - 4 x - 4 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 9$ , $b = - 4$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (9 \cdot - 4)}}{2 \cdot 9}$

Simplify.

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Simplify the numerator.

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Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 9 \cdot - 4}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot - 4$ .

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Multiply $- 4$ by $9$ .

$x = \frac{4 \pm \sqrt{16 - 36 \cdot - 4}}{2 \cdot 9}$

Multiply $- 36$ by $- 4$ .

$x = \frac{4 \pm \sqrt{16 + 144}}{2 \cdot 9}$

Add $16$ and $144$ .

$x = \frac{4 \pm \sqrt{160}}{2 \cdot 9}$

Rewrite $160$ as $4^{2} \cdot 10$ .

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Factor $16$ out of $160$ .

$x = \frac{4 \pm \sqrt{16 (10)}}{2 \cdot 9}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{4 \pm \sqrt{4^{2} \cdot 10}}{2 \cdot 9}$

Pull terms out from under the radical.

$x = \frac{4 \pm 4 \sqrt{10}}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{4 \pm 4 \sqrt{10}}{18}$

Simplify $\frac{4 \pm 4 \sqrt{10}}{18}$ .

$x = \frac{2 \pm 2 \sqrt{10}}{9}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 9 \cdot - 4}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot - 4$ .

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Multiply $- 4$ by $9$ .

$x = \frac{4 \pm \sqrt{16 - 36 \cdot - 4}}{2 \cdot 9}$

Multiply $- 36$ by $- 4$ .

$x = \frac{4 \pm \sqrt{16 + 144}}{2 \cdot 9}$

Add $16$ and $144$ .

$x = \frac{4 \pm \sqrt{160}}{2 \cdot 9}$

Rewrite $160$ as $4^{2} \cdot 10$ .

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Factor $16$ out of $160$ .

$x = \frac{4 \pm \sqrt{16 (10)}}{2 \cdot 9}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{4 \pm \sqrt{4^{2} \cdot 10}}{2 \cdot 9}$

Pull terms out from under the radical.

$x = \frac{4 \pm 4 \sqrt{10}}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{4 \pm 4 \sqrt{10}}{18}$

Simplify $\frac{4 \pm 4 \sqrt{10}}{18}$ .

$x = \frac{2 \pm 2 \sqrt{10}}{9}$

Change the $\pm$ to $+$ .

$x = \frac{2 + 2 \sqrt{10}}{9}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $- 4$ to the power of $2$ .

$x = \frac{4 \pm \sqrt{16 - 4 \cdot 9 \cdot - 4}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot - 4$ .

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Multiply $- 4$ by $9$ .

$x = \frac{4 \pm \sqrt{16 - 36 \cdot - 4}}{2 \cdot 9}$

Multiply $- 36$ by $- 4$ .

$x = \frac{4 \pm \sqrt{16 + 144}}{2 \cdot 9}$

Add $16$ and $144$ .

$x = \frac{4 \pm \sqrt{160}}{2 \cdot 9}$

Rewrite $160$ as $4^{2} \cdot 10$ .

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Factor $16$ out of $160$ .

$x = \frac{4 \pm \sqrt{16 (10)}}{2 \cdot 9}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{4 \pm \sqrt{4^{2} \cdot 10}}{2 \cdot 9}$

Pull terms out from under the radical.

$x = \frac{4 \pm 4 \sqrt{10}}{2 \cdot 9}$

Multiply $2$ by $9$ .

$x = \frac{4 \pm 4 \sqrt{10}}{18}$

Simplify $\frac{4 \pm 4 \sqrt{10}}{18}$ .

$x = \frac{2 \pm 2 \sqrt{10}}{9}$

Change the $\pm$ to $-$ .

$x = \frac{2 - 2 \sqrt{10}}{9}$

The final answer is the combination of both solutions.

$x = \frac{2 + 2 \sqrt{10}}{9}, \frac{2 - 2 \sqrt{10}}{9}$

Step 6

Find the values where the derivative is undefined.

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The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = \frac{2 + 2 \sqrt{10}}{9}, \frac{2 - 2 \sqrt{10}}{9}$

Step 8

Evaluate the second derivative at $x = \frac{2 + 2 \sqrt{10}}{9}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$18 (\frac{2 + 2 \sqrt{10}}{9}) - 4$

Step 9

Evaluate the second derivative.

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Simplify each term.

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Cancel the common factor of $9$ .

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Factor $9$ out of $18$ .

$9 (2) \frac{2 + 2 \sqrt{10}}{9} - 4$

Cancel the common factor.

$9 \cdot 2 \frac{2 + 2 \sqrt{10}}{9} - 4$

Rewrite the expression.

$2 (2 + 2 \sqrt{10}) - 4$

Apply the distributive property.

$2 \cdot 2 + 2 (2 \sqrt{10}) - 4$

Multiply $2$ by $2$ .

$4 + 2 (2 \sqrt{10}) - 4$

Multiply $2$ by $2$ .

$4 + 4 \sqrt{10} - 4$

Simplify by subtracting numbers.

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Subtract $4$ from $4$ .

$0 + 4 \sqrt{10}$

Add $0$ and $4 \sqrt{10}$ .

$4 \sqrt{10}$

Step 10

$x = \frac{2 + 2 \sqrt{10}}{9}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{2 + 2 \sqrt{10}}{9}$ is a local minimum

Step 11

Find the y-value when $x = \frac{2 + 2 \sqrt{10}}{9}$ .

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Replace the variable $x$ with $\frac{2 + 2 \sqrt{10}}{9}$ in the expression.

$f (\frac{2 + 2 \sqrt{10}}{9}) = 3 {(\frac{2 + 2 \sqrt{10}}{9})}^{3} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Simplify the result.

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Simplify each term.

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Apply the product rule to $\frac{2 + 2 \sqrt{10}}{9}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = 3 (\frac{{(2 + 2 \sqrt{10})}^{3}}{9^{3}}) - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Raise $9$ to the power of $3$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = 3 (\frac{{(2 + 2 \sqrt{10})}^{3}}{729}) - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Cancel the common factor of $3$ .

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Factor $3$ out of $729$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = 3 (\frac{{(2 + 2 \sqrt{10})}^{3}}{3 (243)}) - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Cancel the common factor.

$f (\frac{2 + 2 \sqrt{10}}{9}) = 3 (\frac{{(2 + 2 \sqrt{10})}^{3}}{3 \cdot 243}) - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Rewrite the expression.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{{(2 + 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Use the Binomial Theorem.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{2^{3} + 3 \cdot (2^{2} (2 \sqrt{10})) + 3 \cdot (2 {(2 \sqrt{10})}^{2}) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Simplify each term.

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Raise $2$ to the power of $3$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 3 \cdot (2^{2} (2 \sqrt{10})) + 3 \cdot (2 {(2 \sqrt{10})}^{2}) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Multiply $2^{2}$ by $2$ by adding the exponents.

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Move $2$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 3 \cdot ((2 \cdot 2^{2}) \sqrt{10}) + 3 \cdot (2 {(2 \sqrt{10})}^{2}) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Multiply $2$ by $2^{2}$ .

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Raise $2$ to the power of $1$ .

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 3 \cdot (2^{1 + 2} \sqrt{10}) + 3 \cdot (2 {(2 \sqrt{10})}^{2}) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Add $1$ and $2$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 3 \cdot (2^{3} \sqrt{10}) + 3 \cdot (2 {(2 \sqrt{10})}^{2}) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Raise $2$ to the power of $3$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 3 \cdot (8 \sqrt{10}) + 3 \cdot (2 {(2 \sqrt{10})}^{2}) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Multiply $3$ by $8$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 3 \cdot (2 {(2 \sqrt{10})}^{2}) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Multiply $3$ by $2$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 6 {(2 \sqrt{10})}^{2} + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Apply the product rule to $2 \sqrt{10}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 6 (2^{2} {\sqrt{10}}^{2}) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Raise $2$ to the power of $2$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 6 (4 {\sqrt{10}}^{2}) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Rewrite ${\sqrt{10}}^{2}$ as $10$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{10}$ as $10^{\frac{1}{2}}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 6 (4 {(10^{\frac{1}{2}})}^{2}) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 6 (4 \cdot 10^{\frac{1}{2} \cdot 2}) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Combine $\frac{1}{2}$ and $2$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 6 (4 \cdot 10^{\frac{2}{2}}) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Cancel the common factor of $2$ .

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Cancel the common factor.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 6 (4 \cdot 10^{\frac{2}{2}}) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Rewrite the expression.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 6 \cdot 4 \cdot 10 + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Evaluate the exponent.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 6 (4 \cdot 10) + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Multiply $6 (4 \cdot 10)$ .

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Multiply $4$ by $10$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 6 \cdot 40 + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Multiply $6$ by $40$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 240 + {(2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Apply the product rule to $2 \sqrt{10}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 240 + 2^{3} {\sqrt{10}}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Raise $2$ to the power of $3$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 240 + 8 {\sqrt{10}}^{3}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Rewrite ${\sqrt{10}}^{3}$ as ${(10^{3})}^{\frac{1}{2}}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 240 + 8 \sqrt{10^{3}}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Raise $10$ to the power of $3$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 240 + 8 \sqrt{1000}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Rewrite $1000$ as $10^{2} \cdot 10$ .

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Factor $100$ out of $1000$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 240 + 8 \sqrt{100 (10)}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Rewrite $100$ as $10^{2}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 240 + 8 \sqrt{10^{2} \cdot 10}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Pull terms out from under the radical.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 240 + 8 (10 \sqrt{10})}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Multiply $10$ by $8$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{8 + 24 \sqrt{10} + 240 + 80 \sqrt{10}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Add $8$ and $240$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 24 \sqrt{10} + 80 \sqrt{10}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Add $24 \sqrt{10}$ and $80 \sqrt{10}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 {(\frac{2 + 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Apply the product rule to $\frac{2 + 2 \sqrt{10}}{9}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{{(2 + 2 \sqrt{10})}^{2}}{9^{2}} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Raise $9$ to the power of $2$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{{(2 + 2 \sqrt{10})}^{2}}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Rewrite ${(2 + 2 \sqrt{10})}^{2}$ as $(2 + 2 \sqrt{10}) (2 + 2 \sqrt{10})$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{(2 + 2 \sqrt{10}) (2 + 2 \sqrt{10})}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Expand $(2 + 2 \sqrt{10}) (2 + 2 \sqrt{10})$ using the FOIL Method.

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Apply the distributive property.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{2 (2 + 2 \sqrt{10}) + 2 \sqrt{10} (2 + 2 \sqrt{10})}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Apply the distributive property.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{2 \cdot 2 + 2 (2 \sqrt{10}) + 2 \sqrt{10} (2 + 2 \sqrt{10})}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Apply the distributive property.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{2 \cdot 2 + 2 (2 \sqrt{10}) + 2 \sqrt{10} \cdot 2 + 2 \sqrt{10} (2 \sqrt{10})}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Simplify and combine like terms.

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Simplify each term.

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Multiply $2$ by $2$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 2 (2 \sqrt{10}) + 2 \sqrt{10} \cdot 2 + 2 \sqrt{10} (2 \sqrt{10})}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Multiply $2$ by $2$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 2 \sqrt{10} \cdot 2 + 2 \sqrt{10} (2 \sqrt{10})}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Multiply $2$ by $2$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 4 \sqrt{10} + 2 \sqrt{10} (2 \sqrt{10})}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Multiply $2 \sqrt{10} (2 \sqrt{10})$ .

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Multiply $2$ by $2$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 4 \sqrt{10} + 4 \sqrt{10} \sqrt{10}}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Raise $\sqrt{10}$ to the power of $1$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 4 \sqrt{10} + 4 (\sqrt{10} \sqrt{10})}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Raise $\sqrt{10}$ to the power of $1$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 4 \sqrt{10} + 4 (\sqrt{10} \sqrt{10})}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 4 \sqrt{10} + 4 {\sqrt{10}}^{1 + 1}}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Add $1$ and $1$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 4 \sqrt{10} + 4 {\sqrt{10}}^{2}}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Rewrite ${\sqrt{10}}^{2}$ as $10$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{10}$ as $10^{\frac{1}{2}}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 4 \sqrt{10} + 4 {(10^{\frac{1}{2}})}^{2}}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 4 \sqrt{10} + 4 \cdot 10^{\frac{1}{2} \cdot 2}}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Combine $\frac{1}{2}$ and $2$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 4 \sqrt{10} + 4 \cdot 10^{\frac{2}{2}}}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Cancel the common factor of $2$ .

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Cancel the common factor.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 4 \sqrt{10} + 4 \cdot 10^{\frac{2}{2}}}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Rewrite the expression.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 4 \sqrt{10} + 4 \cdot 10}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Evaluate the exponent.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 4 \sqrt{10} + 4 \cdot 10}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Multiply $4$ by $10$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{4 + 4 \sqrt{10} + 4 \sqrt{10} + 40}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Add $4$ and $40$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{44 + 4 \sqrt{10} + 4 \sqrt{10}}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Add $4 \sqrt{10}$ and $4 \sqrt{10}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - 2 \frac{44 + 8 \sqrt{10}}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Combine $- 2$ and $\frac{44 + 8 \sqrt{10}}{81}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} + \frac{- 2 (44 + 8 \sqrt{10})}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Move the negative in front of the fraction.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - \frac{(2) (44 + 8 \sqrt{10})}{81} - 4 \frac{2 + 2 \sqrt{10}}{9} - 3$

Combine $- 4$ and $\frac{2 + 2 \sqrt{10}}{9}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - \frac{2 (44 + 8 \sqrt{10})}{81} + \frac{- 4 (2 + 2 \sqrt{10})}{9} - 3$

Move the negative in front of the fraction.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - \frac{2 (44 + 8 \sqrt{10})}{81} - \frac{4 (2 + 2 \sqrt{10})}{9} - 3$

Find the common denominator.

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Multiply $\frac{2 (44 + 8 \sqrt{10})}{81}$ by $\frac{3}{3}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - (\frac{2 (44 + 8 \sqrt{10})}{81} \cdot \frac{3}{3}) - \frac{4 (2 + 2 \sqrt{10})}{9} - 3$

Multiply $\frac{2 (44 + 8 \sqrt{10})}{81}$ by $\frac{3}{3}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - \frac{2 (44 + 8 \sqrt{10}) \cdot 3}{81 \cdot 3} - \frac{4 (2 + 2 \sqrt{10})}{9} - 3$

Multiply $\frac{4 (2 + 2 \sqrt{10})}{9}$ by $\frac{27}{27}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - \frac{2 (44 + 8 \sqrt{10}) \cdot 3}{81 \cdot 3} - (\frac{4 (2 + 2 \sqrt{10})}{9} \cdot \frac{27}{27}) - 3$

Multiply $\frac{4 (2 + 2 \sqrt{10})}{9}$ by $\frac{27}{27}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - \frac{2 (44 + 8 \sqrt{10}) \cdot 3}{81 \cdot 3} - \frac{4 (2 + 2 \sqrt{10}) \cdot 27}{9 \cdot 27} - 3$

Write $- 3$ as a fraction with denominator $1$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - \frac{2 (44 + 8 \sqrt{10}) \cdot 3}{81 \cdot 3} - \frac{4 (2 + 2 \sqrt{10}) \cdot 27}{9 \cdot 27} + \frac{- 3}{1}$

Multiply $\frac{- 3}{1}$ by $\frac{243}{243}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - \frac{2 (44 + 8 \sqrt{10}) \cdot 3}{81 \cdot 3} - \frac{4 (2 + 2 \sqrt{10}) \cdot 27}{9 \cdot 27} + \frac{- 3}{1} \cdot \frac{243}{243}$

Multiply $\frac{- 3}{1}$ by $\frac{243}{243}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - \frac{2 (44 + 8 \sqrt{10}) \cdot 3}{81 \cdot 3} - \frac{4 (2 + 2 \sqrt{10}) \cdot 27}{9 \cdot 27} + \frac{- 3 \cdot 243}{243}$

Reorder the factors of $81 \cdot 3$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - \frac{2 (44 + 8 \sqrt{10}) \cdot 3}{3 \cdot 81} - \frac{4 (2 + 2 \sqrt{10}) \cdot 27}{9 \cdot 27} + \frac{- 3 \cdot 243}{243}$

Multiply $3$ by $81$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - \frac{2 (44 + 8 \sqrt{10}) \cdot 3}{243} - \frac{4 (2 + 2 \sqrt{10}) \cdot 27}{9 \cdot 27} + \frac{- 3 \cdot 243}{243}$

Multiply $9$ by $27$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10}}{243} - \frac{2 (44 + 8 \sqrt{10}) \cdot 3}{243} - \frac{4 (2 + 2 \sqrt{10}) \cdot 27}{243} + \frac{- 3 \cdot 243}{243}$

Combine the numerators over the common denominator.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} - 2 (44 + 8 \sqrt{10}) \cdot 3 - 4 (2 + 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Simplify each term.

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Apply the distributive property.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} + (- 2 \cdot 44 - 2 (8 \sqrt{10})) \cdot 3 - 4 (2 + 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Multiply $- 2$ by $44$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} + (- 88 - 2 (8 \sqrt{10})) \cdot 3 - 4 (2 + 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Multiply $8$ by $- 2$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} + (- 88 - 16 \sqrt{10}) \cdot 3 - 4 (2 + 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Apply the distributive property.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} - 88 \cdot 3 - 16 \sqrt{10} \cdot 3 - 4 (2 + 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Multiply $- 88$ by $3$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} - 264 - 16 \sqrt{10} \cdot 3 - 4 (2 + 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Multiply $3$ by $- 16$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} - 264 - 48 \sqrt{10} - 4 (2 + 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Apply the distributive property.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} - 264 - 48 \sqrt{10} + (- 4 \cdot 2 - 4 (2 \sqrt{10})) \cdot 27 - 3 \cdot 243}{243}$

Multiply $- 4$ by $2$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} - 264 - 48 \sqrt{10} + (- 8 - 4 (2 \sqrt{10})) \cdot 27 - 3 \cdot 243}{243}$

Multiply $2$ by $- 4$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} - 264 - 48 \sqrt{10} + (- 8 - 8 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Apply the distributive property.

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} - 264 - 48 \sqrt{10} - 8 \cdot 27 - 8 \sqrt{10} \cdot 27 - 3 \cdot 243}{243}$

Multiply $- 8$ by $27$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} - 264 - 48 \sqrt{10} - 216 - 8 \sqrt{10} \cdot 27 - 3 \cdot 243}{243}$

Multiply $27$ by $- 8$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} - 264 - 48 \sqrt{10} - 216 - 216 \sqrt{10} - 3 \cdot 243}{243}$

Multiply $- 3$ by $243$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{248 + 104 \sqrt{10} - 264 - 48 \sqrt{10} - 216 - 216 \sqrt{10} - 729}{243}$

Simplify terms.

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Subtract $264$ from $248$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{- 16 + 104 \sqrt{10} - 48 \sqrt{10} - 216 - 216 \sqrt{10} - 729}{243}$

Simplify by subtracting numbers.

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Subtract $216$ from $- 16$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{- 232 + 104 \sqrt{10} - 48 \sqrt{10} - 216 \sqrt{10} - 729}{243}$

Subtract $729$ from $- 232$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{- 961 + 104 \sqrt{10} - 48 \sqrt{10} - 216 \sqrt{10}}{243}$

Subtract $48 \sqrt{10}$ from $104 \sqrt{10}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{- 961 + 56 \sqrt{10} - 216 \sqrt{10}}{243}$

Subtract $216 \sqrt{10}$ from $56 \sqrt{10}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{- 961 - 160 \sqrt{10}}{243}$

Rewrite $- 961$ as $- 1 (961)$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{- 1 \cdot 961 - 160 \sqrt{10}}{243}$

Factor $- 1$ out of $- 160 \sqrt{10}$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{- 1 \cdot 961 - (160 \sqrt{10})}{243}$

Factor $- 1$ out of $- 1 (961) - (160 \sqrt{10})$ .

$f (\frac{2 + 2 \sqrt{10}}{9}) = \frac{- 1 (961 + 160 \sqrt{10})}{243}$

Move the negative in front of the fraction.

$f (\frac{2 + 2 \sqrt{10}}{9}) = - \frac{961 + 160 \sqrt{10}}{243}$

The final answer is $- \frac{961 + 160 \sqrt{10}}{243}$ .

$y = - \frac{961 + 160 \sqrt{10}}{243}$

Step 12

Evaluate the second derivative at $x = \frac{2 - 2 \sqrt{10}}{9}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$18 (\frac{2 - 2 \sqrt{10}}{9}) - 4$

Step 13

Evaluate the second derivative.

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Simplify each term.

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Cancel the common factor of $9$ .

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Factor $9$ out of $18$ .

$9 (2) \frac{2 - 2 \sqrt{10}}{9} - 4$

Cancel the common factor.

$9 \cdot 2 \frac{2 - 2 \sqrt{10}}{9} - 4$

Rewrite the expression.

$2 (2 - 2 \sqrt{10}) - 4$

Apply the distributive property.

$2 \cdot 2 + 2 (- 2 \sqrt{10}) - 4$

Multiply $2$ by $2$ .

$4 + 2 (- 2 \sqrt{10}) - 4$

Multiply $- 2$ by $2$ .

$4 - 4 \sqrt{10} - 4$

Simplify by subtracting numbers.

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Subtract $4$ from $4$ .

$0 - 4 \sqrt{10}$

Subtract $4 \sqrt{10}$ from $0$ .

$- 4 \sqrt{10}$

Step 14

$x = \frac{2 - 2 \sqrt{10}}{9}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{2 - 2 \sqrt{10}}{9}$ is a local maximum

Step 15

Find the y-value when $x = \frac{2 - 2 \sqrt{10}}{9}$ .

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Replace the variable $x$ with $\frac{2 - 2 \sqrt{10}}{9}$ in the expression.

$f (\frac{2 - 2 \sqrt{10}}{9}) = 3 {(\frac{2 - 2 \sqrt{10}}{9})}^{3} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Simplify the result.

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Simplify each term.

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Apply the product rule to $\frac{2 - 2 \sqrt{10}}{9}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = 3 (\frac{{(2 - 2 \sqrt{10})}^{3}}{9^{3}}) - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Raise $9$ to the power of $3$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = 3 (\frac{{(2 - 2 \sqrt{10})}^{3}}{729}) - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Cancel the common factor of $3$ .

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Factor $3$ out of $729$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = 3 (\frac{{(2 - 2 \sqrt{10})}^{3}}{3 (243)}) - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Cancel the common factor.

$f (\frac{2 - 2 \sqrt{10}}{9}) = 3 (\frac{{(2 - 2 \sqrt{10})}^{3}}{3 \cdot 243}) - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Rewrite the expression.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{{(2 - 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Use the Binomial Theorem.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{2^{3} + 3 \cdot (2^{2} (- 2 \sqrt{10})) + 3 \cdot (2 {(- 2 \sqrt{10})}^{2}) + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Simplify each term.

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Raise $2$ to the power of $3$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 + 3 \cdot (2^{2} (- 2 \sqrt{10})) + 3 \cdot (2 {(- 2 \sqrt{10})}^{2}) + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Raise $2$ to the power of $2$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 + 3 \cdot (4 (- 2 \sqrt{10})) + 3 \cdot (2 {(- 2 \sqrt{10})}^{2}) + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Multiply $3$ by $4$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 + 12 (- 2 \sqrt{10}) + 3 \cdot (2 {(- 2 \sqrt{10})}^{2}) + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Multiply $- 2$ by $12$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 3 \cdot (2 {(- 2 \sqrt{10})}^{2}) + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Multiply $3$ by $2$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 6 {(- 2 \sqrt{10})}^{2} + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Apply the product rule to $- 2 \sqrt{10}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 6 ({(- 2)}^{2} {\sqrt{10}}^{2}) + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Raise $- 2$ to the power of $2$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 6 (4 {\sqrt{10}}^{2}) + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Rewrite ${\sqrt{10}}^{2}$ as $10$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{10}$ as $10^{\frac{1}{2}}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 6 (4 {(10^{\frac{1}{2}})}^{2}) + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 6 (4 \cdot 10^{\frac{1}{2} \cdot 2}) + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Combine $\frac{1}{2}$ and $2$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 6 (4 \cdot 10^{\frac{2}{2}}) + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Cancel the common factor of $2$ .

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Cancel the common factor.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 6 (4 \cdot 10^{\frac{2}{2}}) + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Rewrite the expression.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 6 \cdot 4 \cdot 10 + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Evaluate the exponent.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 6 (4 \cdot 10) + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Multiply $6 (4 \cdot 10)$ .

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Multiply $4$ by $10$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 6 \cdot 40 + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Multiply $6$ by $40$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 240 + {(- 2 \sqrt{10})}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Apply the product rule to $- 2 \sqrt{10}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 240 + {(- 2)}^{3} {\sqrt{10}}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Raise $- 2$ to the power of $3$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 240 - 8 {\sqrt{10}}^{3}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Rewrite ${\sqrt{10}}^{3}$ as ${(10^{3})}^{\frac{1}{2}}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 240 - 8 \sqrt{10^{3}}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Raise $10$ to the power of $3$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 240 - 8 \sqrt{1000}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Rewrite $1000$ as $10^{2} \cdot 10$ .

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Factor $100$ out of $1000$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 240 - 8 \sqrt{100 (10)}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Rewrite $100$ as $10^{2}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 240 - 8 \sqrt{10^{2} \cdot 10}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Pull terms out from under the radical.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 240 - 8 (10 \sqrt{10})}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Multiply $10$ by $- 8$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{8 - 24 \sqrt{10} + 240 - 80 \sqrt{10}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Add $8$ and $240$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 24 \sqrt{10} - 80 \sqrt{10}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Subtract $80 \sqrt{10}$ from $- 24 \sqrt{10}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 {(\frac{2 - 2 \sqrt{10}}{9})}^{2} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Apply the product rule to $\frac{2 - 2 \sqrt{10}}{9}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{{(2 - 2 \sqrt{10})}^{2}}{9^{2}} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Raise $9$ to the power of $2$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{{(2 - 2 \sqrt{10})}^{2}}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Rewrite ${(2 - 2 \sqrt{10})}^{2}$ as $(2 - 2 \sqrt{10}) (2 - 2 \sqrt{10})$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{(2 - 2 \sqrt{10}) (2 - 2 \sqrt{10})}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Expand $(2 - 2 \sqrt{10}) (2 - 2 \sqrt{10})$ using the FOIL Method.

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Apply the distributive property.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{2 (2 - 2 \sqrt{10}) - 2 \sqrt{10} (2 - 2 \sqrt{10})}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Apply the distributive property.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{2 \cdot 2 + 2 (- 2 \sqrt{10}) - 2 \sqrt{10} (2 - 2 \sqrt{10})}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Apply the distributive property.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{2 \cdot 2 + 2 (- 2 \sqrt{10}) - 2 \sqrt{10} \cdot 2 - 2 \sqrt{10} (- 2 \sqrt{10})}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Simplify and combine like terms.

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Simplify each term.

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Multiply $2$ by $2$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 + 2 (- 2 \sqrt{10}) - 2 \sqrt{10} \cdot 2 - 2 \sqrt{10} (- 2 \sqrt{10})}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Multiply $- 2$ by $2$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 2 \sqrt{10} \cdot 2 - 2 \sqrt{10} (- 2 \sqrt{10})}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Multiply $2$ by $- 2$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 4 \sqrt{10} - 2 \sqrt{10} (- 2 \sqrt{10})}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Multiply $- 2 \sqrt{10} (- 2 \sqrt{10})$ .

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Multiply $- 2$ by $- 2$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 4 \sqrt{10} + 4 \sqrt{10} \sqrt{10}}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Raise $\sqrt{10}$ to the power of $1$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 4 \sqrt{10} + 4 (\sqrt{10} \sqrt{10})}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Raise $\sqrt{10}$ to the power of $1$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 4 \sqrt{10} + 4 (\sqrt{10} \sqrt{10})}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 4 \sqrt{10} + 4 {\sqrt{10}}^{1 + 1}}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Add $1$ and $1$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 4 \sqrt{10} + 4 {\sqrt{10}}^{2}}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Rewrite ${\sqrt{10}}^{2}$ as $10$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{10}$ as $10^{\frac{1}{2}}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 4 \sqrt{10} + 4 {(10^{\frac{1}{2}})}^{2}}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 4 \sqrt{10} + 4 \cdot 10^{\frac{1}{2} \cdot 2}}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Combine $\frac{1}{2}$ and $2$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 4 \sqrt{10} + 4 \cdot 10^{\frac{2}{2}}}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Cancel the common factor of $2$ .

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Cancel the common factor.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 4 \sqrt{10} + 4 \cdot 10^{\frac{2}{2}}}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Rewrite the expression.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 4 \sqrt{10} + 4 \cdot 10}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Evaluate the exponent.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 4 \sqrt{10} + 4 \cdot 10}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Multiply $4$ by $10$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{4 - 4 \sqrt{10} - 4 \sqrt{10} + 40}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Add $4$ and $40$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{44 - 4 \sqrt{10} - 4 \sqrt{10}}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Subtract $4 \sqrt{10}$ from $- 4 \sqrt{10}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - 2 \frac{44 - 8 \sqrt{10}}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Combine $- 2$ and $\frac{44 - 8 \sqrt{10}}{81}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} + \frac{- 2 (44 - 8 \sqrt{10})}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Move the negative in front of the fraction.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - \frac{(2) (44 - 8 \sqrt{10})}{81} - 4 \frac{2 - 2 \sqrt{10}}{9} - 3$

Combine $- 4$ and $\frac{2 - 2 \sqrt{10}}{9}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - \frac{2 (44 - 8 \sqrt{10})}{81} + \frac{- 4 (2 - 2 \sqrt{10})}{9} - 3$

Move the negative in front of the fraction.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - \frac{2 (44 - 8 \sqrt{10})}{81} - \frac{4 (2 - 2 \sqrt{10})}{9} - 3$

Find the common denominator.

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Multiply $\frac{2 (44 - 8 \sqrt{10})}{81}$ by $\frac{3}{3}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - (\frac{2 (44 - 8 \sqrt{10})}{81} \cdot \frac{3}{3}) - \frac{4 (2 - 2 \sqrt{10})}{9} - 3$

Multiply $\frac{2 (44 - 8 \sqrt{10})}{81}$ by $\frac{3}{3}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - \frac{2 (44 - 8 \sqrt{10}) \cdot 3}{81 \cdot 3} - \frac{4 (2 - 2 \sqrt{10})}{9} - 3$

Multiply $\frac{4 (2 - 2 \sqrt{10})}{9}$ by $\frac{27}{27}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - \frac{2 (44 - 8 \sqrt{10}) \cdot 3}{81 \cdot 3} - (\frac{4 (2 - 2 \sqrt{10})}{9} \cdot \frac{27}{27}) - 3$

Multiply $\frac{4 (2 - 2 \sqrt{10})}{9}$ by $\frac{27}{27}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - \frac{2 (44 - 8 \sqrt{10}) \cdot 3}{81 \cdot 3} - \frac{4 (2 - 2 \sqrt{10}) \cdot 27}{9 \cdot 27} - 3$

Write $- 3$ as a fraction with denominator $1$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - \frac{2 (44 - 8 \sqrt{10}) \cdot 3}{81 \cdot 3} - \frac{4 (2 - 2 \sqrt{10}) \cdot 27}{9 \cdot 27} + \frac{- 3}{1}$

Multiply $\frac{- 3}{1}$ by $\frac{243}{243}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - \frac{2 (44 - 8 \sqrt{10}) \cdot 3}{81 \cdot 3} - \frac{4 (2 - 2 \sqrt{10}) \cdot 27}{9 \cdot 27} + \frac{- 3}{1} \cdot \frac{243}{243}$

Multiply $\frac{- 3}{1}$ by $\frac{243}{243}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - \frac{2 (44 - 8 \sqrt{10}) \cdot 3}{81 \cdot 3} - \frac{4 (2 - 2 \sqrt{10}) \cdot 27}{9 \cdot 27} + \frac{- 3 \cdot 243}{243}$

Reorder the factors of $81 \cdot 3$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - \frac{2 (44 - 8 \sqrt{10}) \cdot 3}{3 \cdot 81} - \frac{4 (2 - 2 \sqrt{10}) \cdot 27}{9 \cdot 27} + \frac{- 3 \cdot 243}{243}$

Multiply $3$ by $81$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - \frac{2 (44 - 8 \sqrt{10}) \cdot 3}{243} - \frac{4 (2 - 2 \sqrt{10}) \cdot 27}{9 \cdot 27} + \frac{- 3 \cdot 243}{243}$

Multiply $9$ by $27$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10}}{243} - \frac{2 (44 - 8 \sqrt{10}) \cdot 3}{243} - \frac{4 (2 - 2 \sqrt{10}) \cdot 27}{243} + \frac{- 3 \cdot 243}{243}$

Combine the numerators over the common denominator.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} - 2 (44 - 8 \sqrt{10}) \cdot 3 - 4 (2 - 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Simplify each term.

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Apply the distributive property.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} + (- 2 \cdot 44 - 2 (- 8 \sqrt{10})) \cdot 3 - 4 (2 - 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Multiply $- 2$ by $44$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} + (- 88 - 2 (- 8 \sqrt{10})) \cdot 3 - 4 (2 - 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Multiply $- 8$ by $- 2$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} + (- 88 + 16 \sqrt{10}) \cdot 3 - 4 (2 - 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Apply the distributive property.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} - 88 \cdot 3 + 16 \sqrt{10} \cdot 3 - 4 (2 - 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Multiply $- 88$ by $3$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} - 264 + 16 \sqrt{10} \cdot 3 - 4 (2 - 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Multiply $3$ by $16$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} - 264 + 48 \sqrt{10} - 4 (2 - 2 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Apply the distributive property.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} - 264 + 48 \sqrt{10} + (- 4 \cdot 2 - 4 (- 2 \sqrt{10})) \cdot 27 - 3 \cdot 243}{243}$

Multiply $- 4$ by $2$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} - 264 + 48 \sqrt{10} + (- 8 - 4 (- 2 \sqrt{10})) \cdot 27 - 3 \cdot 243}{243}$

Multiply $- 2$ by $- 4$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} - 264 + 48 \sqrt{10} + (- 8 + 8 \sqrt{10}) \cdot 27 - 3 \cdot 243}{243}$

Apply the distributive property.

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} - 264 + 48 \sqrt{10} - 8 \cdot 27 + 8 \sqrt{10} \cdot 27 - 3 \cdot 243}{243}$

Multiply $- 8$ by $27$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} - 264 + 48 \sqrt{10} - 216 + 8 \sqrt{10} \cdot 27 - 3 \cdot 243}{243}$

Multiply $27$ by $8$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} - 264 + 48 \sqrt{10} - 216 + 216 \sqrt{10} - 3 \cdot 243}{243}$

Multiply $- 3$ by $243$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{248 - 104 \sqrt{10} - 264 + 48 \sqrt{10} - 216 + 216 \sqrt{10} - 729}{243}$

Simplify terms.

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Subtract $264$ from $248$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{- 16 - 104 \sqrt{10} + 48 \sqrt{10} - 216 + 216 \sqrt{10} - 729}{243}$

Simplify by subtracting numbers.

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Subtract $216$ from $- 16$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{- 232 - 104 \sqrt{10} + 48 \sqrt{10} + 216 \sqrt{10} - 729}{243}$

Subtract $729$ from $- 232$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{- 961 - 104 \sqrt{10} + 48 \sqrt{10} + 216 \sqrt{10}}{243}$

Add $- 104 \sqrt{10}$ and $48 \sqrt{10}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{- 961 - 56 \sqrt{10} + 216 \sqrt{10}}{243}$

Add $- 56 \sqrt{10}$ and $216 \sqrt{10}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{- 961 + 160 \sqrt{10}}{243}$

Rewrite $- 961$ as $- 1 (961)$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{- 1 \cdot 961 + 160 \sqrt{10}}{243}$

Factor $- 1$ out of $160 \sqrt{10}$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{- 1 \cdot 961 - (- 160 \sqrt{10})}{243}$

Factor $- 1$ out of $- 1 (961) - (- 160 \sqrt{10})$ .

$f (\frac{2 - 2 \sqrt{10}}{9}) = \frac{- 1 (961 - 160 \sqrt{10})}{243}$

Move the negative in front of the fraction.

$f (\frac{2 - 2 \sqrt{10}}{9}) = - \frac{961 - 160 \sqrt{10}}{243}$

The final answer is $- \frac{961 - 160 \sqrt{10}}{243}$ .

$y = - \frac{961 - 160 \sqrt{10}}{243}$

Step 16

These are the local extrema for $f (x) = 3 x^{3} - 2 x^{2} - 4 x - 3$ .

$(\frac{2 + 2 \sqrt{10}}{9}, - \frac{961 + 160 \sqrt{10}}{243})$ is a local minima

$(\frac{2 - 2 \sqrt{10}}{9}, - \frac{961 - 160 \sqrt{10}}{243})$ is a local maxima

Step 17