Algebra Examples

$x^{5} - 10 x^{3} + 9 x$

Step 1

Write $x^{5} - 10 x^{3} + 9 x$ as a function.

$f (x) = x^{5} - 10 x^{3} + 9 x$

Step 2

Find the first derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of $x^{5} - 10 x^{3} + 9 x$ with respect to $x$ is $\frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 10 x^{3}] + \frac{d}{d x} [9 x]$ .

$\frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 10 x^{3}] + \frac{d}{d x} [9 x]$

Step 2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$5 x^{4} + \frac{d}{d x} [- 10 x^{3}] + \frac{d}{d x} [9 x]$

Step 2.2

Evaluate $\frac{d}{d x} [- 10 x^{3}]$ .

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Step 2.2.1

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10 x^{3}$ with respect to $x$ is $- 10 \frac{d}{d x} [x^{3}]$ .

$5 x^{4} - 10 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [9 x]$

Step 2.2.2

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

$5 x^{4} - 10 (3 x^{2}) + \frac{d}{d x} [9 x]$

Step 2.2.3

Multiply $3$ by $- 10$ .

$5 x^{4} - 30 x^{2} + \frac{d}{d x} [9 x]$

Step 2.3

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

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Step 2.3.1

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

$5 x^{4} - 30 x^{2} + 9 \frac{d}{d x} [x]$

Step 2.3.2

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

$5 x^{4} - 30 x^{2} + 9 \cdot 1$

Step 2.3.3

Multiply $9$ by $1$ .

$5 x^{4} - 30 x^{2} + 9$

Step 3

Find the second derivative of the function.

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Step 3.1

By the Sum Rule, the derivative of $5 x^{4} - 30 x^{2} + 9$ with respect to $x$ is $\frac{d}{d x} [5 x^{4}] + \frac{d}{d x} [- 30 x^{2}] + \frac{d}{d x} [9]$ .

$f'' (x) = \frac{d}{d x} (5 x^{4}) + \frac{d}{d x} (- 30 x^{2}) + \frac{d}{d x} (9)$

Step 3.2

Evaluate $\frac{d}{d x} [5 x^{4}]$ .

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Step 3.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{4}$ with respect to $x$ is $5 \frac{d}{d x} [x^{4}]$ .

$f'' (x) = 5 \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 30 x^{2}) + \frac{d}{d x} (9)$

Step 3.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f'' (x) = 5 (4 x^{3}) + \frac{d}{d x} (- 30 x^{2}) + \frac{d}{d x} (9)$

Step 3.2.3

Multiply $4$ by $5$ .

$f'' (x) = 20 x^{3} + \frac{d}{d x} (- 30 x^{2}) + \frac{d}{d x} (9)$

Step 3.3

Evaluate $\frac{d}{d x} [- 30 x^{2}]$ .

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Step 3.3.1

Since $- 30$ is constant with respect to $x$ , the derivative of $- 30 x^{2}$ with respect to $x$ is $- 30 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 20 x^{3} - 30 \frac{d}{d x} x^{2} + \frac{d}{d x} (9)$

Step 3.3.2

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

$f'' (x) = 20 x^{3} - 30 (2 x) + \frac{d}{d x} (9)$

Step 3.3.3

Multiply $2$ by $- 30$ .

$f'' (x) = 20 x^{3} - 60 x + \frac{d}{d x} (9)$

Step 3.4

Differentiate using the Constant Rule.

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Step 3.4.1

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$f'' (x) = 20 x^{3} - 60 x + 0$

Step 3.4.2

Add $20 x^{3} - 60 x$ and $0$ .

$f'' (x) = 20 x^{3} - 60 x$

Step 4

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

$5 x^{4} - 30 x^{2} + 9 = 0$

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

Differentiate.

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Step 5.1.1.1

By the Sum Rule, the derivative of $x^{5} - 10 x^{3} + 9 x$ with respect to $x$ is $\frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 10 x^{3}] + \frac{d}{d x} [9 x]$ .

$f' (x) = \frac{d}{d x} (x^{5}) + \frac{d}{d x} (- 10 x^{3}) + \frac{d}{d x} (9 x)$

Step 5.1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$f' (x) = 5 x^{4} + \frac{d}{d x} (- 10 x^{3}) + \frac{d}{d x} (9 x)$

Step 5.1.2

Evaluate $\frac{d}{d x} [- 10 x^{3}]$ .

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Step 5.1.2.1

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10 x^{3}$ with respect to $x$ is $- 10 \frac{d}{d x} [x^{3}]$ .

$f' (x) = 5 x^{4} - 10 \frac{d}{d x} x^{3} + \frac{d}{d x} (9 x)$

Step 5.1.2.2

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

$f' (x) = 5 x^{4} - 10 (3 x^{2}) + \frac{d}{d x} (9 x)$

Step 5.1.2.3

Multiply $3$ by $- 10$ .

$f' (x) = 5 x^{4} - 30 x^{2} + \frac{d}{d x} (9 x)$

Step 5.1.3

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

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Step 5.1.3.1

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

$f' (x) = 5 x^{4} - 30 x^{2} + 9 \frac{d}{d x} (x)$

Step 5.1.3.2

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

$f' (x) = 5 x^{4} - 30 x^{2} + 9 \cdot 1$

Step 5.1.3.3

Multiply $9$ by $1$ .

$f' (x) = 5 x^{4} - 30 x^{2} + 9$

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $5 x^{4} - 30 x^{2} + 9$ .

$5 x^{4} - 30 x^{2} + 9$

Step 6

Set the first derivative equal to $0$ then solve the equation $5 x^{4} - 30 x^{2} + 9 = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$5 x^{4} - 30 x^{2} + 9 = 0$

Step 6.2

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

$5 u^{2} - 30 u + 9 = 0$

$u = x^{2}$

Step 6.3

Use the quadratic formula to find the solutions.

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

Step 6.4

Substitute the values $a = 5$ , $b = - 30$ , and $c = 9$ into the quadratic formula and solve for $u$ .

$\frac{30 \pm \sqrt{{(- 30)}^{2} - 4 \cdot (5 \cdot 9)}}{2 \cdot 5}$

Step 6.5

Simplify.

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Step 6.5.1

Simplify the numerator.

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Step 6.5.1.1

Raise $- 30$ to the power of $2$ .

$u = \frac{30 \pm \sqrt{900 - 4 \cdot 5 \cdot 9}}{2 \cdot 5}$

Step 6.5.1.2

Multiply $- 4 \cdot 5 \cdot 9$ .

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Step 6.5.1.2.1

Multiply $- 4$ by $5$ .

$u = \frac{30 \pm \sqrt{900 - 20 \cdot 9}}{2 \cdot 5}$

Step 6.5.1.2.2

Multiply $- 20$ by $9$ .

$u = \frac{30 \pm \sqrt{900 - 180}}{2 \cdot 5}$

Step 6.5.1.3

Subtract $180$ from $900$ .

$u = \frac{30 \pm \sqrt{720}}{2 \cdot 5}$

Step 6.5.1.4

Rewrite $720$ as $12^{2} \cdot 5$ .

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Step 6.5.1.4.1

Factor $144$ out of $720$ .

$u = \frac{30 \pm \sqrt{144 (5)}}{2 \cdot 5}$

Step 6.5.1.4.2

Rewrite $144$ as $12^{2}$ .

$u = \frac{30 \pm \sqrt{12^{2} \cdot 5}}{2 \cdot 5}$

Step 6.5.1.5

Pull terms out from under the radical.

$u = \frac{30 \pm 12 \sqrt{5}}{2 \cdot 5}$

Step 6.5.2

Multiply $2$ by $5$ .

$u = \frac{30 \pm 12 \sqrt{5}}{10}$

Step 6.5.3

Simplify $\frac{30 \pm 12 \sqrt{5}}{10}$ .

$u = \frac{15 \pm 6 \sqrt{5}}{5}$

Step 6.6

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

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Step 6.6.1

Simplify the numerator.

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Step 6.6.1.1

Raise $- 30$ to the power of $2$ .

$u = \frac{30 \pm \sqrt{900 - 4 \cdot 5 \cdot 9}}{2 \cdot 5}$

Step 6.6.1.2

Multiply $- 4 \cdot 5 \cdot 9$ .

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Step 6.6.1.2.1

Multiply $- 4$ by $5$ .

$u = \frac{30 \pm \sqrt{900 - 20 \cdot 9}}{2 \cdot 5}$

Step 6.6.1.2.2

Multiply $- 20$ by $9$ .

$u = \frac{30 \pm \sqrt{900 - 180}}{2 \cdot 5}$

Step 6.6.1.3

Subtract $180$ from $900$ .

$u = \frac{30 \pm \sqrt{720}}{2 \cdot 5}$

Step 6.6.1.4

Rewrite $720$ as $12^{2} \cdot 5$ .

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Step 6.6.1.4.1

Factor $144$ out of $720$ .

$u = \frac{30 \pm \sqrt{144 (5)}}{2 \cdot 5}$

Step 6.6.1.4.2

Rewrite $144$ as $12^{2}$ .

$u = \frac{30 \pm \sqrt{12^{2} \cdot 5}}{2 \cdot 5}$

Step 6.6.1.5

Pull terms out from under the radical.

$u = \frac{30 \pm 12 \sqrt{5}}{2 \cdot 5}$

Step 6.6.2

Multiply $2$ by $5$ .

$u = \frac{30 \pm 12 \sqrt{5}}{10}$

Step 6.6.3

Simplify $\frac{30 \pm 12 \sqrt{5}}{10}$ .

$u = \frac{15 \pm 6 \sqrt{5}}{5}$

Step 6.6.4

Change the $\pm$ to $+$ .

$u = \frac{15 + 6 \sqrt{5}}{5}$

Step 6.7

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

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Step 6.7.1

Simplify the numerator.

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Step 6.7.1.1

Raise $- 30$ to the power of $2$ .

$u = \frac{30 \pm \sqrt{900 - 4 \cdot 5 \cdot 9}}{2 \cdot 5}$

Step 6.7.1.2

Multiply $- 4 \cdot 5 \cdot 9$ .

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Step 6.7.1.2.1

Multiply $- 4$ by $5$ .

$u = \frac{30 \pm \sqrt{900 - 20 \cdot 9}}{2 \cdot 5}$

Step 6.7.1.2.2

Multiply $- 20$ by $9$ .

$u = \frac{30 \pm \sqrt{900 - 180}}{2 \cdot 5}$

Step 6.7.1.3

Subtract $180$ from $900$ .

$u = \frac{30 \pm \sqrt{720}}{2 \cdot 5}$

Step 6.7.1.4

Rewrite $720$ as $12^{2} \cdot 5$ .

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Step 6.7.1.4.1

Factor $144$ out of $720$ .

$u = \frac{30 \pm \sqrt{144 (5)}}{2 \cdot 5}$

Step 6.7.1.4.2

Rewrite $144$ as $12^{2}$ .

$u = \frac{30 \pm \sqrt{12^{2} \cdot 5}}{2 \cdot 5}$

Step 6.7.1.5

Pull terms out from under the radical.

$u = \frac{30 \pm 12 \sqrt{5}}{2 \cdot 5}$

Step 6.7.2

Multiply $2$ by $5$ .

$u = \frac{30 \pm 12 \sqrt{5}}{10}$

Step 6.7.3

Simplify $\frac{30 \pm 12 \sqrt{5}}{10}$ .

$u = \frac{15 \pm 6 \sqrt{5}}{5}$

Step 6.7.4

Change the $\pm$ to $-$ .

$u = \frac{15 - 6 \sqrt{5}}{5}$

Step 6.8

The final answer is the combination of both solutions.

$u = \frac{15 + 6 \sqrt{5}}{5}, \frac{15 - 6 \sqrt{5}}{5}$

Step 6.9

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = 5.68328157$

${(x^{2})}^{1} = 0.31671842$

Step 6.10

Solve the first equation for $x$ .

$x^{2} = 5.68328157$

Step 6.11

Solve the equation for $x$ .

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Step 6.11.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{5.68328157}$

Step 6.11.2

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.11.2.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{5.68328157}$

Step 6.11.2.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{5.68328157}$

Step 6.11.2.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{5.68328157}, - \sqrt{5.68328157}$

Step 6.12

Solve the second equation for $x$ .

${(x^{2})}^{1} = 0.31671842$

Step 6.13

Solve the equation for $x$ .

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Step 6.13.1

Remove parentheses.

$x^{2} = 0.31671842$

Step 6.13.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0.31671842}$

Step 6.13.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.13.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{0.31671842}$

Step 6.13.3.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{0.31671842}$

Step 6.13.3.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{0.31671842}, - \sqrt{0.31671842}$

Step 6.14

The solution to $5 x^{4} - 30 x^{2} + 9 = 0$ is $x = \sqrt{5.68328157}, - \sqrt{5.68328157}, \sqrt{0.31671842}, - \sqrt{0.31671842}$ .

$x = \sqrt{5.68328157}, - \sqrt{5.68328157}, \sqrt{0.31671842}, - \sqrt{0.31671842}$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

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 8

Critical points to evaluate.

$x = \sqrt{5.68328157}, - \sqrt{5.68328157}, \sqrt{0.31671842}, - \sqrt{0.31671842}$

Step 9

Evaluate the second derivative at $x = \sqrt{5.68328157}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$20 {(\sqrt{5.68328157})}^{3} - 60 \sqrt{5.68328157}$

Step 10

Simplify each term.

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Step 10.1

Rewrite ${\sqrt{5.68328157}}^{3}$ as $\sqrt{{5.68328157}^{3}}$ .

$20 \sqrt{{5.68328157}^{3}} - 60 \sqrt{5.68328157}$

Step 10.2

Raise $5.68328157$ to the power of $3$ .

$20 \sqrt{183.56822979} - 60 \sqrt{5.68328157}$

Step 11

$x = \sqrt{5.68328157}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \sqrt{5.68328157}$ is a local minimum

Step 12

Find the y-value when $x = \sqrt{5.68328157}$ .

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Step 12.1

Replace the variable $x$ with $\sqrt{5.68328157}$ in the expression.

$f (\sqrt{5.68328157}) = {(\sqrt{5.68328157})}^{5} - 10 {(\sqrt{5.68328157})}^{3} + 9 (\sqrt{5.68328157})$

Step 12.2

Simplify the result.

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Step 12.2.1

Simplify each term.

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Step 12.2.1.1

Rewrite ${\sqrt{5.68328157}}^{5}$ as $\sqrt{{5.68328157}^{5}}$ .

$f (\sqrt{5.68328157}) = \sqrt{{5.68328157}^{5}} - 10 {(\sqrt{5.68328157})}^{3} + 9 (\sqrt{5.68328157})$

Step 12.2.1.2

Raise $5.68328157$ to the power of $5$ .

$f (\sqrt{5.68328157}) = \sqrt{5929.19681311} - 10 {(\sqrt{5.68328157})}^{3} + 9 (\sqrt{5.68328157})$

Step 12.2.1.3

Rewrite ${\sqrt{5.68328157}}^{3}$ as $\sqrt{{5.68328157}^{3}}$ .

$f (\sqrt{5.68328157}) = \sqrt{5929.19681311} - 10 \sqrt{{5.68328157}^{3}} + 9 (\sqrt{5.68328157})$

Step 12.2.1.4

Raise $5.68328157$ to the power of $3$ .

$f (\sqrt{5.68328157}) = \sqrt{5929.19681311} - 10 \sqrt{183.56822979} + 9 \sqrt{5.68328157}$

Step 12.2.2

The final answer is $\sqrt{5929.19681311} - 10 \sqrt{183.56822979} + 9 \sqrt{5.68328157}$ .

$y = \sqrt{5929.19681311} - 10 \sqrt{183.56822979} + 9 \sqrt{5.68328157}$

Step 13

Evaluate the second derivative at $x = - \sqrt{5.68328157}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$20 {(- \sqrt{5.68328157})}^{3} - 60 (- \sqrt{5.68328157})$

Step 14

Simplify each term.

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Step 14.1

Apply the product rule to $- \sqrt{5.68328157}$ .

$20 ({(- 1)}^{3} {\sqrt{5.68328157}}^{3}) - 60 (- \sqrt{5.68328157})$

Step 14.2

Raise $- 1$ to the power of $3$ .

$20 (- {\sqrt{5.68328157}}^{3}) - 60 (- \sqrt{5.68328157})$

Step 14.3

Rewrite ${\sqrt{5.68328157}}^{3}$ as $\sqrt{{5.68328157}^{3}}$ .

$20 (- \sqrt{{5.68328157}^{3}}) - 60 (- \sqrt{5.68328157})$

Step 14.4

Raise $5.68328157$ to the power of $3$ .

$20 (- \sqrt{183.56822979}) - 60 (- \sqrt{5.68328157})$

Step 14.5

Multiply $- 1$ by $20$ .

$- 20 \sqrt{183.56822979} - 60 (- \sqrt{5.68328157})$

Step 14.6

Multiply $- 1$ by $- 60$ .

$- 20 \sqrt{183.56822979} + 60 \sqrt{5.68328157}$

Step 15

$x = - \sqrt{5.68328157}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - \sqrt{5.68328157}$ is a local maximum

Step 16

Find the y-value when $x = - \sqrt{5.68328157}$ .

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Step 16.1

Replace the variable $x$ with $- \sqrt{5.68328157}$ in the expression.

$f (- \sqrt{5.68328157}) = {(- \sqrt{5.68328157})}^{5} - 10 {(- \sqrt{5.68328157})}^{3} + 9 (- \sqrt{5.68328157})$

Step 16.2

Simplify the result.

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Step 16.2.1

Simplify each term.

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Step 16.2.1.1

Apply the product rule to $- \sqrt{5.68328157}$ .

$f (- \sqrt{5.68328157}) = {(- 1)}^{5} {\sqrt{5.68328157}}^{5} - 10 {(- \sqrt{5.68328157})}^{3} + 9 (- \sqrt{5.68328157})$

Step 16.2.1.2

Raise $- 1$ to the power of $5$ .

$f (- \sqrt{5.68328157}) = - {\sqrt{5.68328157}}^{5} - 10 {(- \sqrt{5.68328157})}^{3} + 9 (- \sqrt{5.68328157})$

Step 16.2.1.3

Rewrite ${\sqrt{5.68328157}}^{5}$ as $\sqrt{{5.68328157}^{5}}$ .

$f (- \sqrt{5.68328157}) = - \sqrt{{5.68328157}^{5}} - 10 {(- \sqrt{5.68328157})}^{3} + 9 (- \sqrt{5.68328157})$

Step 16.2.1.4

Raise $5.68328157$ to the power of $5$ .

$f (- \sqrt{5.68328157}) = - \sqrt{5929.19681311} - 10 {(- \sqrt{5.68328157})}^{3} + 9 (- \sqrt{5.68328157})$

Step 16.2.1.5

Apply the product rule to $- \sqrt{5.68328157}$ .

$f (- \sqrt{5.68328157}) = - \sqrt{5929.19681311} - 10 ({(- 1)}^{3} {\sqrt{5.68328157}}^{3}) + 9 (- \sqrt{5.68328157})$

Step 16.2.1.6

Raise $- 1$ to the power of $3$ .

$f (- \sqrt{5.68328157}) = - \sqrt{5929.19681311} - 10 (- {\sqrt{5.68328157}}^{3}) + 9 (- \sqrt{5.68328157})$

Step 16.2.1.7

Rewrite ${\sqrt{5.68328157}}^{3}$ as $\sqrt{{5.68328157}^{3}}$ .

$f (- \sqrt{5.68328157}) = - \sqrt{5929.19681311} - 10 (- \sqrt{{5.68328157}^{3}}) + 9 (- \sqrt{5.68328157})$

Step 16.2.1.8

Raise $5.68328157$ to the power of $3$ .

$f (- \sqrt{5.68328157}) = - \sqrt{5929.19681311} - 10 (- \sqrt{183.56822979}) + 9 (- \sqrt{5.68328157})$

Step 16.2.1.9

Multiply $- 1$ by $- 10$ .

$f (- \sqrt{5.68328157}) = - \sqrt{5929.19681311} + 10 \sqrt{183.56822979} + 9 (- \sqrt{5.68328157})$

Step 16.2.1.10

Multiply $- 1$ by $9$ .

$f (- \sqrt{5.68328157}) = - \sqrt{5929.19681311} + 10 \sqrt{183.56822979} - 9 \sqrt{5.68328157}$

Step 16.2.2

The final answer is $- \sqrt{5929.19681311} + 10 \sqrt{183.56822979} - 9 \sqrt{5.68328157}$ .

$y = - \sqrt{5929.19681311} + 10 \sqrt{183.56822979} - 9 \sqrt{5.68328157}$

Step 17

Evaluate the second derivative at $x = \sqrt{0.31671842}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$20 {(\sqrt{0.31671842})}^{3} - 60 \sqrt{0.31671842}$

Step 18

Simplify each term.

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Step 18.1

Rewrite ${\sqrt{0.31671842}}^{3}$ as $\sqrt{{0.31671842}^{3}}$ .

$20 \sqrt{{0.31671842}^{3}} - 60 \sqrt{0.31671842}$

Step 18.2

Raise $0.31671842$ to the power of $3$ .

$20 \sqrt{0.0317702} - 60 \sqrt{0.31671842}$

Step 19

$x = \sqrt{0.31671842}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \sqrt{0.31671842}$ is a local maximum

Step 20

Find the y-value when $x = \sqrt{0.31671842}$ .

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Step 20.1

Replace the variable $x$ with $\sqrt{0.31671842}$ in the expression.

$f (\sqrt{0.31671842}) = {(\sqrt{0.31671842})}^{5} - 10 {(\sqrt{0.31671842})}^{3} + 9 (\sqrt{0.31671842})$

Step 20.2

Simplify the result.

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Step 20.2.1

Simplify each term.

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Step 20.2.1.1

Rewrite ${\sqrt{0.31671842}}^{5}$ as $\sqrt{{0.31671842}^{5}}$ .

$f (\sqrt{0.31671842}) = \sqrt{{0.31671842}^{5}} - 10 {(\sqrt{0.31671842})}^{3} + 9 (\sqrt{0.31671842})$

Step 20.2.1.2

Raise $0.31671842$ to the power of $5$ .

$f (\sqrt{0.31671842}) = \sqrt{0.00318688} - 10 {(\sqrt{0.31671842})}^{3} + 9 (\sqrt{0.31671842})$

Step 20.2.1.3

Rewrite ${\sqrt{0.31671842}}^{3}$ as $\sqrt{{0.31671842}^{3}}$ .

$f (\sqrt{0.31671842}) = \sqrt{0.00318688} - 10 \sqrt{{0.31671842}^{3}} + 9 (\sqrt{0.31671842})$

Step 20.2.1.4

Raise $0.31671842$ to the power of $3$ .

$f (\sqrt{0.31671842}) = \sqrt{0.00318688} - 10 \sqrt{0.0317702} + 9 \sqrt{0.31671842}$

Step 20.2.2

The final answer is $\sqrt{0.00318688} - 10 \sqrt{0.0317702} + 9 \sqrt{0.31671842}$ .

$y = \sqrt{0.00318688} - 10 \sqrt{0.0317702} + 9 \sqrt{0.31671842}$

Step 21

Evaluate the second derivative at $x = - \sqrt{0.31671842}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$20 {(- \sqrt{0.31671842})}^{3} - 60 (- \sqrt{0.31671842})$

Step 22

Simplify each term.

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Step 22.1

Apply the product rule to $- \sqrt{0.31671842}$ .

$20 ({(- 1)}^{3} {\sqrt{0.31671842}}^{3}) - 60 (- \sqrt{0.31671842})$

Step 22.2

Raise $- 1$ to the power of $3$ .

$20 (- {\sqrt{0.31671842}}^{3}) - 60 (- \sqrt{0.31671842})$

Step 22.3

Rewrite ${\sqrt{0.31671842}}^{3}$ as $\sqrt{{0.31671842}^{3}}$ .

$20 (- \sqrt{{0.31671842}^{3}}) - 60 (- \sqrt{0.31671842})$

Step 22.4

Raise $0.31671842$ to the power of $3$ .

$20 (- \sqrt{0.0317702}) - 60 (- \sqrt{0.31671842})$

Step 22.5

Multiply $- 1$ by $20$ .

$- 20 \sqrt{0.0317702} - 60 (- \sqrt{0.31671842})$

Step 22.6

Multiply $- 1$ by $- 60$ .

$- 20 \sqrt{0.0317702} + 60 \sqrt{0.31671842}$

Step 23

$x = - \sqrt{0.31671842}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - \sqrt{0.31671842}$ is a local minimum

Step 24

Find the y-value when $x = - \sqrt{0.31671842}$ .

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Step 24.1

Replace the variable $x$ with $- \sqrt{0.31671842}$ in the expression.

$f (- \sqrt{0.31671842}) = {(- \sqrt{0.31671842})}^{5} - 10 {(- \sqrt{0.31671842})}^{3} + 9 (- \sqrt{0.31671842})$

Step 24.2

Simplify the result.

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Step 24.2.1

Simplify each term.

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Step 24.2.1.1

Apply the product rule to $- \sqrt{0.31671842}$ .

$f (- \sqrt{0.31671842}) = {(- 1)}^{5} {\sqrt{0.31671842}}^{5} - 10 {(- \sqrt{0.31671842})}^{3} + 9 (- \sqrt{0.31671842})$

Step 24.2.1.2

Raise $- 1$ to the power of $5$ .

$f (- \sqrt{0.31671842}) = - {\sqrt{0.31671842}}^{5} - 10 {(- \sqrt{0.31671842})}^{3} + 9 (- \sqrt{0.31671842})$

Step 24.2.1.3

Rewrite ${\sqrt{0.31671842}}^{5}$ as $\sqrt{{0.31671842}^{5}}$ .

$f (- \sqrt{0.31671842}) = - \sqrt{{0.31671842}^{5}} - 10 {(- \sqrt{0.31671842})}^{3} + 9 (- \sqrt{0.31671842})$

Step 24.2.1.4

Raise $0.31671842$ to the power of $5$ .

$f (- \sqrt{0.31671842}) = - \sqrt{0.00318688} - 10 {(- \sqrt{0.31671842})}^{3} + 9 (- \sqrt{0.31671842})$

Step 24.2.1.5

Apply the product rule to $- \sqrt{0.31671842}$ .

$f (- \sqrt{0.31671842}) = - \sqrt{0.00318688} - 10 ({(- 1)}^{3} {\sqrt{0.31671842}}^{3}) + 9 (- \sqrt{0.31671842})$

Step 24.2.1.6

Raise $- 1$ to the power of $3$ .

$f (- \sqrt{0.31671842}) = - \sqrt{0.00318688} - 10 (- {\sqrt{0.31671842}}^{3}) + 9 (- \sqrt{0.31671842})$

Step 24.2.1.7

Rewrite ${\sqrt{0.31671842}}^{3}$ as $\sqrt{{0.31671842}^{3}}$ .

$f (- \sqrt{0.31671842}) = - \sqrt{0.00318688} - 10 (- \sqrt{{0.31671842}^{3}}) + 9 (- \sqrt{0.31671842})$

Step 24.2.1.8

Raise $0.31671842$ to the power of $3$ .

$f (- \sqrt{0.31671842}) = - \sqrt{0.00318688} - 10 (- \sqrt{0.0317702}) + 9 (- \sqrt{0.31671842})$

Step 24.2.1.9

Multiply $- 1$ by $- 10$ .

$f (- \sqrt{0.31671842}) = - \sqrt{0.00318688} + 10 \sqrt{0.0317702} + 9 (- \sqrt{0.31671842})$

Step 24.2.1.10

Multiply $- 1$ by $9$ .

$f (- \sqrt{0.31671842}) = - \sqrt{0.00318688} + 10 \sqrt{0.0317702} - 9 \sqrt{0.31671842}$

Step 24.2.2

The final answer is $- \sqrt{0.00318688} + 10 \sqrt{0.0317702} - 9 \sqrt{0.31671842}$ .

$y = - \sqrt{0.00318688} + 10 \sqrt{0.0317702} - 9 \sqrt{0.31671842}$

Step 25

These are the local extrema for $f (x) = x^{5} - 10 x^{3} + 9 x$ .

$(\sqrt{5.68328157}, \sqrt{5929.19681311} - 10 \sqrt{183.56822979} + 9 \sqrt{5.68328157})$ is a local minima

$(- \sqrt{5.68328157}, - \sqrt{5929.19681311} + 10 \sqrt{183.56822979} - 9 \sqrt{5.68328157})$ is a local maxima

$(\sqrt{0.31671842}, \sqrt{0.00318688} - 10 \sqrt{0.0317702} + 9 \sqrt{0.31671842})$ is a local maxima

$(- \sqrt{0.31671842}, - \sqrt{0.00318688} + 10 \sqrt{0.0317702} - 9 \sqrt{0.31671842})$ is a local minima

Step 26