Calculus Examples

$y = x^{3} + x^{2} - x$

Step 1

Write $y = x^{3} + x^{2} - x$ as a function.

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

Step 2

Find the first derivative of the function.

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Differentiate.

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

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

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

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

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

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

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

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Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

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

$3 x^{2} + 2 x - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$3 x^{2} + 2 x - 1$

Step 3

Find the second derivative of the function.

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By the Sum Rule, the derivative of $3 x^{2} + 2 x - 1$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

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

Evaluate $\frac{d}{d x} [3 x^{2}]$ .

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Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

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

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

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

Multiply $2$ by $3$ .

$f'' (x) = 6 x + \frac{d}{d x} (2 x) + \frac{d}{d x} (- 1)$

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

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

$f'' (x) = 6 x + 2 \frac{d}{d x} (x) + \frac{d}{d x} (- 1)$

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

$f'' (x) = 6 x + 2 \cdot 1 + \frac{d}{d x} (- 1)$

Multiply $2$ by $1$ .

$f'' (x) = 6 x + 2 + \frac{d}{d x} (- 1)$

Differentiate using the Constant Rule.

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Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f'' (x) = 6 x + 2 + 0$

Add $6 x + 2$ and $0$ .

$f'' (x) = 6 x + 2$

Step 4

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

$3 x^{2} + 2 x - 1 = 0$

Step 5

Find the first derivative.

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Find the first derivative.

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Differentiate.

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

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

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

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

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

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

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

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Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

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

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

Multiply $- 1$ by $1$ .

$f' (x) = 3 x^{2} + 2 x - 1$

The first derivative of $f (x)$ with respect to $x$ is $3 x^{2} + 2 x - 1$ .

$3 x^{2} + 2 x - 1$

Step 6

Set the first derivative equal to $0$ then solve the equation $3 x^{2} + 2 x - 1 = 0$ .

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Set the first derivative equal to $0$ .

$3 x^{2} + 2 x - 1 = 0$

Factor by grouping.

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For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 1 = - 3$ and whose sum is $b = 2$ .

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Factor $2$ out of $2 x$ .

$3 x^{2} + 2 (x) - 1 = 0$

Rewrite $2$ as $- 1$ plus $3$

$3 x^{2} + (- 1 + 3) x - 1 = 0$

Apply the distributive property.

$3 x^{2} - 1 x + 3 x - 1 = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$(3 x^{2} - 1 x) + 3 x - 1 = 0$

Factor out the greatest common factor (GCF) from each group.

$x (3 x - 1) + 1 (3 x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x - 1$ .

$(3 x - 1) (x + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$3 x - 1 = 0$

$x + 1 = 0$

Set $3 x - 1$ equal to $0$ and solve for $x$ .

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Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Solve $3 x - 1 = 0$ for $x$ .

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Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

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Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Simplify the left side.

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Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Set $x + 1$ equal to $0$ and solve for $x$ .

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Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(3 x - 1) (x + 1) = 0$ true.

$x = \frac{1}{3}, - 1$

Step 7

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 8

Critical points to evaluate.

$x = \frac{1}{3}, - 1$

Step 9

Evaluate the second derivative at $x = \frac{1}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$6 (\frac{1}{3}) + 2$

Step 10

Evaluate the second derivative.

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Cancel the common factor of $3$ .

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Factor $3$ out of $6$ .

$3 (2) \frac{1}{3} + 2$

Cancel the common factor.

$3 \cdot 2 \frac{1}{3} + 2$

Rewrite the expression.

$2 + 2$

Add $2$ and $2$ .

$4$

Step 11

$x = \frac{1}{3}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{1}{3}$ is a local minimum

Step 12

Find the y-value when $x = \frac{1}{3}$ .

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Replace the variable $x$ with $\frac{1}{3}$ in the expression.

$f (\frac{1}{3}) = {(\frac{1}{3})}^{3} + {(\frac{1}{3})}^{2} - (\frac{1}{3})$

Simplify the result.

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Simplify each term.

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Apply the product rule to $\frac{1}{3}$ .

$f (\frac{1}{3}) = \frac{1^{3}}{3^{3}} + {(\frac{1}{3})}^{2} - (\frac{1}{3})$

One to any power is one.

$f (\frac{1}{3}) = \frac{1}{3^{3}} + {(\frac{1}{3})}^{2} - (\frac{1}{3})$

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

$f (\frac{1}{3}) = \frac{1}{27} + {(\frac{1}{3})}^{2} - (\frac{1}{3})$

Apply the product rule to $\frac{1}{3}$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{1^{2}}{3^{2}} - (\frac{1}{3})$

One to any power is one.

$f (\frac{1}{3}) = \frac{1}{27} + \frac{1}{3^{2}} - (\frac{1}{3})$

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

$f (\frac{1}{3}) = \frac{1}{27} + \frac{1}{9} - \frac{1}{3}$

Find the common denominator.

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Multiply $\frac{1}{9}$ by $\frac{3}{3}$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{1}{9} \cdot \frac{3}{3} - \frac{1}{3}$

Multiply $\frac{1}{9}$ by $\frac{3}{3}$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{9 \cdot 3} - \frac{1}{3}$

Multiply $\frac{1}{3}$ by $\frac{9}{9}$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{9 \cdot 3} - (\frac{1}{3} \cdot \frac{9}{9})$

Multiply $\frac{1}{3}$ by $\frac{9}{9}$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{9 \cdot 3} - \frac{9}{3 \cdot 9}$

Reorder the factors of $9 \cdot 3$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{3 \cdot 9} - \frac{9}{3 \cdot 9}$

Multiply $3$ by $9$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{27} - \frac{9}{3 \cdot 9}$

Multiply $3$ by $9$ .

$f (\frac{1}{3}) = \frac{1}{27} + \frac{3}{27} - \frac{9}{27}$

Combine the numerators over the common denominator.

$f (\frac{1}{3}) = \frac{1 + 3 - 9}{27}$

Simplify the expression.

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Add $1$ and $3$ .

$f (\frac{1}{3}) = \frac{4 - 9}{27}$

Subtract $9$ from $4$ .

$f (\frac{1}{3}) = \frac{- 5}{27}$

Move the negative in front of the fraction.

$f (\frac{1}{3}) = - \frac{5}{27}$

The final answer is $- \frac{5}{27}$ .

$y = - \frac{5}{27}$

Step 13

Evaluate the second derivative at $x = - 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$6 (- 1) + 2$

Step 14

Evaluate the second derivative.

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Multiply $6$ by $- 1$ .

$- 6 + 2$

Add $- 6$ and $2$ .

$- 4$

Step 15

$x = - 1$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 1$ is a local maximum

Step 16

Find the y-value when $x = - 1$ .

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Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = {(- 1)}^{3} + {(- 1)}^{2} - (- 1)$

Simplify the result.

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Simplify each term.

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

$f (- 1) = - 1 + {(- 1)}^{2} - (- 1)$

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

$f (- 1) = - 1 + 1 - (- 1)$

Multiply $- 1$ by $- 1$ .

$f (- 1) = - 1 + 1 + 1$

Simplify by adding numbers.

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Add $- 1$ and $1$ .

$f (- 1) = 0 + 1$

Add $0$ and $1$ .

$f (- 1) = 1$

The final answer is $1$ .

$y = 1$

Step 17

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

$(\frac{1}{3}, - \frac{5}{27})$ is a local minima

$(- 1, 1)$ is a local maxima

Step 18