Calculus Examples

$f (x) = {(x + 1)}^{5} - 5 x - 2$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of ${(x + 1)}^{5} - 5 x - 2$ with respect to $x$ is $\frac{d}{d x} [{(x + 1)}^{5}] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$ .

$\frac{d}{d x} [{(x + 1)}^{5}] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 1.2

Evaluate $\frac{d}{d x} [{(x + 1)}^{5}]$ .

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Step 1.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{5}$ and $g (x) = x + 1$ .

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Step 1.2.1.1

To apply the Chain Rule, set $u$ as $x + 1$ .

$\frac{d}{d u} [u^{5}] \frac{d}{d x} [x + 1] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 1.2.1.2

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

$5 u^{4} \frac{d}{d x} [x + 1] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 1.2.1.3

Replace all occurrences of $u$ with $x + 1$ .

$5 {(x + 1)}^{4} \frac{d}{d x} [x + 1] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 1.2.2

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

$5 {(x + 1)}^{4} (\frac{d}{d x} [x] + \frac{d}{d x} [1]) + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 1.2.3

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

$5 {(x + 1)}^{4} (1 + \frac{d}{d x} [1]) + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 1.2.4

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

$5 {(x + 1)}^{4} (1 + 0) + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 1.2.5

Add $1$ and $0$ .

$5 {(x + 1)}^{4} \cdot 1 + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 1.2.6

Multiply $5$ by $1$ .

$5 {(x + 1)}^{4} + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 1.3

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

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Step 1.3.1

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

$5 {(x + 1)}^{4} - 5 \frac{d}{d x} [x] + \frac{d}{d x} [- 2]$

Step 1.3.2

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

$5 {(x + 1)}^{4} - 5 \cdot 1 + \frac{d}{d x} [- 2]$

Step 1.3.3

Multiply $- 5$ by $1$ .

$5 {(x + 1)}^{4} - 5 + \frac{d}{d x} [- 2]$

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$5 {(x + 1)}^{4} - 5 + 0$

Step 1.4.2

Add $5 {(x + 1)}^{4} - 5$ and $0$ .

$5 {(x + 1)}^{4} - 5$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $5 {(x + 1)}^{4} - 5$ with respect to $x$ is $\frac{d}{d x} [5 {(x + 1)}^{4}] + \frac{d}{d x} [- 5]$ .

$f'' (x) = \frac{d}{d x} (5 {(x + 1)}^{4}) + \frac{d}{d x} (- 5)$

Step 2.2

Evaluate $\frac{d}{d x} [5 {(x + 1)}^{4}]$ .

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Step 2.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 {(x + 1)}^{4}$ with respect to $x$ is $5 \frac{d}{d x} [{(x + 1)}^{4}]$ .

$f'' (x) = 5 \frac{d}{d x} ({(x + 1)}^{4}) + \frac{d}{d x} (- 5)$

Step 2.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = x + 1$ .

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Step 2.2.2.1

To apply the Chain Rule, set $u$ as $x + 1$ .

$f'' (x) = 5 (\frac{d}{d u} (u^{4}) \frac{d}{d x} (x + 1)) + \frac{d}{d x} (- 5)$

Step 2.2.2.2

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

$f'' (x) = 5 (4 u^{3} \frac{d}{d x} (x + 1)) + \frac{d}{d x} (- 5)$

Step 2.2.2.3

Replace all occurrences of $u$ with $x + 1$ .

$f'' (x) = 5 (4 {(x + 1)}^{3} \frac{d}{d x} (x + 1)) + \frac{d}{d x} (- 5)$

Step 2.2.3

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

$f'' (x) = 5 (4 {(x + 1)}^{3} (\frac{d}{d x} (x) + \frac{d}{d x} (1))) + \frac{d}{d x} (- 5)$

Step 2.2.4

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

$f'' (x) = 5 (4 {(x + 1)}^{3} (1 + \frac{d}{d x} (1))) + \frac{d}{d x} (- 5)$

Step 2.2.5

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

$f'' (x) = 5 (4 {(x + 1)}^{3} (1 + 0)) + \frac{d}{d x} (- 5)$

Step 2.2.6

Add $1$ and $0$ .

$f'' (x) = 5 (4 {(x + 1)}^{3} \cdot 1) + \frac{d}{d x} (- 5)$

Step 2.2.7

Multiply $4$ by $1$ .

$f'' (x) = 5 (4 {(x + 1)}^{3}) + \frac{d}{d x} (- 5)$

Step 2.2.8

Multiply $4$ by $5$ .

$f'' (x) = 20 {(x + 1)}^{3} + \frac{d}{d x} (- 5)$

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$f'' (x) = 20 {(x + 1)}^{3} + 0$

Step 2.3.2

Add $20 {(x + 1)}^{3}$ and $0$ .

$f'' (x) = 20 {(x + 1)}^{3}$

Step 3

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

$5 {(x + 1)}^{4} - 5 = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

By the Sum Rule, the derivative of ${(x + 1)}^{5} - 5 x - 2$ with respect to $x$ is $\frac{d}{d x} [{(x + 1)}^{5}] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$ .

$\frac{d}{d x} [{(x + 1)}^{5}] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 4.1.2

Evaluate $\frac{d}{d x} [{(x + 1)}^{5}]$ .

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Step 4.1.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{5}$ and $g (x) = x + 1$ .

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Step 4.1.2.1.1

To apply the Chain Rule, set $u$ as $x + 1$ .

$\frac{d}{d u} [u^{5}] \frac{d}{d x} [x + 1] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 4.1.2.1.2

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

$5 u^{4} \frac{d}{d x} [x + 1] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 4.1.2.1.3

Replace all occurrences of $u$ with $x + 1$ .

$5 {(x + 1)}^{4} \frac{d}{d x} [x + 1] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 4.1.2.2

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

$5 {(x + 1)}^{4} (\frac{d}{d x} [x] + \frac{d}{d x} [1]) + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 4.1.2.3

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

$5 {(x + 1)}^{4} (1 + \frac{d}{d x} [1]) + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 4.1.2.4

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

$5 {(x + 1)}^{4} (1 + 0) + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 4.1.2.5

Add $1$ and $0$ .

$5 {(x + 1)}^{4} \cdot 1 + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 4.1.2.6

Multiply $5$ by $1$ .

$5 {(x + 1)}^{4} + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [- 2]$

Step 4.1.3

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

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Step 4.1.3.1

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

$5 {(x + 1)}^{4} - 5 \frac{d}{d x} [x] + \frac{d}{d x} [- 2]$

Step 4.1.3.2

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

$5 {(x + 1)}^{4} - 5 \cdot 1 + \frac{d}{d x} [- 2]$

Step 4.1.3.3

Multiply $- 5$ by $1$ .

$5 {(x + 1)}^{4} - 5 + \frac{d}{d x} [- 2]$

Step 4.1.4

Differentiate using the Constant Rule.

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Step 4.1.4.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$5 {(x + 1)}^{4} - 5 + 0$

Step 4.1.4.2

Add $5 {(x + 1)}^{4} - 5$ and $0$ .

$f' (x) = 5 {(x + 1)}^{4} - 5$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $5 {(x + 1)}^{4} - 5$ .

$5 {(x + 1)}^{4} - 5$

Step 5

Set the first derivative equal to $0$ then solve the equation $5 {(x + 1)}^{4} - 5 = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$5 {(x + 1)}^{4} - 5 = 0$

Step 5.2

Simplify $5 {(x + 1)}^{4} - 5$ .

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Use the Binomial Theorem.

$5 (x^{4} + 4 x^{3} \cdot 1 + 6 x^{2} \cdot 1^{2} + 4 x \cdot 1^{3} + 1^{4}) - 5 = 0$

Step 5.2.1.2

Simplify each term.

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Step 5.2.1.2.1

Multiply $4$ by $1$ .

$5 (x^{4} + 4 x^{3} + 6 x^{2} \cdot 1^{2} + 4 x \cdot 1^{3} + 1^{4}) - 5 = 0$

Step 5.2.1.2.2

One to any power is one.

$5 (x^{4} + 4 x^{3} + 6 x^{2} \cdot 1 + 4 x \cdot 1^{3} + 1^{4}) - 5 = 0$

Step 5.2.1.2.3

Multiply $6$ by $1$ .

$5 (x^{4} + 4 x^{3} + 6 x^{2} + 4 x \cdot 1^{3} + 1^{4}) - 5 = 0$

Step 5.2.1.2.4

One to any power is one.

$5 (x^{4} + 4 x^{3} + 6 x^{2} + 4 x \cdot 1 + 1^{4}) - 5 = 0$

Step 5.2.1.2.5

Multiply $4$ by $1$ .

$5 (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1^{4}) - 5 = 0$

Step 5.2.1.2.6

One to any power is one.

$5 (x^{4} + 4 x^{3} + 6 x^{2} + 4 x + 1) - 5 = 0$

Step 5.2.1.3

Apply the distributive property.

$5 x^{4} + 5 (4 x^{3}) + 5 (6 x^{2}) + 5 (4 x) + 5 \cdot 1 - 5 = 0$

Step 5.2.1.4

Simplify.

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Step 5.2.1.4.1

Multiply $4$ by $5$ .

$5 x^{4} + 20 x^{3} + 5 (6 x^{2}) + 5 (4 x) + 5 \cdot 1 - 5 = 0$

Step 5.2.1.4.2

Multiply $6$ by $5$ .

$5 x^{4} + 20 x^{3} + 30 x^{2} + 5 (4 x) + 5 \cdot 1 - 5 = 0$

Step 5.2.1.4.3

Multiply $4$ by $5$ .

$5 x^{4} + 20 x^{3} + 30 x^{2} + 20 x + 5 \cdot 1 - 5 = 0$

Step 5.2.1.4.4

Multiply $5$ by $1$ .

$5 x^{4} + 20 x^{3} + 30 x^{2} + 20 x + 5 - 5 = 0$

Step 5.2.2

Combine the opposite terms in $5 x^{4} + 20 x^{3} + 30 x^{2} + 20 x + 5 - 5$ .

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Step 5.2.2.1

Subtract $5$ from $5$ .

$5 x^{4} + 20 x^{3} + 30 x^{2} + 20 x + 0 = 0$

Step 5.2.2.2

Add $5 x^{4} + 20 x^{3} + 30 x^{2} + 20 x$ and $0$ .

$5 x^{4} + 20 x^{3} + 30 x^{2} + 20 x = 0$

Step 5.3

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x = - 2, 0$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

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 7

Critical points to evaluate.

$x = - 2, 0$

Step 8

Evaluate the second derivative at $x = - 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$20 {((- 2) + 1)}^{3}$

Step 9

Evaluate the second derivative.

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Step 9.1

Add $- 2$ and $1$ .

$20 {(- 1)}^{3}$

Step 9.2

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

$20 \cdot - 1$

Step 9.3

Multiply $20$ by $- 1$ .

$- 20$

Step 10

$x = - 2$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 2$ is a local maximum

Step 11

Find the y-value when $x = - 2$ .

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Step 11.1

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = {((- 2) + 1)}^{5} - 5 \cdot - 2 - 2$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Add $- 2$ and $1$ .

$f (- 2) = {(- 1)}^{5} - 5 \cdot - 2 - 2$

Step 11.2.1.2

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

$f (- 2) = - 1 - 5 \cdot - 2 - 2$

Step 11.2.1.3

Multiply $- 5$ by $- 2$ .

$f (- 2) = - 1 + 10 - 2$

Step 11.2.2

Simplify by adding and subtracting.

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Step 11.2.2.1

Add $- 1$ and $10$ .

$f (- 2) = 9 - 2$

Step 11.2.2.2

Subtract $2$ from $9$ .

$f (- 2) = 7$

Step 11.2.3

The final answer is $7$ .

$y = 7$

Step 12

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$20 {((0) + 1)}^{3}$

Step 13

Evaluate the second derivative.

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Step 13.1

Add $0$ and $1$ .

$20 \cdot 1^{3}$

Step 13.2

One to any power is one.

$20 \cdot 1$

Step 13.3

Multiply $20$ by $1$ .

$20$

Step 14

$x = 0$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 0$ is a local minimum

Step 15

Find the y-value when $x = 0$ .

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Step 15.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = {((0) + 1)}^{5} - 5 \cdot 0 - 2$

Step 15.2

Simplify the result.

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Step 15.2.1

Simplify each term.

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Step 15.2.1.1

Add $0$ and $1$ .

$f (0) = 1^{5} - 5 \cdot 0 - 2$

Step 15.2.1.2

One to any power is one.

$f (0) = 1 - 5 \cdot 0 - 2$

Step 15.2.1.3

Multiply $- 5$ by $0$ .

$f (0) = 1 + 0 - 2$

Step 15.2.2

Simplify by adding and subtracting.

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Step 15.2.2.1

Add $1$ and $0$ .

$f (0) = 1 - 2$

Step 15.2.2.2

Subtract $2$ from $1$ .

$f (0) = - 1$

Step 15.2.3

The final answer is $- 1$ .

$y = - 1$

Step 16

These are the local extrema for $f (x) = {(x + 1)}^{5} - 5 x - 2$ .

$(- 2, 7)$ is a local maxima

$(0, - 1)$ is a local minima

Step 17