Calculus Examples

$f (x) = - | x | + 7$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $- | x | + 7$ with respect to $x$ is $\frac{d}{d x} [- | x |] + \frac{d}{d x} [7]$ .

$\frac{d}{d x} [- | x |] + \frac{d}{d x} [7]$

Step 1.2

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

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Step 1.2.1

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

$- \frac{d}{d x} [| x |] + \frac{d}{d x} [7]$

Step 1.2.2

The derivative of $| x |$ with respect to $x$ is $\frac{x}{| x |}$ .

$- \frac{x}{| x |} + \frac{d}{d x} [7]$

Step 1.3

Differentiate using the Constant Rule.

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Step 1.3.1

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$- \frac{x}{| x |} + 0$

Step 1.3.2

Add $- \frac{x}{| x |}$ and $0$ .

$- \frac{x}{| x |}$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{x}{| x |}$ .

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

Step 2.2

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = | x |$ .

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

Step 2.3

Differentiate using the Power Rule.

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Step 2.3.1

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

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

Step 2.3.2

Multiply $| x |$ by $1$ .

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

Step 2.4

The derivative of $| x |$ with respect to $x$ is $\frac{x}{| x |}$ .

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

Step 2.5

Combine $\frac{x}{| x |}$ and $x$ .

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

Step 2.6

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

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

Step 2.7

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

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

Step 2.8

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

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

Step 2.9

Add $1$ and $1$ .

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

Step 2.10

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

$f'' (x) = - \frac{| x | - \frac{x^{2}}{| x |}}{{| x |}^{2}} + \frac{x}{| x |} \cdot 0$

Step 2.11

Simplify the expression.

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Step 2.11.1

Multiply $\frac{x}{| x |}$ by $0$ .

$f'' (x) = - \frac{| x | - \frac{x^{2}}{| x |}}{{| x |}^{2}} + 0$

Step 2.11.2

Add $- \frac{| x | - \frac{x^{2}}{| x |}}{{| x |}^{2}}$ and $0$ .

$f'' (x) = - \frac{| x | - \frac{x^{2}}{| x |}}{{| x |}^{2}}$

Step 2.12

Simplify.

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Step 2.12.1

Simplify the numerator.

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Step 2.12.1.1

To write $| x |$ as a fraction with a common denominator, multiply by $\frac{| x |}{| x |}$ .

$f'' (x) = - \frac{\frac{| x | | x |}{| x |} - \frac{x^{2}}{| x |}}{{| x |}^{2}}$

Step 2.12.1.2

Combine the numerators over the common denominator.

$f'' (x) = - \frac{\frac{| x | | x | - x^{2}}{| x |}}{{| x |}^{2}}$

Step 2.12.1.3

Simplify the numerator.

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Step 2.12.1.3.1

Multiply $| x | | x |$ .

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Step 2.12.1.3.1.1

To multiply absolute values, multiply the terms inside each absolute value.

$f'' (x) = - \frac{\frac{| x \cdot x | - x^{2}}{| x |}}{{| x |}^{2}}$

Step 2.12.1.3.1.2

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

$f'' (x) = - \frac{\frac{| x \cdot x | - x^{2}}{| x |}}{{| x |}^{2}}$

Step 2.12.1.3.1.3

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

$f'' (x) = - \frac{\frac{| x \cdot x | - x^{2}}{| x |}}{{| x |}^{2}}$

Step 2.12.1.3.1.4

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

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

Step 2.12.1.3.1.5

Add $1$ and $1$ .

$f'' (x) = - \frac{\frac{| x^{2} | - x^{2}}{| x |}}{{| x |}^{2}}$

Step 2.12.1.3.2

Remove non-negative terms from the absolute value.

$f'' (x) = - \frac{\frac{x^{2} - x^{2}}{| x |}}{{| x |}^{2}}$

Step 2.12.1.3.3

Subtract $x^{2}$ from $x^{2}$ .

$f'' (x) = - \frac{\frac{0}{| x |}}{{| x |}^{2}}$

Step 2.12.1.4

Divide $0$ by $| x |$ .

$f'' (x) = - \frac{0}{{| x |}^{2}}$

Step 2.12.2

Remove the absolute value in ${| x |}^{2}$ because exponentiations with even powers are always positive.

$f'' (x) = - \frac{0}{x^{2}}$

Step 2.12.3

Divide $0$ by $x^{2}$ .

$f'' (x) = - 0$

Step 2.12.4

Multiply $- 1$ by $0$ .

$f'' (x) = 0$

Step 3

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

$- \frac{x}{| x |} = 0$

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 $- | x | + 7$ with respect to $x$ is $\frac{d}{d x} [- | x |] + \frac{d}{d x} [7]$ .

$\frac{d}{d x} [- | x |] + \frac{d}{d x} [7]$

Step 4.1.2

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

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Step 4.1.2.1

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

$- \frac{d}{d x} [| x |] + \frac{d}{d x} [7]$

Step 4.1.2.2

The derivative of $| x |$ with respect to $x$ is $\frac{x}{| x |}$ .

$- \frac{x}{| x |} + \frac{d}{d x} [7]$

Step 4.1.3

Differentiate using the Constant Rule.

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Step 4.1.3.1

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$- \frac{x}{| x |} + 0$

Step 4.1.3.2

Add $- \frac{x}{| x |}$ and $0$ .

$f' (x) = - \frac{x}{| x |}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $- \frac{x}{| x |}$ .

$- \frac{x}{| x |}$

Step 5

Set the first derivative equal to $0$ then solve the equation $- \frac{x}{| x |} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$- \frac{x}{| x |} = 0$

Step 5.2

Set the numerator equal to zero.

$x = 0$

Step 5.3

Exclude the solutions that do not make $- \frac{x}{| x |} = 0$ true.

$No solution$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{x}{| x |}$ equal to $0$ to find where the expression is undefined.

$| x | = 0$

Step 6.2

Solve for $x$ .

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Step 6.2.1

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x = \pm 0$

Step 6.2.2

Plus or minus $0$ is $0$ .

$x = 0$

Step 7

Critical points to evaluate.

$x = 0$

Step 8

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.

$0$

Step 9

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

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Step 9.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, 0) \cup (0, \infty)$

Step 9.2

Substitute any number, such as $- 2$ , from the interval $(- \infty, 0)$ in the first derivative $- \frac{x}{| x |}$ to check if the result is negative or positive.

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Step 9.2.1

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

$f' (- 2) = - \frac{- 2}{| - 2 |}$

Step 9.2.2

Simplify the result.

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Step 9.2.2.1

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$f' (- 2) = - \frac{- 2}{2}$

Step 9.2.2.2

Divide $- 2$ by $2$ .

$f' (- 2) = 1$

Step 9.2.2.3

The final answer is $1$ .

$1$

Step 9.3

Substitute any number, such as $2$ , from the interval $(0, \infty)$ in the first derivative $- \frac{x}{| x |}$ to check if the result is negative or positive.

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Step 9.3.1

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

$f' (2) = - \frac{2}{| 2 |}$

Step 9.3.2

Simplify the result.

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Step 9.3.2.1

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$f' (2) = - \frac{2}{2}$

Step 9.3.2.2

Cancel the common factor of $2$ .

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Step 9.3.2.2.1

Cancel the common factor.

$f' (2) = - \frac{2}{2}$

Step 9.3.2.2.2

Rewrite the expression.

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

Step 9.3.2.3

Multiply $- 1$ by $1$ .

$f' (2) = - 1$

Step 9.3.2.4

The final answer is $- 1$ .

$- 1$

Step 9.4

Since the first derivative changed signs from positive to negative around $x = 0$ , then $x = 0$ is a local maximum.

$x = 0$ is a local maximum

Step 10