Calculus Examples

$f (x) = \frac{x - 2}{x + 7}$

Step 1

Find the $x$ values where the second derivative is equal to $0$ .

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Step 1.1

Find the second derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

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 $f (x) = x - 2$ and $g (x) = x + 7$ .

$f' (x) = \frac{(x + 7) \frac{d}{d x} (x - 2) - (x - 2) \frac{d}{d x} (x + 7)}{{(x + 7)}^{2}}$

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

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

$f' (x) = \frac{(x + 7) (\frac{d}{d x} (x) + \frac{d}{d x} (- 2)) - (x - 2) \frac{d}{d x} (x + 7)}{{(x + 7)}^{2}}$

Step 1.1.1.2.2

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

$f' (x) = \frac{(x + 7) (1 + \frac{d}{d x} (- 2)) - (x - 2) \frac{d}{d x} (x + 7)}{{(x + 7)}^{2}}$

Step 1.1.1.2.3

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

$f' (x) = \frac{(x + 7) (1 + 0) - (x - 2) \frac{d}{d x} (x + 7)}{{(x + 7)}^{2}}$

Step 1.1.1.2.4

Simplify the expression.

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Step 1.1.1.2.4.1

Add $1$ and $0$ .

$f' (x) = \frac{(x + 7) \cdot 1 - (x - 2) \frac{d}{d x} (x + 7)}{{(x + 7)}^{2}}$

Step 1.1.1.2.4.2

Multiply $x + 7$ by $1$ .

$f' (x) = \frac{x + 7 - (x - 2) \frac{d}{d x} (x + 7)}{{(x + 7)}^{2}}$

Step 1.1.1.2.5

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

$f' (x) = \frac{x + 7 - (x - 2) (\frac{d}{d x} (x) + \frac{d}{d x} (7))}{{(x + 7)}^{2}}$

Step 1.1.1.2.6

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

$f' (x) = \frac{x + 7 - (x - 2) (1 + \frac{d}{d x} (7))}{{(x + 7)}^{2}}$

Step 1.1.1.2.7

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

$f' (x) = \frac{x + 7 - (x - 2) (1 + 0)}{{(x + 7)}^{2}}$

Step 1.1.1.2.8

Simplify the expression.

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Step 1.1.1.2.8.1

Add $1$ and $0$ .

$f' (x) = \frac{x + 7 - (x - 2) \cdot 1}{{(x + 7)}^{2}}$

Step 1.1.1.2.8.2

Multiply $- 1$ by $1$ .

$f' (x) = \frac{x + 7 - (x - 2)}{{(x + 7)}^{2}}$

Step 1.1.1.3

Simplify.

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Step 1.1.1.3.1

Apply the distributive property.

$f' (x) = \frac{x + 7 - x + 2}{{(x + 7)}^{2}}$

Step 1.1.1.3.2

Simplify the numerator.

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Step 1.1.1.3.2.1

Combine the opposite terms in $x + 7 - x - - 2$ .

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Step 1.1.1.3.2.1.1

Subtract $x$ from $x$ .

$f' (x) = \frac{0 + 7 + 2}{{(x + 7)}^{2}}$

Step 1.1.1.3.2.1.2

Add $0$ and $7$ .

$f' (x) = \frac{7 + 2}{{(x + 7)}^{2}}$

Step 1.1.1.3.2.2

Multiply $- 1$ by $- 2$ .

$f' (x) = \frac{7 + 2}{{(x + 7)}^{2}}$

Step 1.1.1.3.2.3

Add $7$ and $2$ .

$f' (x) = \frac{9}{{(x + 7)}^{2}}$

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

Differentiate using the Constant Multiple Rule.

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Step 1.1.2.1.1

Since $9$ is constant with respect to $x$ , the derivative of $\frac{9}{{(x + 7)}^{2}}$ with respect to $x$ is $9 \frac{d}{d x} [\frac{1}{{(x + 7)}^{2}}]$ .

$f'' (x) = 9 \frac{d}{d x} (\frac{1}{{(x + 7)}^{2}})$

Step 1.1.2.1.2

Apply basic rules of exponents.

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Step 1.1.2.1.2.1

Rewrite $\frac{1}{{(x + 7)}^{2}}$ as ${({(x + 7)}^{2})}^{- 1}$ .

$f'' (x) = 9 \frac{d}{d x} ({({(x + 7)}^{2})}^{- 1})$

Step 1.1.2.1.2.2

Multiply the exponents in ${({(x + 7)}^{2})}^{- 1}$ .

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Step 1.1.2.1.2.2.1

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

$f'' (x) = 9 \frac{d}{d x} ({(x + 7)}^{2 \cdot - 1})$

Step 1.1.2.1.2.2.2

Multiply $2$ by $- 1$ .

$f'' (x) = 9 \frac{d}{d x} ({(x + 7)}^{- 2})$

Step 1.1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 2}$ and $g (x) = x + 7$ .

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Step 1.1.2.2.1

To apply the Chain Rule, set $u$ as $x + 7$ .

$f'' (x) = 9 (\frac{d}{d u} (u^{- 2}) \frac{d}{d x} (x + 7))$

Step 1.1.2.2.2

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

$f'' (x) = 9 (- 2 u^{- 3} \frac{d}{d x} (x + 7))$

Step 1.1.2.2.3

Replace all occurrences of $u$ with $x + 7$ .

$f'' (x) = 9 (- 2 {(x + 7)}^{- 3} \frac{d}{d x} (x + 7))$

Step 1.1.2.3

Differentiate.

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Step 1.1.2.3.1

Multiply $- 2$ by $9$ .

$f'' (x) = - 18 ({(x + 7)}^{- 3} \frac{d}{d x} (x + 7))$

Step 1.1.2.3.2

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

$f'' (x) = - 18 {(x + 7)}^{- 3} (\frac{d}{d x} (x) + \frac{d}{d x} (7))$

Step 1.1.2.3.3

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

$f'' (x) = - 18 {(x + 7)}^{- 3} (1 + \frac{d}{d x} (7))$

Step 1.1.2.3.4

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

$f'' (x) = - 18 {(x + 7)}^{- 3} (1 + 0)$

Step 1.1.2.3.5

Simplify the expression.

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Step 1.1.2.3.5.1

Add $1$ and $0$ .

$f'' (x) = - 18 {(x + 7)}^{- 3} \cdot 1$

Step 1.1.2.3.5.2

Multiply $- 18$ by $1$ .

$f'' (x) = - 18 {(x + 7)}^{- 3}$

Step 1.1.2.4

Simplify.

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Step 1.1.2.4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = - 18 \frac{1}{{(x + 7)}^{3}}$

Step 1.1.2.4.2

Combine terms.

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Step 1.1.2.4.2.1

Combine $- 18$ and $\frac{1}{{(x + 7)}^{3}}$ .

$f'' (x) = \frac{- 18}{{(x + 7)}^{3}}$

Step 1.1.2.4.2.2

Move the negative in front of the fraction.

$f'' (x) = - \frac{18}{{(x + 7)}^{3}}$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $- \frac{18}{{(x + 7)}^{3}}$ .

$- \frac{18}{{(x + 7)}^{3}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $- \frac{18}{{(x + 7)}^{3}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$- \frac{18}{{(x + 7)}^{3}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$18 = 0$

Step 1.2.3

Since $18 \neq 0$ , there are no solutions.

No solution

Step 2

Find the domain of $f (x) = \frac{x - 2}{x + 7}$ .

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Step 2.1

Set the denominator in $\frac{x - 2}{x + 7}$ equal to $0$ to find where the expression is undefined.

$x + 7 = 0$

Step 2.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 2.3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 7) \cup (- 7, \infty)$

Set-Builder Notation:

${x | x \neq - 7}$

Interval Notation:

$(- \infty, - 7) \cup (- 7, \infty)$

Set-Builder Notation:

${x | x \neq - 7}$

Step 3

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, - 7) \cup (- 7, \infty)$

Step 4

Substitute any number from the interval $(- \infty, - 7)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

Replace the variable $x$ with $- 10$ in the expression.

$f'' (- 10) = - \frac{18}{{((- 10) + 7)}^{3}}$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify the denominator.

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Step 4.2.1.1

Add $- 10$ and $7$ .

$f'' (- 10) = - \frac{18}{{(- 3)}^{3}}$

Step 4.2.1.2

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

$f'' (- 10) = - \frac{18}{- 27}$

Step 4.2.2

Reduce the expression by cancelling the common factors.

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Step 4.2.2.1

Cancel the common factor of $18$ and $- 27$ .

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Step 4.2.2.1.1

Factor $9$ out of $18$ .

$f'' (- 10) = - \frac{9 (2)}{- 27}$

Step 4.2.2.1.2

Cancel the common factors.

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Step 4.2.2.1.2.1

Factor $9$ out of $- 27$ .

$f'' (- 10) = - \frac{9 \cdot 2}{9 \cdot - 3}$

Step 4.2.2.1.2.2

Cancel the common factor.

$f'' (- 10) = - \frac{9 \cdot 2}{9 \cdot - 3}$

Step 4.2.2.1.2.3

Rewrite the expression.

$f'' (- 10) = - \frac{2}{- 3}$

Step 4.2.2.2

Move the negative in front of the fraction.

$f'' (- 10) = \frac{2}{3}$

Step 4.2.3

The final answer is $\frac{2}{3}$ .

$\frac{2}{3}$

Step 4.3

The graph is concave up on the interval $(- \infty, - 7)$ because $f'' (- 10)$ is positive.

Concave up on $(- \infty, - 7)$ since $f'' (x)$ is positive

Step 5

Substitute any number from the interval $(- 7, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

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

$f'' (0) = - \frac{18}{{((0) + 7)}^{3}}$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify the denominator.

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Step 5.2.1.1

Add $0$ and $7$ .

$f'' (0) = - \frac{18}{7^{3}}$

Step 5.2.1.2

Raise $7$ to the power of $3$ .

$f'' (0) = - \frac{18}{343}$

Step 5.2.2

The final answer is $- \frac{18}{343}$ .

$- \frac{18}{343}$

Step 5.3

The graph is concave down on the interval $(- 7, \infty)$ because $f'' (0)$ is negative.

Concave down on $(- 7, \infty)$ since $f'' (x)$ is negative

Step 6

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave up on $(- \infty, - 7)$ since $f'' (x)$ is positive

Concave down on $(- 7, \infty)$ since $f'' (x)$ is negative

Step 7