Calculus Examples

$\frac{x + 4}{x^{2} - 3 x - 28}$

Step 1

Write $\frac{x + 4}{x^{2} - 3 x - 28}$ as a function.

$f (x) = \frac{x + 4}{x^{2} - 3 x - 28}$

Step 2

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

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Step 2.1

Find the second derivative.

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Step 2.1.1

Find the first derivative.

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Step 2.1.1.1

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 $f (x) = x + 4$ and $g (x) = x^{2} - 3 x - 28$ .

$\frac{(x^{2} - 3 x - 28) \frac{d}{d x} [x + 4] - (x + 4) \frac{d}{d x} [x^{2} - 3 x - 28]}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.2

Differentiate.

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Step 2.1.1.2.1

By the Sum Rule, the derivative of $x + 4$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [4]$ .

$\frac{(x^{2} - 3 x - 28) (\frac{d}{d x} [x] + \frac{d}{d x} [4]) - (x + 4) \frac{d}{d x} [x^{2} - 3 x - 28]}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.2.2

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

$\frac{(x^{2} - 3 x - 28) (1 + \frac{d}{d x} [4]) - (x + 4) \frac{d}{d x} [x^{2} - 3 x - 28]}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.2.3

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$\frac{(x^{2} - 3 x - 28) (1 + 0) - (x + 4) \frac{d}{d x} [x^{2} - 3 x - 28]}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.2.4

Simplify the expression.

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Step 2.1.1.2.4.1

Add $1$ and $0$ .

$\frac{(x^{2} - 3 x - 28) \cdot 1 - (x + 4) \frac{d}{d x} [x^{2} - 3 x - 28]}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.2.4.2

Multiply $x^{2} - 3 x - 28$ by $1$ .

$\frac{x^{2} - 3 x - 28 - (x + 4) \frac{d}{d x} [x^{2} - 3 x - 28]}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.2.5

By the Sum Rule, the derivative of $x^{2} - 3 x - 28$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [- 28]$ .

$\frac{x^{2} - 3 x - 28 - (x + 4) (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [- 28])}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.2.6

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

$\frac{x^{2} - 3 x - 28 - (x + 4) (2 x + \frac{d}{d x} [- 3 x] + \frac{d}{d x} [- 28])}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.2.7

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

$\frac{x^{2} - 3 x - 28 - (x + 4) (2 x - 3 \frac{d}{d x} [x] + \frac{d}{d x} [- 28])}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.2.8

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

$\frac{x^{2} - 3 x - 28 - (x + 4) (2 x - 3 \cdot 1 + \frac{d}{d x} [- 28])}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.2.9

Multiply $- 3$ by $1$ .

$\frac{x^{2} - 3 x - 28 - (x + 4) (2 x - 3 + \frac{d}{d x} [- 28])}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.2.10

Since $- 28$ is constant with respect to $x$ , the derivative of $- 28$ with respect to $x$ is $0$ .

$\frac{x^{2} - 3 x - 28 - (x + 4) (2 x - 3 + 0)}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.2.11

Add $2 x - 3$ and $0$ .

$\frac{x^{2} - 3 x - 28 - (x + 4) (2 x - 3)}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3

Simplify.

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Step 2.1.1.3.1

Apply the distributive property.

$\frac{x^{2} - 3 x - 28 + (- x - 1 \cdot 4) (2 x - 3)}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2

Simplify the numerator.

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Step 2.1.1.3.2.1

Simplify each term.

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Step 2.1.1.3.2.1.1

Multiply $- 1$ by $4$ .

$\frac{x^{2} - 3 x - 28 + (- x - 4) (2 x - 3)}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.1.2

Expand $(- x - 4) (2 x - 3)$ using the FOIL Method.

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Step 2.1.1.3.2.1.2.1

Apply the distributive property.

$\frac{x^{2} - 3 x - 28 - x (2 x - 3) - 4 (2 x - 3)}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.1.2.2

Apply the distributive property.

$\frac{x^{2} - 3 x - 28 - x (2 x) - x \cdot - 3 - 4 (2 x - 3)}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.1.2.3

Apply the distributive property.

$\frac{x^{2} - 3 x - 28 - x (2 x) - x \cdot - 3 - 4 (2 x) - 4 \cdot - 3}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.1.3

Simplify and combine like terms.

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Step 2.1.1.3.2.1.3.1

Simplify each term.

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Step 2.1.1.3.2.1.3.1.1

Rewrite using the commutative property of multiplication.

$\frac{x^{2} - 3 x - 28 - 1 \cdot 2 x \cdot x - x \cdot - 3 - 4 (2 x) - 4 \cdot - 3}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.1.3.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 2.1.1.3.2.1.3.1.2.1

Move $x$ .

$\frac{x^{2} - 3 x - 28 - 1 \cdot 2 (x \cdot x) - x \cdot - 3 - 4 (2 x) - 4 \cdot - 3}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.1.3.1.2.2

Multiply $x$ by $x$ .

$\frac{x^{2} - 3 x - 28 - 1 \cdot 2 x^{2} - x \cdot - 3 - 4 (2 x) - 4 \cdot - 3}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.1.3.1.3

Multiply $- 1$ by $2$ .

$\frac{x^{2} - 3 x - 28 - 2 x^{2} - x \cdot - 3 - 4 (2 x) - 4 \cdot - 3}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.1.3.1.4

Multiply $- 3$ by $- 1$ .

$\frac{x^{2} - 3 x - 28 - 2 x^{2} + 3 x - 4 (2 x) - 4 \cdot - 3}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.1.3.1.5

Multiply $2$ by $- 4$ .

$\frac{x^{2} - 3 x - 28 - 2 x^{2} + 3 x - 8 x - 4 \cdot - 3}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.1.3.1.6

Multiply $- 4$ by $- 3$ .

$\frac{x^{2} - 3 x - 28 - 2 x^{2} + 3 x - 8 x + 12}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.1.3.2

Subtract $8 x$ from $3 x$ .

$\frac{x^{2} - 3 x - 28 - 2 x^{2} - 5 x + 12}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.2

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

$\frac{- x^{2} - 3 x - 28 - 5 x + 12}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.3

Subtract $5 x$ from $- 3 x$ .

$\frac{- x^{2} - 8 x - 28 + 12}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.2.4

Add $- 28$ and $12$ .

$\frac{- x^{2} - 8 x - 16}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.3

Factor by grouping.

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Step 2.1.1.3.3.1

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 $a \cdot c = - 1 \cdot - 16 = 16$ and whose sum is $b = - 8$ .

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Step 2.1.1.3.3.1.1

Factor $- 8$ out of $- 8 x$ .

$\frac{- x^{2} - 8 (x) - 16}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.3.1.2

Rewrite $- 8$ as $- 4$ plus $- 4$

$\frac{- x^{2} + (- 4 - 4) x - 16}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.3.1.3

Apply the distributive property.

$\frac{- x^{2} - 4 x - 4 x - 16}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.3.2

Factor out the greatest common factor from each group.

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Step 2.1.1.3.3.2.1

Group the first two terms and the last two terms.

$\frac{(- x^{2} - 4 x) - 4 x - 16}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.3.2.2

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

$\frac{x (- x - 4) + 4 (- x - 4)}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.3.3

Factor the polynomial by factoring out the greatest common factor, $- x - 4$ .

$\frac{(- x - 4) (x + 4)}{{(x^{2} - 3 x - 28)}^{2}}$

Step 2.1.1.3.4

Simplify the denominator.

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Step 2.1.1.3.4.1

Factor $x^{2} - 3 x - 28$ using the AC method.

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Step 2.1.1.3.4.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 28$ and whose sum is $- 3$ .

$- 7, 4$

Step 2.1.1.3.4.1.2

Write the factored form using these integers.

$\frac{(- x - 4) (x + 4)}{{((x - 7) (x + 4))}^{2}}$

Step 2.1.1.3.4.2

Apply the product rule to $(x - 7) (x + 4)$ .

$\frac{(- x - 4) (x + 4)}{{(x - 7)}^{2} {(x + 4)}^{2}}$

Step 2.1.1.3.5

Simplify the numerator.

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Step 2.1.1.3.5.1

Factor $- 1$ out of $- x$ .

$\frac{(- (x) - 4) (x + 4)}{{(x - 7)}^{2} {(x + 4)}^{2}}$

Step 2.1.1.3.5.2

Rewrite $- 4$ as $- 1 (4)$ .

$\frac{(- (x) - 1 (4)) (x + 4)}{{(x - 7)}^{2} {(x + 4)}^{2}}$

Step 2.1.1.3.5.3

Factor $- 1$ out of $- (x) - 1 (4)$ .

$\frac{- (x + 4) (x + 4)}{{(x - 7)}^{2} {(x + 4)}^{2}}$

Step 2.1.1.3.5.4

Rewrite $- (x + 4)$ as $- 1 (x + 4)$ .

$\frac{- 1 (x + 4) (x + 4)}{{(x - 7)}^{2} {(x + 4)}^{2}}$

Step 2.1.1.3.5.5

Raise $x + 4$ to the power of $1$ .

$\frac{- 1 ({(x + 4)}^{1} (x + 4))}{{(x - 7)}^{2} {(x + 4)}^{2}}$

Step 2.1.1.3.5.6

Raise $x + 4$ to the power of $1$ .

$\frac{- 1 ({(x + 4)}^{1} {(x + 4)}^{1})}{{(x - 7)}^{2} {(x + 4)}^{2}}$

Step 2.1.1.3.5.7

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

$\frac{- 1 {(x + 4)}^{1 + 1}}{{(x - 7)}^{2} {(x + 4)}^{2}}$

Step 2.1.1.3.5.8

Add $1$ and $1$ .

$\frac{- 1 {(x + 4)}^{2}}{{(x - 7)}^{2} {(x + 4)}^{2}}$

Step 2.1.1.3.6

Cancel the common factor of ${(x + 4)}^{2}$ .

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Step 2.1.1.3.6.1

Cancel the common factor.

$\frac{- 1 {(x + 4)}^{2}}{{(x - 7)}^{2} {(x + 4)}^{2}}$

Step 2.1.1.3.6.2

Rewrite the expression.

$\frac{- 1}{{(x - 7)}^{2}}$

Step 2.1.1.3.7

Move the negative in front of the fraction.

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

Step 2.1.2

Find the second derivative.

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Step 2.1.2.1

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 $g (x) = \frac{1}{{(x - 7)}^{2}}$ .

$- \frac{d}{d x} [\frac{1}{{(x - 7)}^{2}}] + \frac{1}{{(x - 7)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.2

Apply basic rules of exponents.

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Step 2.1.2.2.1

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

$- \frac{d}{d x} [{({(x - 7)}^{2})}^{- 1}] + \frac{1}{{(x - 7)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.2.2

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

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Step 2.1.2.2.2.1

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

$- \frac{d}{d x} [{(x - 7)}^{2 \cdot - 1}] + \frac{1}{{(x - 7)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.2.2.2

Multiply $2$ by $- 1$ .

$- \frac{d}{d x} [{(x - 7)}^{- 2}] + \frac{1}{{(x - 7)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.3

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 2.1.2.3.1

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

$- (\frac{d}{d u} [u^{- 2}] \frac{d}{d x} [x - 7]) + \frac{1}{{(x - 7)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.3.2

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

$- (- 2 u^{- 3} \frac{d}{d x} [x - 7]) + \frac{1}{{(x - 7)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.3.3

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

$- (- 2 {(x - 7)}^{- 3} \frac{d}{d x} [x - 7]) + \frac{1}{{(x - 7)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.4

Differentiate.

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Step 2.1.2.4.1

Multiply $- 2$ by $- 1$ .

$2 ({(x - 7)}^{- 3} \frac{d}{d x} [x - 7]) + \frac{1}{{(x - 7)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.4.2

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

$2 {(x - 7)}^{- 3} (\frac{d}{d x} [x] + \frac{d}{d x} [- 7]) + \frac{1}{{(x - 7)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.4.3

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

$2 {(x - 7)}^{- 3} (1 + \frac{d}{d x} [- 7]) + \frac{1}{{(x - 7)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.4.4

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

$2 {(x - 7)}^{- 3} (1 + 0) + \frac{1}{{(x - 7)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.4.5

Simplify the expression.

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Step 2.1.2.4.5.1

Add $1$ and $0$ .

$2 {(x - 7)}^{- 3} \cdot 1 + \frac{1}{{(x - 7)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.4.5.2

Multiply $2$ by $1$ .

$2 {(x - 7)}^{- 3} + \frac{1}{{(x - 7)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.4.6

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

$2 {(x - 7)}^{- 3} + \frac{1}{{(x - 7)}^{2}} \cdot 0$

Step 2.1.2.4.7

Simplify the expression.

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Step 2.1.2.4.7.1

Multiply $\frac{1}{{(x - 7)}^{2}}$ by $0$ .

$2 {(x - 7)}^{- 3} + 0$

Step 2.1.2.4.7.2

Add $2 {(x - 7)}^{- 3}$ and $0$ .

$2 {(x - 7)}^{- 3}$

Step 2.1.2.5

Simplify.

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Step 2.1.2.5.1

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

$2 \frac{1}{{(x - 7)}^{3}}$

Step 2.1.2.5.2

Combine $2$ and $\frac{1}{{(x - 7)}^{3}}$ .

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

Step 2.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{2}{{(x - 7)}^{3}}$ .

$\frac{2}{{(x - 7)}^{3}}$

Step 2.2

Set the second derivative equal to $0$ then solve the equation $\frac{2}{{(x - 7)}^{3}} = 0$ .

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Step 2.2.1

Set the second derivative equal to $0$ .

$\frac{2}{{(x - 7)}^{3}} = 0$

Step 2.2.2

Set the numerator equal to zero.

$2 = 0$

Step 2.2.3

Since $2 \neq 0$ , there are no solutions.

No solution

Step 3

Find the domain of $f (x) = \frac{x + 4}{x^{2} - 3 x - 28}$ .

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Step 3.1

Set the denominator in $\frac{x + 4}{x^{2} - 3 x - 28}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 3 x - 28 = 0$

Step 3.2

Solve for $x$ .

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Step 3.2.1

Factor $x^{2} - 3 x - 28$ using the AC method.

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Step 3.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 28$ and whose sum is $- 3$ .

$- 7, 4$

Step 3.2.1.2

Write the factored form using these integers.

$(x - 7) (x + 4) = 0$

Step 3.2.2

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

$x - 7 = 0$

$x + 4 = 0$

Step 3.2.3

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 3.2.3.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 3.2.3.2

Add $7$ to both sides of the equation.

$x = 7$

Step 3.2.4

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 3.2.4.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 3.2.4.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 3.2.5

The final solution is all the values that make $(x - 7) (x + 4) = 0$ true.

$x = 7, - 4$

Step 3.3

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

Interval Notation:

$(- \infty, - 4) \cup (- 4, 7) \cup (7, \infty)$

Set-Builder Notation:

${x | x \neq - 4, 7}$

Interval Notation:

$(- \infty, - 4) \cup (- 4, 7) \cup (7, \infty)$

Set-Builder Notation:

${x | x \neq - 4, 7}$

Step 4

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

$(- \infty, - 4) \cup (- 4, 7) \cup (7, \infty)$

Step 5

Substitute any number from the interval $(- \infty, - 4)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

Replace the variable $x$ with $- 6$ in the expression.

$f'' (- 6) = \frac{2}{{((- 6) - 7)}^{3}}$

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

Subtract $7$ from $- 6$ .

$f'' (- 6) = \frac{2}{{(- 13)}^{3}}$

Step 5.2.1.2

Raise $- 13$ to the power of $3$ .

$f'' (- 6) = \frac{2}{- 2197}$

Step 5.2.2

Move the negative in front of the fraction.

$f'' (- 6) = - \frac{2}{2197}$

Step 5.2.3

The final answer is $- \frac{2}{2197}$ .

$- \frac{2}{2197}$

Step 5.3

The graph is concave down on the interval $(- \infty, - 4)$ because $f'' (- 6)$ is negative.

Concave down on $(- \infty, - 4)$ since $f'' (x)$ is negative

Step 6

Substitute any number from the interval $(- 4, 7)$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

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

$f'' (0) = \frac{2}{{((0) - 7)}^{3}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the denominator.

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Step 6.2.1.1

Subtract $7$ from $0$ .

$f'' (0) = \frac{2}{{(- 7)}^{3}}$

Step 6.2.1.2

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

$f'' (0) = \frac{2}{- 343}$

Step 6.2.2

Move the negative in front of the fraction.

$f'' (0) = - \frac{2}{343}$

Step 6.2.3

The final answer is $- \frac{2}{343}$ .

$- \frac{2}{343}$

Step 6.3

The graph is concave down on the interval $(- 4, 7)$ because $f'' (0)$ is negative.

Concave down on $(- 4, 7)$ since $f'' (x)$ is negative

Step 7

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

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Step 7.1

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

$f'' (10) = \frac{2}{{((10) - 7)}^{3}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify the denominator.

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Step 7.2.1.1

Subtract $7$ from $10$ .

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

Step 7.2.1.2

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

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

Step 7.2.2

The final answer is $\frac{2}{27}$ .

$\frac{2}{27}$

Step 7.3

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

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

Step 8

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

Concave down on $(- \infty, - 4)$ since $f'' (x)$ is negative

Concave down on $(- 4, 7)$ since $f'' (x)$ is negative

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

Step 9