Calculus Examples

$\frac{x + 5}{x^{2} - 25}$

Step 1

Write $\frac{x + 5}{x^{2} - 25}$ as a function.

$f (x) = \frac{x + 5}{x^{2} - 25}$

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 $f (x) = x + 5$ and $g (x) = x^{2} - 25$ .

$\frac{(x^{2} - 25) \frac{d}{d x} [x + 5] - (x + 5) \frac{d}{d x} [x^{2} - 25]}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.2

Differentiate.

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Step 2.1.1.2.1

By the Sum Rule, the derivative of $x + 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [5]$ .

$\frac{(x^{2} - 25) (\frac{d}{d x} [x] + \frac{d}{d x} [5]) - (x + 5) \frac{d}{d x} [x^{2} - 25]}{{(x^{2} - 25)}^{2}}$

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 $n = 1$ .

$\frac{(x^{2} - 25) (1 + \frac{d}{d x} [5]) - (x + 5) \frac{d}{d x} [x^{2} - 25]}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.2.3

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

$\frac{(x^{2} - 25) (1 + 0) - (x + 5) \frac{d}{d x} [x^{2} - 25]}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.2.4

Simplify the expression.

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Step 2.1.1.2.4.1

Add $1$ and $0$ .

$\frac{(x^{2} - 25) \cdot 1 - (x + 5) \frac{d}{d x} [x^{2} - 25]}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.2.4.2

Multiply $x^{2} - 25$ by $1$ .

$\frac{x^{2} - 25 - (x + 5) \frac{d}{d x} [x^{2} - 25]}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.2.5

By the Sum Rule, the derivative of $x^{2} - 25$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 25]$ .

$\frac{x^{2} - 25 - (x + 5) (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 25])}{{(x^{2} - 25)}^{2}}$

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 $n = 2$ .

$\frac{x^{2} - 25 - (x + 5) (2 x + \frac{d}{d x} [- 25])}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.2.7

Since $- 25$ is constant with respect to $x$ , the derivative of $- 25$ with respect to $x$ is $0$ .

$\frac{x^{2} - 25 - (x + 5) (2 x + 0)}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.2.8

Simplify the expression.

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Step 2.1.1.2.8.1

Add $2 x$ and $0$ .

$\frac{x^{2} - 25 - (x + 5) (2 x)}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.2.8.2

Multiply $2$ by $- 1$ .

$\frac{x^{2} - 25 - 2 (x + 5) x}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3

Simplify.

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Step 2.1.1.3.1

Apply the distributive property.

$\frac{x^{2} - 25 + (- 2 x - 2 \cdot 5) x}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3.2

Apply the distributive property.

$\frac{x^{2} - 25 - 2 x \cdot x - 2 \cdot 5 x}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3.3

Simplify the numerator.

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Step 2.1.1.3.3.1

Simplify each term.

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Step 2.1.1.3.3.1.1

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

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Step 2.1.1.3.3.1.1.1

Move $x$ .

$\frac{x^{2} - 25 - 2 (x \cdot x) - 2 \cdot 5 x}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3.3.1.1.2

Multiply $x$ by $x$ .

$\frac{x^{2} - 25 - 2 x^{2} - 2 \cdot 5 x}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3.3.1.2

Multiply $- 2$ by $5$ .

$\frac{x^{2} - 25 - 2 x^{2} - 10 x}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3.3.2

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

$\frac{- x^{2} - 25 - 10 x}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3.4

Reorder terms.

$\frac{- x^{2} - 10 x - 25}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3.5

Factor by grouping.

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Step 2.1.1.3.5.1

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 $a \cdot c = - 1 \cdot - 25 = 25$ and whose sum is $b = - 10$ .

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Step 2.1.1.3.5.1.1

Factor $- 10$ out of $- 10 x$ .

$\frac{- x^{2} - 10 (x) - 25}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3.5.1.2

Rewrite $- 10$ as $- 5$ plus $- 5$

$\frac{- x^{2} + (- 5 - 5) x - 25}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3.5.1.3

Apply the distributive property.

$\frac{- x^{2} - 5 x - 5 x - 25}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3.5.2

Factor out the greatest common factor from each group.

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Step 2.1.1.3.5.2.1

Group the first two terms and the last two terms.

$\frac{(- x^{2} - 5 x) - 5 x - 25}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3.5.2.2

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

$\frac{x (- x - 5) + 5 (- x - 5)}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3.5.3

Factor the polynomial by factoring out the greatest common factor, $- x - 5$ .

$\frac{(- x - 5) (x + 5)}{{(x^{2} - 25)}^{2}}$

Step 2.1.1.3.6

Simplify the denominator.

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Step 2.1.1.3.6.1

Rewrite $25$ as $5^{2}$ .

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

Step 2.1.1.3.6.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

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

Step 2.1.1.3.6.3

Apply the product rule to $(x + 5) (x - 5)$ .

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

Step 2.1.1.3.7

Simplify the numerator.

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Step 2.1.1.3.7.1

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

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

Step 2.1.1.3.7.2

Rewrite $- 5$ as $- 1 (5)$ .

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

Step 2.1.1.3.7.3

Factor $- 1$ out of $- (x) - 1 (5)$ .

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

Step 2.1.1.3.7.4

Rewrite $- (x + 5)$ as $- 1 (x + 5)$ .

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

Step 2.1.1.3.7.5

Raise $x + 5$ to the power of $1$ .

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

Step 2.1.1.3.7.6

Raise $x + 5$ to the power of $1$ .

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

Step 2.1.1.3.7.7

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

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

Step 2.1.1.3.7.8

Add $1$ and $1$ .

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

Step 2.1.1.3.8

Cancel the common factor of ${(x + 5)}^{2}$ .

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Step 2.1.1.3.8.1

Cancel the common factor.

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

Step 2.1.1.3.8.2

Rewrite the expression.

$\frac{- 1}{{(x - 5)}^{2}}$

Step 2.1.1.3.9

Move the negative in front of the fraction.

$f' (x) = - \frac{1}{{(x - 5)}^{2}}$

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 $g (x) = \frac{1}{{(x - 5)}^{2}}$ .

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

Step 2.1.2.2

Apply basic rules of exponents.

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Step 2.1.2.2.1

Rewrite $\frac{1}{{(x - 5)}^{2}}$ as ${({(x - 5)}^{2})}^{- 1}$ .

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

Step 2.1.2.2.2

Multiply the exponents in ${({(x - 5)}^{2})}^{- 1}$ .

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Step 2.1.2.2.2.1

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

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

Step 2.1.2.2.2.2

Multiply $2$ by $- 1$ .

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

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 $g (x) = x - 5$ .

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Step 2.1.2.3.1

To apply the Chain Rule, set $u$ as $x - 5$ .

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

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 $n = - 2$ .

$- (- 2 u^{- 3} \frac{d}{d x} [x - 5]) + \frac{1}{{(x - 5)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.3.3

Replace all occurrences of $u$ with $x - 5$ .

$- (- 2 {(x - 5)}^{- 3} \frac{d}{d x} [x - 5]) + \frac{1}{{(x - 5)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.4

Differentiate.

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Step 2.1.2.4.1

Multiply $- 2$ by $- 1$ .

$2 ({(x - 5)}^{- 3} \frac{d}{d x} [x - 5]) + \frac{1}{{(x - 5)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.4.2

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

$2 {(x - 5)}^{- 3} (\frac{d}{d x} [x] + \frac{d}{d x} [- 5]) + \frac{1}{{(x - 5)}^{2}} \frac{d}{d x} [- 1]$

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 $n = 1$ .

$2 {(x - 5)}^{- 3} (1 + \frac{d}{d x} [- 5]) + \frac{1}{{(x - 5)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.4.4

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

$2 {(x - 5)}^{- 3} (1 + 0) + \frac{1}{{(x - 5)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.4.5

Simplify the expression.

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Step 2.1.2.4.5.1

Add $1$ and $0$ .

$2 {(x - 5)}^{- 3} \cdot 1 + \frac{1}{{(x - 5)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.4.5.2

Multiply $2$ by $1$ .

$2 {(x - 5)}^{- 3} + \frac{1}{{(x - 5)}^{2}} \frac{d}{d x} [- 1]$

Step 2.1.2.4.6

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

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

Step 2.1.2.4.7

Simplify the expression.

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Step 2.1.2.4.7.1

Multiply $\frac{1}{{(x - 5)}^{2}}$ by $0$ .

$2 {(x - 5)}^{- 3} + 0$

Step 2.1.2.4.7.2

Add $2 {(x - 5)}^{- 3}$ and $0$ .

$2 {(x - 5)}^{- 3}$

Step 2.1.2.5

Simplify.

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Step 2.1.2.5.1

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

$2 \frac{1}{{(x - 5)}^{3}}$

Step 2.1.2.5.2

Combine $2$ and $\frac{1}{{(x - 5)}^{3}}$ .

$f'' (x) = \frac{2}{{(x - 5)}^{3}}$

Step 2.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{2}{{(x - 5)}^{3}}$ .

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

Step 2.2

Set the second derivative equal to $0$ then solve the equation $\frac{2}{{(x - 5)}^{3}} = 0$ .

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Step 2.2.1

Set the second derivative equal to $0$ .

$\frac{2}{{(x - 5)}^{3}} = 0$

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 $f (x) = \frac{x + 5}{x^{2} - 25}$ .

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Step 3.1

Set the denominator in $\frac{x + 5}{x^{2} - 25}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 25 = 0$

Step 3.2

Solve for $x$ .

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Step 3.2.1

Add $25$ to both sides of the equation.

$x^{2} = 25$

Step 3.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{25}$

Step 3.2.3

Simplify $\pm \sqrt{25}$ .

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Step 3.2.3.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Step 3.2.3.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 5$

Step 3.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 5$

Step 3.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 5$

Step 3.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 5, - 5$

Step 3.3

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

Interval Notation:

$(- \infty, - 5) \cup (- 5, 5) \cup (5, \infty)$

Set-Builder Notation:

${x | x \neq 5, - 5}$

Interval Notation:

$(- \infty, - 5) \cup (- 5, 5) \cup (5, \infty)$

Set-Builder Notation:

${x | x \neq 5, - 5}$

Step 4

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

$(- \infty, - 5) \cup (- 5, 5) \cup (5, \infty)$

Step 5

Substitute any number from the interval $(- \infty, - 5)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

Replace the variable $x$ with $- 8$ in the expression.

$f'' (- 8) = \frac{2}{{((- 8) - 5)}^{3}}$

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 $5$ from $- 8$ .

$f'' (- 8) = \frac{2}{{(- 13)}^{3}}$

Step 5.2.1.2

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

$f'' (- 8) = \frac{2}{- 2197}$

Step 5.2.2

Move the negative in front of the fraction.

$f'' (- 8) = - \frac{2}{2197}$

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 $(- \infty, - 5)$ because $f'' (- 8)$ is negative.

Concave down on $(- \infty, - 5)$ since $f'' (x)$ is negative

Step 6

Substitute any number from the interval $(- 5, 5)$ 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.

$f'' (0) = \frac{2}{{((0) - 5)}^{3}}$

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 $5$ from $0$ .

$f'' (0) = \frac{2}{{(- 5)}^{3}}$

Step 6.2.1.2

Raise $- 5$ to the power of $3$ .

$f'' (0) = \frac{2}{- 125}$

Step 6.2.2

Move the negative in front of the fraction.

$f'' (0) = - \frac{2}{125}$

Step 6.2.3

The final answer is $- \frac{2}{125}$ .

$- \frac{2}{125}$

Step 6.3

The graph is concave down on the interval $(- 5, 5)$ because $f'' (0)$ is negative.

Concave down on $(- 5, 5)$ since $f'' (x)$ is negative

Step 7

Substitute any number from the interval $(5, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 7.1

Replace the variable $x$ with $8$ in the expression.

$f'' (8) = \frac{2}{{((8) - 5)}^{3}}$

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 $5$ from $8$ .

$f'' (8) = \frac{2}{3^{3}}$

Step 7.2.1.2

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

$f'' (8) = \frac{2}{27}$

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 $(5, \infty)$ because $f'' (8)$ is positive.

Concave up on $(5, \infty)$ 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 $(- \infty, - 5)$ since $f'' (x)$ is negative

Concave down on $(- 5, 5)$ since $f'' (x)$ is negative

Concave up on $(5, \infty)$ since $f'' (x)$ is positive

Step 9