Calculus Examples

$f (x) = \frac{1}{9 - x^{2}}$

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

Rewrite $\frac{1}{9 - x^{2}}$ as ${(9 - x^{2})}^{- 1}$ .

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

Step 1.1.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = 9 - x^{2}$ .

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Step 1.1.1.2.1

To apply the Chain Rule, set $u$ as $9 - x^{2}$ .

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

Step 1.1.1.2.2

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

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

Step 1.1.1.2.3

Replace all occurrences of $u$ with $9 - x^{2}$ .

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

Step 1.1.1.3

Differentiate.

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Step 1.1.1.3.1

By the Sum Rule, the derivative of $9 - x^{2}$ with respect to $x$ is $\frac{d}{d x} [9] + \frac{d}{d x} [- x^{2}]$ .

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

Step 1.1.1.3.2

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$f' (x) = - {(9 - x^{2})}^{- 2} (0 + \frac{d}{d x} (- x^{2}))$

Step 1.1.1.3.3

Add $0$ and $\frac{d}{d x} [- x^{2}]$ .

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

Step 1.1.1.3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

Step 1.1.1.3.5

Multiply.

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Step 1.1.1.3.5.1

Multiply $- 1$ by $- 1$ .

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

Step 1.1.1.3.5.2

Multiply ${(9 - x^{2})}^{- 2}$ by $1$ .

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

Step 1.1.1.3.6

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

$f' (x) = {(9 - x^{2})}^{- 2} (2 x)$

Step 1.1.1.3.7

Move $2$ to the left of ${(9 - x^{2})}^{- 2}$ .

$f' (x) = 2 \cdot ({(9 - x^{2})}^{- 2} x)$

Step 1.1.1.4

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

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

Step 1.1.1.5

Simplify.

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Step 1.1.1.5.1

Combine terms.

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Step 1.1.1.5.1.1

Combine $2$ and $\frac{1}{{(9 - x^{2})}^{2}}$ .

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

Step 1.1.1.5.1.2

Combine $\frac{2}{{(9 - x^{2})}^{2}}$ and $x$ .

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

Step 1.1.1.5.2

Reorder terms.

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

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

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

Step 1.1.2.2

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 $g (x) = {(- x^{2} + 9)}^{2}$ .

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

Step 1.1.2.3

Differentiate using the Power Rule.

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Step 1.1.2.3.1

Multiply the exponents in ${({(- x^{2} + 9)}^{2})}^{2}$ .

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Step 1.1.2.3.1.1

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

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

Step 1.1.2.3.1.2

Multiply $2$ by $2$ .

$f'' (x) = 2 (\frac{{(- x^{2} + 9)}^{2} \frac{d}{d x} (x) - x \frac{d}{d x} {(- x^{2} + 9)}^{2}}{{(- x^{2} + 9)}^{4}})$

Step 1.1.2.3.2

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

$f'' (x) = 2 (\frac{{(- x^{2} + 9)}^{2} \cdot 1 - x \frac{d}{d x} {(- x^{2} + 9)}^{2}}{{(- x^{2} + 9)}^{4}})$

Step 1.1.2.3.3

Multiply ${(- x^{2} + 9)}^{2}$ by $1$ .

$f'' (x) = 2 (\frac{{(- x^{2} + 9)}^{2} - x \frac{d}{d x} {(- x^{2} + 9)}^{2}}{{(- x^{2} + 9)}^{4}})$

Step 1.1.2.4

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 $g (x) = - x^{2} + 9$ .

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Step 1.1.2.4.1

To apply the Chain Rule, set $u$ as $- x^{2} + 9$ .

$f'' (x) = 2 (\frac{{(- x^{2} + 9)}^{2} - x (\frac{d}{d u} (u^{2}) \frac{d}{d x} (- x^{2} + 9))}{{(- x^{2} + 9)}^{4}})$

Step 1.1.2.4.2

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

$f'' (x) = 2 (\frac{{(- x^{2} + 9)}^{2} - x (2 u \frac{d}{d x} (- x^{2} + 9))}{{(- x^{2} + 9)}^{4}})$

Step 1.1.2.4.3

Replace all occurrences of $u$ with $- x^{2} + 9$ .

$f'' (x) = 2 (\frac{{(- x^{2} + 9)}^{2} - x (2 (- x^{2} + 9) \frac{d}{d x} (- x^{2} + 9))}{{(- x^{2} + 9)}^{4}})$

Step 1.1.2.5

Simplify with factoring out.

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Step 1.1.2.5.1

Multiply $2$ by $- 1$ .

$f'' (x) = 2 (\frac{{(- x^{2} + 9)}^{2} - 2 x ((- x^{2} + 9) \frac{d}{d x} (- x^{2} + 9))}{{(- x^{2} + 9)}^{4}})$

Step 1.1.2.5.2

Factor $- x^{2} + 9$ out of ${(- x^{2} + 9)}^{2} - 2 x ((- x^{2} + 9) \frac{d}{d x} [- x^{2} + 9])$ .

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Step 1.1.2.5.2.1

Factor $- x^{2} + 9$ out of ${(- x^{2} + 9)}^{2}$ .

$f'' (x) = 2 (\frac{(- x^{2} + 9) (- x^{2} + 9) - 2 x ((- x^{2} + 9) \frac{d}{d x} (- x^{2} + 9))}{{(- x^{2} + 9)}^{4}})$

Step 1.1.2.5.2.2

Factor $- x^{2} + 9$ out of $- 2 x ((- x^{2} + 9) \frac{d}{d x} [- x^{2} + 9])$ .

$f'' (x) = 2 (\frac{(- x^{2} + 9) (- x^{2} + 9) + (- x^{2} + 9) (- 2 x (\frac{d}{d x} (- x^{2} + 9)))}{{(- x^{2} + 9)}^{4}})$

Step 1.1.2.5.2.3

Factor $- x^{2} + 9$ out of $(- x^{2} + 9) (- x^{2} + 9) + (- x^{2} + 9) (- 2 x (\frac{d}{d x} [- x^{2} + 9]))$ .

$f'' (x) = 2 (\frac{(- x^{2} + 9) (- x^{2} + 9 - 2 x (\frac{d}{d x} (- x^{2} + 9)))}{{(- x^{2} + 9)}^{4}})$

Step 1.1.2.6

Cancel the common factors.

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Step 1.1.2.6.1

Factor $- x^{2} + 9$ out of ${(- x^{2} + 9)}^{4}$ .

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

Step 1.1.2.6.2

Cancel the common factor.

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

Step 1.1.2.6.3

Rewrite the expression.

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

Step 1.1.2.7

By the Sum Rule, the derivative of $- x^{2} + 9$ with respect to $x$ is $\frac{d}{d x} [- x^{2}] + \frac{d}{d x} [9]$ .

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

Step 1.1.2.8

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

Step 1.1.2.9

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

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

Step 1.1.2.10

Multiply $2$ by $- 1$ .

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

Step 1.1.2.11

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$f'' (x) = 2 (\frac{- x^{2} + 9 - 2 x (- 2 x + 0)}{{(- x^{2} + 9)}^{3}})$

Step 1.1.2.12

Simplify the expression.

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Step 1.1.2.12.1

Add $- 2 x$ and $0$ .

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

Step 1.1.2.12.2

Multiply $- 2$ by $- 2$ .

$f'' (x) = 2 (\frac{- x^{2} + 9 + 4 x \cdot x}{{(- x^{2} + 9)}^{3}})$

Step 1.1.2.13

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

$f'' (x) = 2 (\frac{- x^{2} + 9 + 4 (x \cdot x)}{{(- x^{2} + 9)}^{3}})$

Step 1.1.2.14

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

$f'' (x) = 2 (\frac{- x^{2} + 9 + 4 (x \cdot x)}{{(- x^{2} + 9)}^{3}})$

Step 1.1.2.15

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

$f'' (x) = 2 (\frac{- x^{2} + 9 + 4 x^{1 + 1}}{{(- x^{2} + 9)}^{3}})$

Step 1.1.2.16

Add $1$ and $1$ .

$f'' (x) = 2 (\frac{- x^{2} + 9 + 4 x^{2}}{{(- x^{2} + 9)}^{3}})$

Step 1.1.2.17

Add $- x^{2}$ and $4 x^{2}$ .

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

Step 1.1.2.18

Combine $2$ and $\frac{3 x^{2} + 9}{{(- x^{2} + 9)}^{3}}$ .

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

Step 1.1.2.19

Simplify.

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Step 1.1.2.19.1

Apply the distributive property.

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

Step 1.1.2.19.2

Simplify each term.

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Step 1.1.2.19.2.1

Multiply $3$ by $2$ .

$f'' (x) = \frac{6 x^{2} + 2 \cdot 9}{{(- x^{2} + 9)}^{3}}$

Step 1.1.2.19.2.2

Multiply $2$ by $9$ .

$f'' (x) = \frac{6 x^{2} + 18}{{(- x^{2} + 9)}^{3}}$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{6 x^{2} + 18}{{(- x^{2} + 9)}^{3}}$ .

$\frac{6 x^{2} + 18}{{(- x^{2} + 9)}^{3}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $\frac{6 x^{2} + 18}{{(- x^{2} + 9)}^{3}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$\frac{6 x^{2} + 18}{{(- x^{2} + 9)}^{3}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$6 x^{2} + 18 = 0$

Step 1.2.3

Solve the equation for $x$ .

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Step 1.2.3.1

Subtract $18$ from both sides of the equation.

$6 x^{2} = - 18$

Step 1.2.3.2

Divide each term in $6 x^{2} = - 18$ by $6$ and simplify.

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Step 1.2.3.2.1

Divide each term in $6 x^{2} = - 18$ by $6$ .

$\frac{6 x^{2}}{6} = \frac{- 18}{6}$

Step 1.2.3.2.2

Simplify the left side.

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Step 1.2.3.2.2.1

Cancel the common factor of $6$ .

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Step 1.2.3.2.2.1.1

Cancel the common factor.

$\frac{6 x^{2}}{6} = \frac{- 18}{6}$

Step 1.2.3.2.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 18}{6}$

Step 1.2.3.2.3

Simplify the right side.

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Step 1.2.3.2.3.1

Divide $- 18$ by $6$ .

$x^{2} = - 3$

Step 1.2.3.3

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

$x = \pm \sqrt{- 3}$

Step 1.2.3.4

Simplify $\pm \sqrt{- 3}$ .

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Step 1.2.3.4.1

Rewrite $- 3$ as $- 1 (3)$ .

$x = \pm \sqrt{- 1 (3)}$

Step 1.2.3.4.2

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{3}$

Step 1.2.3.4.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \sqrt{3}$

Step 1.2.3.5

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

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Step 1.2.3.5.1

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

$x = i \sqrt{3}$

Step 1.2.3.5.2

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

$x = - i \sqrt{3}$

Step 1.2.3.5.3

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

$x = i \sqrt{3}, - i \sqrt{3}$

Step 2

Find the domain of $f (x) = \frac{1}{9 - x^{2}}$ .

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Step 2.1

Set the denominator in $\frac{1}{9 - x^{2}}$ equal to $0$ to find where the expression is undefined.

$9 - x^{2} = 0$

Step 2.2

Solve for $x$ .

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Step 2.2.1

Subtract $9$ from both sides of the equation.

$- x^{2} = - 9$

Step 2.2.2

Divide each term in $- x^{2} = - 9$ by $- 1$ and simplify.

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Step 2.2.2.1

Divide each term in $- x^{2} = - 9$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 9}{- 1}$

Step 2.2.2.2

Simplify the left side.

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Step 2.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 9}{- 1}$

Step 2.2.2.2.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 9}{- 1}$

Step 2.2.2.3

Simplify the right side.

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Step 2.2.2.3.1

Divide $- 9$ by $- 1$ .

$x^{2} = 9$

Step 2.2.3

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

$x = \pm \sqrt{9}$

Step 2.2.4

Simplify $\pm \sqrt{9}$ .

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Step 2.2.4.1

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

Step 2.2.4.2

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

$x = \pm 3$

Step 2.2.5

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

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Step 2.2.5.1

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

$x = 3$

Step 2.2.5.2

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

$x = - 3$

Step 2.2.5.3

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

$x = 3, - 3$

Step 2.3

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

Interval Notation:

$(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \neq 3, - 3}$

Interval Notation:

$(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \neq 3, - 3}$

Step 3

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

$(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$

Step 4

Substitute any number from the interval $(- \infty, - 3)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

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

$f'' (- 6) = \frac{6 {(- 6)}^{2} + 18}{{(- {(- 6)}^{2} + 9)}^{3}}$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify the numerator.

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Step 4.2.1.1

Raise $- 6$ to the power of $2$ .

$f'' (- 6) = \frac{6 \cdot 36 + 18}{{(- {(- 6)}^{2} + 9)}^{3}}$

Step 4.2.1.2

Multiply $6$ by $36$ .

$f'' (- 6) = \frac{216 + 18}{{(- {(- 6)}^{2} + 9)}^{3}}$

Step 4.2.1.3

Add $216$ and $18$ .

$f'' (- 6) = \frac{234}{{(- {(- 6)}^{2} + 9)}^{3}}$

Step 4.2.2

Simplify the denominator.

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Step 4.2.2.1

Raise $- 6$ to the power of $2$ .

$f'' (- 6) = \frac{234}{{(- 1 \cdot 36 + 9)}^{3}}$

Step 4.2.2.2

Multiply $- 1$ by $36$ .

$f'' (- 6) = \frac{234}{{(- 36 + 9)}^{3}}$

Step 4.2.2.3

Add $- 36$ and $9$ .

$f'' (- 6) = \frac{234}{{(- 27)}^{3}}$

Step 4.2.2.4

Raise $- 27$ to the power of $3$ .

$f'' (- 6) = \frac{234}{- 19683}$

Step 4.2.3

Reduce the expression by cancelling the common factors.

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Step 4.2.3.1

Cancel the common factor of $234$ and $- 19683$ .

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Step 4.2.3.1.1

Factor $9$ out of $234$ .

$f'' (- 6) = \frac{9 (26)}{- 19683}$

Step 4.2.3.1.2

Cancel the common factors.

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Step 4.2.3.1.2.1

Factor $9$ out of $- 19683$ .

$f'' (- 6) = \frac{9 \cdot 26}{9 \cdot - 2187}$

Step 4.2.3.1.2.2

Cancel the common factor.

$f'' (- 6) = \frac{9 \cdot 26}{9 \cdot - 2187}$

Step 4.2.3.1.2.3

Rewrite the expression.

$f'' (- 6) = \frac{26}{- 2187}$

Step 4.2.3.2

Move the negative in front of the fraction.

$f'' (- 6) = - \frac{26}{2187}$

Step 4.2.4

The final answer is $- \frac{26}{2187}$ .

$- \frac{26}{2187}$

Step 4.3

The graph is concave down on the interval $(- \infty, - 3)$ because $f'' (- 6)$ is negative.

Concave down on $(- \infty, - 3)$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(- 3, 3)$ 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.

$f'' (0) = \frac{6 {(0)}^{2} + 18}{{(- {(0)}^{2} + 9)}^{3}}$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify the numerator.

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Step 5.2.1.1

Raising $0$ to any positive power yields $0$ .

$f'' (0) = \frac{6 \cdot 0 + 18}{{(- 0^{2} + 9)}^{3}}$

Step 5.2.1.2

Multiply $6$ by $0$ .

$f'' (0) = \frac{0 + 18}{{(- 0^{2} + 9)}^{3}}$

Step 5.2.1.3

Add $0$ and $18$ .

$f'' (0) = \frac{18}{{(- 0^{2} + 9)}^{3}}$

Step 5.2.2

Simplify the denominator.

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Step 5.2.2.1

Raising $0$ to any positive power yields $0$ .

$f'' (0) = \frac{18}{{(- 0 + 9)}^{3}}$

Step 5.2.2.2

Multiply $- 1$ by $0$ .

$f'' (0) = \frac{18}{{(0 + 9)}^{3}}$

Step 5.2.2.3

Add $0$ and $9$ .

$f'' (0) = \frac{18}{9^{3}}$

Step 5.2.2.4

Raise $9$ to the power of $3$ .

$f'' (0) = \frac{18}{729}$

Step 5.2.3

Cancel the common factor of $18$ and $729$ .

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Step 5.2.3.1

Factor $9$ out of $18$ .

$f'' (0) = \frac{9 (2)}{729}$

Step 5.2.3.2

Cancel the common factors.

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Step 5.2.3.2.1

Factor $9$ out of $729$ .

$f'' (0) = \frac{9 \cdot 2}{9 \cdot 81}$

Step 5.2.3.2.2

Cancel the common factor.

$f'' (0) = \frac{9 \cdot 2}{9 \cdot 81}$

Step 5.2.3.2.3

Rewrite the expression.

$f'' (0) = \frac{2}{81}$

Step 5.2.4

The final answer is $\frac{2}{81}$ .

$\frac{2}{81}$

Step 5.3

The graph is concave up on the interval $(- 3, 3)$ because $f'' (0)$ is positive.

Concave up on $(- 3, 3)$ since $f'' (x)$ is positive

Step 6

Substitute any number from the interval $(3, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

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

$f'' (6) = \frac{6 {(6)}^{2} + 18}{{(- {(6)}^{2} + 9)}^{3}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the numerator.

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Step 6.2.1.1

Multiply $6$ by $6^{2}$ by adding the exponents.

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Step 6.2.1.1.1

Multiply $6$ by $6^{2}$ .

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Step 6.2.1.1.1.1

Raise $6$ to the power of $1$ .

$f'' (6) = \frac{6 \cdot 6^{2} + 18}{{(- 6^{2} + 9)}^{3}}$

Step 6.2.1.1.1.2

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

$f'' (6) = \frac{6^{1 + 2} + 18}{{(- 6^{2} + 9)}^{3}}$

Step 6.2.1.1.2

Add $1$ and $2$ .

$f'' (6) = \frac{6^{3} + 18}{{(- 6^{2} + 9)}^{3}}$

Step 6.2.1.2

Raise $6$ to the power of $3$ .

$f'' (6) = \frac{216 + 18}{{(- 6^{2} + 9)}^{3}}$

Step 6.2.1.3

Add $216$ and $18$ .

$f'' (6) = \frac{234}{{(- 6^{2} + 9)}^{3}}$

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Raise $6$ to the power of $2$ .

$f'' (6) = \frac{234}{{(- 1 \cdot 36 + 9)}^{3}}$

Step 6.2.2.2

Multiply $- 1$ by $36$ .

$f'' (6) = \frac{234}{{(- 36 + 9)}^{3}}$

Step 6.2.2.3

Add $- 36$ and $9$ .

$f'' (6) = \frac{234}{{(- 27)}^{3}}$

Step 6.2.2.4

Raise $- 27$ to the power of $3$ .

$f'' (6) = \frac{234}{- 19683}$

Step 6.2.3

Reduce the expression by cancelling the common factors.

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Step 6.2.3.1

Cancel the common factor of $234$ and $- 19683$ .

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Step 6.2.3.1.1

Factor $9$ out of $234$ .

$f'' (6) = \frac{9 (26)}{- 19683}$

Step 6.2.3.1.2

Cancel the common factors.

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Step 6.2.3.1.2.1

Factor $9$ out of $- 19683$ .

$f'' (6) = \frac{9 \cdot 26}{9 \cdot - 2187}$

Step 6.2.3.1.2.2

Cancel the common factor.

$f'' (6) = \frac{9 \cdot 26}{9 \cdot - 2187}$

Step 6.2.3.1.2.3

Rewrite the expression.

$f'' (6) = \frac{26}{- 2187}$

Step 6.2.3.2

Move the negative in front of the fraction.

$f'' (6) = - \frac{26}{2187}$

Step 6.2.4

The final answer is $- \frac{26}{2187}$ .

$- \frac{26}{2187}$

Step 6.3

The graph is concave down on the interval $(3, \infty)$ because $f'' (6)$ is negative.

Concave down on $(3, \infty)$ since $f'' (x)$ is negative

Step 7

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

Concave down on $(- \infty, - 3)$ since $f'' (x)$ is negative

Concave up on $(- 3, 3)$ since $f'' (x)$ is positive

Concave down on $(3, \infty)$ since $f'' (x)$ is negative

Step 8