Calculus Examples

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

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 $g (x) = x^{2} + 3$ .

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

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

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

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

Step 1.1.1.2.2

Move $2$ to the left of $x^{2} + 3$ .

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

Step 1.1.1.2.3

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

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

Step 1.1.1.2.4

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

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

Step 1.1.1.2.5

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

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

Step 1.1.1.2.6

Simplify the expression.

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Step 1.1.1.2.6.1

Add $2 x$ and $0$ .

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

Step 1.1.1.2.6.2

Multiply $2$ by $- 1$ .

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

Step 1.1.1.3

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

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

Step 1.1.1.4

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

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

Step 1.1.1.5

Add $1$ and $2$ .

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

Step 1.1.1.6

Simplify.

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Step 1.1.1.6.1

Apply the distributive property.

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

Step 1.1.1.6.2

Apply the distributive property.

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

Step 1.1.1.6.3

Simplify the numerator.

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Step 1.1.1.6.3.1

Simplify each term.

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Step 1.1.1.6.3.1.1

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

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Step 1.1.1.6.3.1.1.1

Move $x$ .

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

Step 1.1.1.6.3.1.1.2

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

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Step 1.1.1.6.3.1.1.2.1

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

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

Step 1.1.1.6.3.1.1.2.2

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

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

Step 1.1.1.6.3.1.1.3

Add $1$ and $2$ .

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

Step 1.1.1.6.3.1.2

Multiply $2$ by $3$ .

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

Step 1.1.1.6.3.2

Combine the opposite terms in $2 x^{3} + 6 x - 2 x^{3}$ .

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Step 1.1.1.6.3.2.1

Subtract $2 x^{3}$ from $2 x^{3}$ .

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

Step 1.1.1.6.3.2.2

Add $6 x$ and $0$ .

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

Since $6$ is constant with respect to $x$ , the derivative of $\frac{6 x}{{(x^{2} + 3)}^{2}}$ with respect to $x$ is $6 \frac{d}{d x} [\frac{x}{{(x^{2} + 3)}^{2}}]$ .

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

Step 1.1.2.2

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

Step 1.1.2.3

Differentiate using the Power Rule.

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Step 1.1.2.3.1

Multiply the exponents in ${({(x^{2} + 3)}^{2})}^{2}$ .

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Step 1.1.2.3.1.1

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

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

Step 1.1.2.3.1.2

Multiply $2$ by $2$ .

$f'' (x) = 6 (\frac{{(x^{2} + 3)}^{2} \frac{d}{d x} (x) - x \frac{d}{d x} {(x^{2} + 3)}^{2}}{{(x^{2} + 3)}^{4}})$

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 $n = 1$ .

$f'' (x) = 6 (\frac{{(x^{2} + 3)}^{2} \cdot 1 - x \frac{d}{d x} {(x^{2} + 3)}^{2}}{{(x^{2} + 3)}^{4}})$

Step 1.1.2.3.3

Multiply ${(x^{2} + 3)}^{2}$ by $1$ .

$f'' (x) = 6 (\frac{{(x^{2} + 3)}^{2} - x \frac{d}{d x} {(x^{2} + 3)}^{2}}{{(x^{2} + 3)}^{4}})$

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 $g (x) = x^{2} + 3$ .

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Step 1.1.2.4.1

To apply the Chain Rule, set $u$ as $x^{2} + 3$ .

$f'' (x) = 6 (\frac{{(x^{2} + 3)}^{2} - x (\frac{d}{d u} (u^{2}) \frac{d}{d x} (x^{2} + 3))}{{(x^{2} + 3)}^{4}})$

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 $n = 2$ .

$f'' (x) = 6 (\frac{{(x^{2} + 3)}^{2} - x (2 u \frac{d}{d x} (x^{2} + 3))}{{(x^{2} + 3)}^{4}})$

Step 1.1.2.4.3

Replace all occurrences of $u$ with $x^{2} + 3$ .

$f'' (x) = 6 (\frac{{(x^{2} + 3)}^{2} - x (2 (x^{2} + 3) \frac{d}{d x} (x^{2} + 3))}{{(x^{2} + 3)}^{4}})$

Step 1.1.2.5

Simplify with factoring out.

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Step 1.1.2.5.1

Multiply $2$ by $- 1$ .

$f'' (x) = 6 (\frac{{(x^{2} + 3)}^{2} - 2 x ((x^{2} + 3) \frac{d}{d x} (x^{2} + 3))}{{(x^{2} + 3)}^{4}})$

Step 1.1.2.5.2

Factor $x^{2} + 3$ out of ${(x^{2} + 3)}^{2} - 2 x ((x^{2} + 3) \frac{d}{d x} [x^{2} + 3])$ .

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Step 1.1.2.5.2.1

Factor $x^{2} + 3$ out of ${(x^{2} + 3)}^{2}$ .

$f'' (x) = 6 (\frac{(x^{2} + 3) (x^{2} + 3) - 2 x ((x^{2} + 3) \frac{d}{d x} (x^{2} + 3))}{{(x^{2} + 3)}^{4}})$

Step 1.1.2.5.2.2

Factor $x^{2} + 3$ out of $- 2 x ((x^{2} + 3) \frac{d}{d x} [x^{2} + 3])$ .

$f'' (x) = 6 (\frac{(x^{2} + 3) (x^{2} + 3) + (x^{2} + 3) (- 2 x (\frac{d}{d x} (x^{2} + 3)))}{{(x^{2} + 3)}^{4}})$

Step 1.1.2.5.2.3

Factor $x^{2} + 3$ out of $(x^{2} + 3) (x^{2} + 3) + (x^{2} + 3) (- 2 x (\frac{d}{d x} [x^{2} + 3]))$ .

$f'' (x) = 6 (\frac{(x^{2} + 3) (x^{2} + 3 - 2 x (\frac{d}{d x} (x^{2} + 3)))}{{(x^{2} + 3)}^{4}})$

Step 1.1.2.6

Cancel the common factors.

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Step 1.1.2.6.1

Factor $x^{2} + 3$ out of ${(x^{2} + 3)}^{4}$ .

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

Step 1.1.2.6.2

Cancel the common factor.

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

Step 1.1.2.6.3

Rewrite the expression.

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

Step 1.1.2.7

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

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

Step 1.1.2.8

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

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

Step 1.1.2.9

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

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

Step 1.1.2.10

Simplify the expression.

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Step 1.1.2.10.1

Add $2 x$ and $0$ .

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

Step 1.1.2.10.2

Multiply $2$ by $- 2$ .

$f'' (x) = 6 (\frac{x^{2} + 3 - 4 x \cdot x}{{(x^{2} + 3)}^{3}})$

Step 1.1.2.11

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

$f'' (x) = 6 (\frac{x^{2} + 3 - 4 (x \cdot x)}{{(x^{2} + 3)}^{3}})$

Step 1.1.2.12

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

$f'' (x) = 6 (\frac{x^{2} + 3 - 4 (x \cdot x)}{{(x^{2} + 3)}^{3}})$

Step 1.1.2.13

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

$f'' (x) = 6 (\frac{x^{2} + 3 - 4 x^{1 + 1}}{{(x^{2} + 3)}^{3}})$

Step 1.1.2.14

Add $1$ and $1$ .

$f'' (x) = 6 (\frac{x^{2} + 3 - 4 x^{2}}{{(x^{2} + 3)}^{3}})$

Step 1.1.2.15

Subtract $4 x^{2}$ from $x^{2}$ .

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

Step 1.1.2.16

Combine $6$ and $\frac{- 3 x^{2} + 3}{{(x^{2} + 3)}^{3}}$ .

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

Step 1.1.2.17

Simplify.

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Step 1.1.2.17.1

Apply the distributive property.

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

Step 1.1.2.17.2

Simplify each term.

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Step 1.1.2.17.2.1

Multiply $- 3$ by $6$ .

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

Step 1.1.2.17.2.2

Multiply $6$ by $3$ .

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

Step 1.1.3

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

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

Step 1.2

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

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Step 1.2.1

Set the second derivative equal to $0$ .

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

Step 1.2.2

Set the numerator equal to zero.

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

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.

$- 18 x^{2} = - 18$

Step 1.2.3.2

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

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Step 1.2.3.2.1

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

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

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 $- 18$ .

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Step 1.2.3.2.2.1.1

Cancel the common factor.

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

Step 1.2.3.2.2.1.2

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

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

Step 1.2.3.2.3

Simplify the right side.

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Step 1.2.3.2.3.1

Divide $- 18$ by $- 18$ .

$x^{2} = 1$

Step 1.2.3.3

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

$x = \pm \sqrt{1}$

Step 1.2.3.4

Any root of $1$ is $1$ .

$x = \pm 1$

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.

$x = 1$

Step 1.2.3.5.2

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

$x = - 1$

Step 1.2.3.5.3

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

$x = 1, - 1$

Step 2

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

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Step 2.1

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

$x^{2} + 3 = 0$

Step 2.2

Solve for $x$ .

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Step 2.2.1

Subtract $3$ from both sides of the equation.

$x^{2} = - 3$

Step 2.2.2

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

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

Step 2.2.3

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

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Step 2.2.3.1

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

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

Step 2.2.3.2

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

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

Step 2.2.3.3

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

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

Step 2.2.4

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

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Step 2.2.4.1

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

$x = i \sqrt{3}$

Step 2.2.4.2

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

$x = - i \sqrt{3}$

Step 2.2.4.3

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

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

Step 2.3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3

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

$(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)$

Step 4

Substitute any number from the interval $(- \infty, - 1)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

Replace the variable $x$ with $- 4$ in the expression.

$f'' (- 4) = \frac{- 18 {(- 4)}^{2} + 18}{{({(- 4)}^{2} + 3)}^{3}}$

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 $- 4$ to the power of $2$ .

$f'' (- 4) = \frac{- 18 \cdot 16 + 18}{{({(- 4)}^{2} + 3)}^{3}}$

Step 4.2.1.2

Multiply $- 18$ by $16$ .

$f'' (- 4) = \frac{- 288 + 18}{{({(- 4)}^{2} + 3)}^{3}}$

Step 4.2.1.3

Add $- 288$ and $18$ .

$f'' (- 4) = \frac{- 270}{{({(- 4)}^{2} + 3)}^{3}}$

Step 4.2.2

Simplify the denominator.

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Step 4.2.2.1

Raise $- 4$ to the power of $2$ .

$f'' (- 4) = \frac{- 270}{{(16 + 3)}^{3}}$

Step 4.2.2.2

Add $16$ and $3$ .

$f'' (- 4) = \frac{- 270}{19^{3}}$

Step 4.2.2.3

Raise $19$ to the power of $3$ .

$f'' (- 4) = \frac{- 270}{6859}$

Step 4.2.3

Move the negative in front of the fraction.

$f'' (- 4) = - \frac{270}{6859}$

Step 4.2.4

The final answer is $- \frac{270}{6859}$ .

$- \frac{270}{6859}$

Step 4.3

The graph is concave down on the interval $(- \infty, - 1)$ because $f'' (- 4)$ is negative.

Concave down on $(- \infty, - 1)$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(- 1, 1)$ 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.

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

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 $0$ .

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

Step 5.2.1.2

Multiply $- 18$ by $0$ .

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

Step 5.2.1.3

Add $0$ and $18$ .

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

Step 5.2.2

Simplify the denominator.

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Step 5.2.2.1

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

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

Step 5.2.2.2

Add $0$ and $3$ .

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

Step 5.2.2.3

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

$f'' (0) = \frac{18}{27}$

Step 5.2.3

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

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Step 5.2.3.1

Factor $9$ out of $18$ .

$f'' (0) = \frac{9 (2)}{27}$

Step 5.2.3.2

Cancel the common factors.

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Step 5.2.3.2.1

Factor $9$ out of $27$ .

$f'' (0) = \frac{9 \cdot 2}{9 \cdot 3}$

Step 5.2.3.2.2

Cancel the common factor.

$f'' (0) = \frac{9 \cdot 2}{9 \cdot 3}$

Step 5.2.3.2.3

Rewrite the expression.

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

Step 5.2.4

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

$\frac{2}{3}$

Step 5.3

The graph is concave up on the interval $(- 1, 1)$ because $f'' (0)$ is positive.

Concave up on $(- 1, 1)$ since $f'' (x)$ is positive

Step 6

Substitute any number from the interval $(1, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

Replace the variable $x$ with $4$ in the expression.

$f'' (4) = \frac{- 18 {(4)}^{2} + 18}{{({(4)}^{2} + 3)}^{3}}$

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

Raise $4$ to the power of $2$ .

$f'' (4) = \frac{- 18 \cdot 16 + 18}{{(4^{2} + 3)}^{3}}$

Step 6.2.1.2

Multiply $- 18$ by $16$ .

$f'' (4) = \frac{- 288 + 18}{{(4^{2} + 3)}^{3}}$

Step 6.2.1.3

Add $- 288$ and $18$ .

$f'' (4) = \frac{- 270}{{(4^{2} + 3)}^{3}}$

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Raise $4$ to the power of $2$ .

$f'' (4) = \frac{- 270}{{(16 + 3)}^{3}}$

Step 6.2.2.2

Add $16$ and $3$ .

$f'' (4) = \frac{- 270}{19^{3}}$

Step 6.2.2.3

Raise $19$ to the power of $3$ .

$f'' (4) = \frac{- 270}{6859}$

Step 6.2.3

Move the negative in front of the fraction.

$f'' (4) = - \frac{270}{6859}$

Step 6.2.4

The final answer is $- \frac{270}{6859}$ .

$- \frac{270}{6859}$

Step 6.3

The graph is concave down on the interval $(1, \infty)$ because $f'' (4)$ is negative.

Concave down on $(1, \infty)$ 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 $(- \infty, - 1)$ since $f'' (x)$ is negative

Concave up on $(- 1, 1)$ since $f'' (x)$ is positive

Concave down on $(1, \infty)$ since $f'' (x)$ is negative

Step 8