Calculus Examples

$f (x) = x^{4} - 8 x^{2} + 5$

Step 1

Find the second derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate.

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Step 1.1.1.1

By the Sum Rule, the derivative of $x^{4} - 8 x^{2} + 5$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 8 x^{2}] + \frac{d}{d x} [5]$ .

$f' (x) = \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 8 x^{2}) + \frac{d}{d x} (5)$

Step 1.1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f' (x) = 4 x^{3} + \frac{d}{d x} (- 8 x^{2}) + \frac{d}{d x} (5)$

Step 1.1.2

Evaluate $\frac{d}{d x} [- 8 x^{2}]$ .

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Step 1.1.2.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x^{2}$ with respect to $x$ is $- 8 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 4 x^{3} - 8 \frac{d}{d x} x^{2} + \frac{d}{d x} (5)$

Step 1.1.2.2

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

$f' (x) = 4 x^{3} - 8 (2 x) + \frac{d}{d x} (5)$

Step 1.1.2.3

Multiply $2$ by $- 8$ .

$f' (x) = 4 x^{3} - 16 x + \frac{d}{d x} (5)$

Step 1.1.3

Differentiate using the Constant Rule.

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Step 1.1.3.1

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

$f' (x) = 4 x^{3} - 16 x + 0$

Step 1.1.3.2

Add $4 x^{3} - 16 x$ and $0$ .

$f' (x) = 4 x^{3} - 16 x$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $4 x^{3} - 16 x$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 16 x]$ .

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

Step 1.2.2

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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Step 1.2.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3}$ with respect to $x$ is $4 \frac{d}{d x} [x^{3}]$ .

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

Step 1.2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

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

Step 1.2.2.3

Multiply $3$ by $4$ .

$f'' (x) = 12 x^{2} + \frac{d}{d x} (- 16 x)$

Step 1.2.3

Evaluate $\frac{d}{d x} [- 16 x]$ .

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Step 1.2.3.1

Since $- 16$ is constant with respect to $x$ , the derivative of $- 16 x$ with respect to $x$ is $- 16 \frac{d}{d x} [x]$ .

$f'' (x) = 12 x^{2} - 16 \frac{d}{d x} x$

Step 1.2.3.2

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

$f'' (x) = 12 x^{2} - 16 \cdot 1$

Step 1.2.3.3

Multiply $- 16$ by $1$ .

$f'' (x) = 12 x^{2} - 16$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $12 x^{2} - 16$ .

$12 x^{2} - 16$

Step 2

Set the second derivative equal to $0$ then solve the equation $12 x^{2} - 16 = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$12 x^{2} - 16 = 0$

Step 2.2

Add $16$ to both sides of the equation.

$12 x^{2} = 16$

Step 2.3

Divide each term in $12 x^{2} = 16$ by $12$ and simplify.

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Step 2.3.1

Divide each term in $12 x^{2} = 16$ by $12$ .

$\frac{12 x^{2}}{12} = \frac{16}{12}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $12$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{12 x^{2}}{12} = \frac{16}{12}$

Step 2.3.2.1.2

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

$x^{2} = \frac{16}{12}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Cancel the common factor of $16$ and $12$ .

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Step 2.3.3.1.1

Factor $4$ out of $16$ .

$x^{2} = \frac{4 (4)}{12}$

Step 2.3.3.1.2

Cancel the common factors.

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Step 2.3.3.1.2.1

Factor $4$ out of $12$ .

$x^{2} = \frac{4 \cdot 4}{4 \cdot 3}$

Step 2.3.3.1.2.2

Cancel the common factor.

$x^{2} = \frac{4 \cdot 4}{4 \cdot 3}$

Step 2.3.3.1.2.3

Rewrite the expression.

$x^{2} = \frac{4}{3}$

Step 2.4

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

$x = \pm \sqrt{\frac{4}{3}}$

Step 2.5

Simplify $\pm \sqrt{\frac{4}{3}}$ .

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Step 2.5.1

Rewrite $\sqrt{\frac{4}{3}}$ as $\frac{\sqrt{4}}{\sqrt{3}}$ .

$x = \pm \frac{\sqrt{4}}{\sqrt{3}}$

Step 2.5.2

Simplify the numerator.

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Step 2.5.2.1

Rewrite $4$ as $2^{2}$ .

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

Step 2.5.2.2

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

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

Step 2.5.3

Multiply $\frac{2}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$x = \pm \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}$

Step 2.5.4

Combine and simplify the denominator.

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Step 2.5.4.1

Multiply $\frac{2}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

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

Step 2.5.4.2

Raise $\sqrt{3}$ to the power of $1$ .

$x = \pm \frac{2 \sqrt{3}}{{\sqrt{3}}^{1} \sqrt{3}}$

Step 2.5.4.3

Raise $\sqrt{3}$ to the power of $1$ .

$x = \pm \frac{2 \sqrt{3}}{{\sqrt{3}}^{1} {\sqrt{3}}^{1}}$

Step 2.5.4.4

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

$x = \pm \frac{2 \sqrt{3}}{{\sqrt{3}}^{1 + 1}}$

Step 2.5.4.5

Add $1$ and $1$ .

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

Step 2.5.4.6

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 2.5.4.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$x = \pm \frac{2 \sqrt{3}}{{(3^{\frac{1}{2}})}^{2}}$

Step 2.5.4.6.2

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

$x = \pm \frac{2 \sqrt{3}}{3^{\frac{1}{2} \cdot 2}}$

Step 2.5.4.6.3

Combine $\frac{1}{2}$ and $2$ .

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

Step 2.5.4.6.4

Cancel the common factor of $2$ .

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Step 2.5.4.6.4.1

Cancel the common factor.

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

Step 2.5.4.6.4.2

Rewrite the expression.

$x = \pm \frac{2 \sqrt{3}}{3^{1}}$

Step 2.5.4.6.5

Evaluate the exponent.

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

Step 2.6

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

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Step 2.6.1

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

$x = \frac{2 \sqrt{3}}{3}$

Step 2.6.2

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

$x = - \frac{2 \sqrt{3}}{3}$

Step 2.6.3

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

$x = \frac{2 \sqrt{3}}{3}, - \frac{2 \sqrt{3}}{3}$

Step 3

Find the points where the second derivative is $0$ .

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Step 3.1

Substitute $\frac{2 \sqrt{3}}{3}$ in $f (x) = x^{4} - 8 x^{2} + 5$ to find the value of $y$ .

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Step 3.1.1

Replace the variable $x$ with $\frac{2 \sqrt{3}}{3}$ in the expression.

$f (\frac{2 \sqrt{3}}{3}) = {(\frac{2 \sqrt{3}}{3})}^{4} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.1.2.1.1.1

Apply the product rule to $\frac{2 \sqrt{3}}{3}$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{{(2 \sqrt{3})}^{4}}{3^{4}} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.1.2

Apply the product rule to $2 \sqrt{3}$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{2^{4} {\sqrt{3}}^{4}}{3^{4}} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.2

Simplify the numerator.

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Step 3.1.2.1.2.1

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

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 {\sqrt{3}}^{4}}{3^{4}} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.2.2

Rewrite ${\sqrt{3}}^{4}$ as $3^{2}$ .

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Step 3.1.2.1.2.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 {(3^{\frac{1}{2}})}^{4}}{3^{4}} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.2.2.2

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

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{\frac{1}{2} \cdot 4}}{3^{4}} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.2.2.3

Combine $\frac{1}{2}$ and $4$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{\frac{4}{2}}}{3^{4}} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.2.2.4

Cancel the common factor of $4$ and $2$ .

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Step 3.1.2.1.2.2.4.1

Factor $2$ out of $4$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{\frac{2 \cdot 2}{2}}}{3^{4}} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.2.2.4.2

Cancel the common factors.

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Step 3.1.2.1.2.2.4.2.1

Factor $2$ out of $2$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{\frac{2 \cdot 2}{2 (1)}}}{3^{4}} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.2.2.4.2.2

Cancel the common factor.

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{\frac{2 \cdot 2}{2 \cdot 1}}}{3^{4}} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.2.2.4.2.3

Rewrite the expression.

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{\frac{2}{1}}}{3^{4}} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.2.2.4.2.4

Divide $2$ by $1$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{2}}{3^{4}} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.2.3

Raise $3$ to the power of $2$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 9}{3^{4}} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.3

Raise $3$ to the power of $4$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 9}{81} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.4

Multiply $16$ by $9$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{144}{81} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.5

Cancel the common factor of $144$ and $81$ .

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Step 3.1.2.1.5.1

Factor $9$ out of $144$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{9 (16)}{81} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.5.2

Cancel the common factors.

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Step 3.1.2.1.5.2.1

Factor $9$ out of $81$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{9 \cdot 16}{9 \cdot 9} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.5.2.2

Cancel the common factor.

$f (\frac{2 \sqrt{3}}{3}) = \frac{9 \cdot 16}{9 \cdot 9} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.5.2.3

Rewrite the expression.

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 {(\frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.1.2.1.6

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.1.2.1.6.1

Apply the product rule to $\frac{2 \sqrt{3}}{3}$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{{(2 \sqrt{3})}^{2}}{3^{2}} + 5$

Step 3.1.2.1.6.2

Apply the product rule to $2 \sqrt{3}$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{2^{2} {\sqrt{3}}^{2}}{3^{2}} + 5$

Step 3.1.2.1.7

Simplify the numerator.

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Step 3.1.2.1.7.1

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

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 {\sqrt{3}}^{2}}{3^{2}} + 5$

Step 3.1.2.1.7.2

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 3.1.2.1.7.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 {(3^{\frac{1}{2}})}^{2}}{3^{2}} + 5$

Step 3.1.2.1.7.2.2

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

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 \cdot 3^{\frac{1}{2} \cdot 2}}{3^{2}} + 5$

Step 3.1.2.1.7.2.3

Combine $\frac{1}{2}$ and $2$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 \cdot 3^{\frac{2}{2}}}{3^{2}} + 5$

Step 3.1.2.1.7.2.4

Cancel the common factor of $2$ .

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Step 3.1.2.1.7.2.4.1

Cancel the common factor.

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 \cdot 3^{\frac{2}{2}}}{3^{2}} + 5$

Step 3.1.2.1.7.2.4.2

Rewrite the expression.

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 \cdot 3}{3^{2}} + 5$

Step 3.1.2.1.7.2.5

Evaluate the exponent.

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 \cdot 3}{3^{2}} + 5$

Step 3.1.2.1.8

Raise $3$ to the power of $2$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 \cdot 3}{9} + 5$

Step 3.1.2.1.9

Multiply $4$ by $3$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 (\frac{12}{9}) + 5$

Step 3.1.2.1.10

Cancel the common factor of $12$ and $9$ .

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Step 3.1.2.1.10.1

Factor $3$ out of $12$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{3 (4)}{9} + 5$

Step 3.1.2.1.10.2

Cancel the common factors.

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Step 3.1.2.1.10.2.1

Factor $3$ out of $9$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{3 \cdot 4}{3 \cdot 3} + 5$

Step 3.1.2.1.10.2.2

Cancel the common factor.

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{3 \cdot 4}{3 \cdot 3} + 5$

Step 3.1.2.1.10.2.3

Rewrite the expression.

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 (\frac{4}{3}) + 5$

Step 3.1.2.1.11

Multiply $- 8 (\frac{4}{3})$ .

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Step 3.1.2.1.11.1

Combine $- 8$ and $\frac{4}{3}$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} + \frac{- 8 \cdot 4}{3} + 5$

Step 3.1.2.1.11.2

Multiply $- 8$ by $4$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} + \frac{- 32}{3} + 5$

Step 3.1.2.1.12

Move the negative in front of the fraction.

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - \frac{32}{3} + 5$

Step 3.1.2.2

Find the common denominator.

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Step 3.1.2.2.1

Multiply $\frac{32}{3}$ by $\frac{3}{3}$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - (\frac{32}{3} \cdot \frac{3}{3}) + 5$

Step 3.1.2.2.2

Multiply $\frac{32}{3}$ by $\frac{3}{3}$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - \frac{32 \cdot 3}{3 \cdot 3} + 5$

Step 3.1.2.2.3

Write $5$ as a fraction with denominator $1$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - \frac{32 \cdot 3}{3 \cdot 3} + \frac{5}{1}$

Step 3.1.2.2.4

Multiply $\frac{5}{1}$ by $\frac{9}{9}$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - \frac{32 \cdot 3}{3 \cdot 3} + \frac{5}{1} \cdot \frac{9}{9}$

Step 3.1.2.2.5

Multiply $\frac{5}{1}$ by $\frac{9}{9}$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - \frac{32 \cdot 3}{3 \cdot 3} + \frac{5 \cdot 9}{9}$

Step 3.1.2.2.6

Multiply $3$ by $3$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16}{9} - \frac{32 \cdot 3}{9} + \frac{5 \cdot 9}{9}$

Step 3.1.2.3

Combine the numerators over the common denominator.

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 - 32 \cdot 3 + 5 \cdot 9}{9}$

Step 3.1.2.4

Simplify each term.

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Step 3.1.2.4.1

Multiply $- 32$ by $3$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 - 96 + 5 \cdot 9}{9}$

Step 3.1.2.4.2

Multiply $5$ by $9$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{16 - 96 + 45}{9}$

Step 3.1.2.5

Simplify the expression.

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Step 3.1.2.5.1

Subtract $96$ from $16$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{- 80 + 45}{9}$

Step 3.1.2.5.2

Add $- 80$ and $45$ .

$f (\frac{2 \sqrt{3}}{3}) = \frac{- 35}{9}$

Step 3.1.2.5.3

Move the negative in front of the fraction.

$f (\frac{2 \sqrt{3}}{3}) = - \frac{35}{9}$

Step 3.1.2.6

The final answer is $- \frac{35}{9}$ .

$- \frac{35}{9}$

Step 3.2

The point found by substituting $\frac{2 \sqrt{3}}{3}$ in $f (x) = x^{4} - 8 x^{2} + 5$ is $(\frac{2 \sqrt{3}}{3}, - \frac{35}{9})$ . This point can be an inflection point.

$(\frac{2 \sqrt{3}}{3}, - \frac{35}{9})$

Step 3.3

Substitute $- \frac{2 \sqrt{3}}{3}$ in $f (x) = x^{4} - 8 x^{2} + 5$ to find the value of $y$ .

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Step 3.3.1

Replace the variable $x$ with $- \frac{2 \sqrt{3}}{3}$ in the expression.

$f (- \frac{2 \sqrt{3}}{3}) = {(- \frac{2 \sqrt{3}}{3})}^{4} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.3.2.1.1.1

Apply the product rule to $- \frac{2 \sqrt{3}}{3}$ .

$f (- \frac{2 \sqrt{3}}{3}) = {(- 1)}^{4} {(\frac{2 \sqrt{3}}{3})}^{4} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.1.2

Apply the product rule to $\frac{2 \sqrt{3}}{3}$ .

$f (- \frac{2 \sqrt{3}}{3}) = {(- 1)}^{4} (\frac{{(2 \sqrt{3})}^{4}}{3^{4}}) - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.1.3

Apply the product rule to $2 \sqrt{3}$ .

$f (- \frac{2 \sqrt{3}}{3}) = {(- 1)}^{4} (\frac{2^{4} {\sqrt{3}}^{4}}{3^{4}}) - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.2

Raise $- 1$ to the power of $4$ .

$f (- \frac{2 \sqrt{3}}{3}) = 1 (\frac{2^{4} {\sqrt{3}}^{4}}{3^{4}}) - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.3

Multiply $\frac{2^{4} {\sqrt{3}}^{4}}{3^{4}}$ by $1$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{2^{4} {\sqrt{3}}^{4}}{3^{4}} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.4

Simplify the numerator.

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Step 3.3.2.1.4.1

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

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 {\sqrt{3}}^{4}}{3^{4}} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.4.2

Rewrite ${\sqrt{3}}^{4}$ as $3^{2}$ .

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Step 3.3.2.1.4.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 {(3^{\frac{1}{2}})}^{4}}{3^{4}} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.4.2.2

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

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{\frac{1}{2} \cdot 4}}{3^{4}} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.4.2.3

Combine $\frac{1}{2}$ and $4$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{\frac{4}{2}}}{3^{4}} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.4.2.4

Cancel the common factor of $4$ and $2$ .

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Step 3.3.2.1.4.2.4.1

Factor $2$ out of $4$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{\frac{2 \cdot 2}{2}}}{3^{4}} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.4.2.4.2

Cancel the common factors.

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Step 3.3.2.1.4.2.4.2.1

Factor $2$ out of $2$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{\frac{2 \cdot 2}{2 (1)}}}{3^{4}} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.4.2.4.2.2

Cancel the common factor.

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{\frac{2 \cdot 2}{2 \cdot 1}}}{3^{4}} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.4.2.4.2.3

Rewrite the expression.

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{\frac{2}{1}}}{3^{4}} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.4.2.4.2.4

Divide $2$ by $1$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 3^{2}}{3^{4}} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.4.3

Raise $3$ to the power of $2$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 9}{3^{4}} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.5

Raise $3$ to the power of $4$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 \cdot 9}{81} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.6

Multiply $16$ by $9$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{144}{81} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.7

Cancel the common factor of $144$ and $81$ .

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Step 3.3.2.1.7.1

Factor $9$ out of $144$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{9 (16)}{81} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.7.2

Cancel the common factors.

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Step 3.3.2.1.7.2.1

Factor $9$ out of $81$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{9 \cdot 16}{9 \cdot 9} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.7.2.2

Cancel the common factor.

$f (- \frac{2 \sqrt{3}}{3}) = \frac{9 \cdot 16}{9 \cdot 9} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.7.2.3

Rewrite the expression.

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 {(- \frac{2 \sqrt{3}}{3})}^{2} + 5$

Step 3.3.2.1.8

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.3.2.1.8.1

Apply the product rule to $- \frac{2 \sqrt{3}}{3}$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 ({(- 1)}^{2} {(\frac{2 \sqrt{3}}{3})}^{2}) + 5$

Step 3.3.2.1.8.2

Apply the product rule to $\frac{2 \sqrt{3}}{3}$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 ({(- 1)}^{2} (\frac{{(2 \sqrt{3})}^{2}}{3^{2}})) + 5$

Step 3.3.2.1.8.3

Apply the product rule to $2 \sqrt{3}$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 ({(- 1)}^{2} (\frac{2^{2} {\sqrt{3}}^{2}}{3^{2}})) + 5$

Step 3.3.2.1.9

Raise $- 1$ to the power of $2$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 (1 (\frac{2^{2} {\sqrt{3}}^{2}}{3^{2}})) + 5$

Step 3.3.2.1.10

Multiply $\frac{2^{2} {\sqrt{3}}^{2}}{3^{2}}$ by $1$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{2^{2} {\sqrt{3}}^{2}}{3^{2}} + 5$

Step 3.3.2.1.11

Simplify the numerator.

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Step 3.3.2.1.11.1

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

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 {\sqrt{3}}^{2}}{3^{2}} + 5$

Step 3.3.2.1.11.2

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 3.3.2.1.11.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 {(3^{\frac{1}{2}})}^{2}}{3^{2}} + 5$

Step 3.3.2.1.11.2.2

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

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 \cdot 3^{\frac{1}{2} \cdot 2}}{3^{2}} + 5$

Step 3.3.2.1.11.2.3

Combine $\frac{1}{2}$ and $2$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 \cdot 3^{\frac{2}{2}}}{3^{2}} + 5$

Step 3.3.2.1.11.2.4

Cancel the common factor of $2$ .

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Step 3.3.2.1.11.2.4.1

Cancel the common factor.

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 \cdot 3^{\frac{2}{2}}}{3^{2}} + 5$

Step 3.3.2.1.11.2.4.2

Rewrite the expression.

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 \cdot 3}{3^{2}} + 5$

Step 3.3.2.1.11.2.5

Evaluate the exponent.

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 \cdot 3}{3^{2}} + 5$

Step 3.3.2.1.12

Raise $3$ to the power of $2$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{4 \cdot 3}{9} + 5$

Step 3.3.2.1.13

Multiply $4$ by $3$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 (\frac{12}{9}) + 5$

Step 3.3.2.1.14

Cancel the common factor of $12$ and $9$ .

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Step 3.3.2.1.14.1

Factor $3$ out of $12$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{3 (4)}{9} + 5$

Step 3.3.2.1.14.2

Cancel the common factors.

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Step 3.3.2.1.14.2.1

Factor $3$ out of $9$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{3 \cdot 4}{3 \cdot 3} + 5$

Step 3.3.2.1.14.2.2

Cancel the common factor.

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 \frac{3 \cdot 4}{3 \cdot 3} + 5$

Step 3.3.2.1.14.2.3

Rewrite the expression.

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - 8 (\frac{4}{3}) + 5$

Step 3.3.2.1.15

Multiply $- 8 (\frac{4}{3})$ .

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Step 3.3.2.1.15.1

Combine $- 8$ and $\frac{4}{3}$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} + \frac{- 8 \cdot 4}{3} + 5$

Step 3.3.2.1.15.2

Multiply $- 8$ by $4$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} + \frac{- 32}{3} + 5$

Step 3.3.2.1.16

Move the negative in front of the fraction.

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - \frac{32}{3} + 5$

Step 3.3.2.2

Find the common denominator.

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Step 3.3.2.2.1

Multiply $\frac{32}{3}$ by $\frac{3}{3}$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - (\frac{32}{3} \cdot \frac{3}{3}) + 5$

Step 3.3.2.2.2

Multiply $\frac{32}{3}$ by $\frac{3}{3}$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - \frac{32 \cdot 3}{3 \cdot 3} + 5$

Step 3.3.2.2.3

Write $5$ as a fraction with denominator $1$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - \frac{32 \cdot 3}{3 \cdot 3} + \frac{5}{1}$

Step 3.3.2.2.4

Multiply $\frac{5}{1}$ by $\frac{9}{9}$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - \frac{32 \cdot 3}{3 \cdot 3} + \frac{5}{1} \cdot \frac{9}{9}$

Step 3.3.2.2.5

Multiply $\frac{5}{1}$ by $\frac{9}{9}$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - \frac{32 \cdot 3}{3 \cdot 3} + \frac{5 \cdot 9}{9}$

Step 3.3.2.2.6

Multiply $3$ by $3$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16}{9} - \frac{32 \cdot 3}{9} + \frac{5 \cdot 9}{9}$

Step 3.3.2.3

Combine the numerators over the common denominator.

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 - 32 \cdot 3 + 5 \cdot 9}{9}$

Step 3.3.2.4

Simplify each term.

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Step 3.3.2.4.1

Multiply $- 32$ by $3$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 - 96 + 5 \cdot 9}{9}$

Step 3.3.2.4.2

Multiply $5$ by $9$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{16 - 96 + 45}{9}$

Step 3.3.2.5

Simplify the expression.

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Step 3.3.2.5.1

Subtract $96$ from $16$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{- 80 + 45}{9}$

Step 3.3.2.5.2

Add $- 80$ and $45$ .

$f (- \frac{2 \sqrt{3}}{3}) = \frac{- 35}{9}$

Step 3.3.2.5.3

Move the negative in front of the fraction.

$f (- \frac{2 \sqrt{3}}{3}) = - \frac{35}{9}$

Step 3.3.2.6

The final answer is $- \frac{35}{9}$ .

$- \frac{35}{9}$

Step 3.4

The point found by substituting $- \frac{2 \sqrt{3}}{3}$ in $f (x) = x^{4} - 8 x^{2} + 5$ is $(- \frac{2 \sqrt{3}}{3}, - \frac{35}{9})$ . This point can be an inflection point.

$(- \frac{2 \sqrt{3}}{3}, - \frac{35}{9})$

Step 3.5

Determine the points that could be inflection points.

$(\frac{2 \sqrt{3}}{3}, - \frac{35}{9}), (- \frac{2 \sqrt{3}}{3}, - \frac{35}{9})$

Step 4

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, - \frac{2 \sqrt{3}}{3}) \cup (- \frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3}) \cup (\frac{2 \sqrt{3}}{3}, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - \frac{2 \sqrt{3}}{3})$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 1.25470053$ in the expression.

$f'' (- 1.25470053) = 12 {(- 1.25470053)}^{2} - 16$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Raise $- 1.25470053$ to the power of $2$ .

$f'' (- 1.25470053) = 12 \cdot 1.57427344 - 16$

Step 5.2.1.2

Multiply $12$ by $1.57427344$ .

$f'' (- 1.25470053) = 18.89128129 - 16$

Step 5.2.2

Subtract $16$ from $18.89128129$ .

$f'' (- 1.25470053) = 2.89128129$

Step 5.2.3

The final answer is $2.89128129$ .

$2.89128129$

Step 5.3

At $- 1.25470053$ , the second derivative is $2.89128129$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, - \frac{2 \sqrt{3}}{3})$ .

Increasing on $(- \infty, - \frac{2 \sqrt{3}}{3})$ since $f'' (x) > 0$

Step 6

Substitute a value from the interval $(- \frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3})$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

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

$f'' (0) = 12 {(0)}^{2} - 16$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

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

$f'' (0) = 12 \cdot 0 - 16$

Step 6.2.1.2

Multiply $12$ by $0$ .

$f'' (0) = 0 - 16$

Step 6.2.2

Subtract $16$ from $0$ .

$f'' (0) = - 16$

Step 6.2.3

The final answer is $- 16$ .

$- 16$

Step 6.3

At $0$ , the second derivative is $- 16$ . Since this is negative, the second derivative is decreasing on the interval $(- \frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3})$

Decreasing on $(- \frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3})$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(\frac{2 \sqrt{3}}{3}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $1.25470053$ in the expression.

$f'' (1.25470053) = 12 {(1.25470053)}^{2} - 16$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Raise $1.25470053$ to the power of $2$ .

$f'' (1.25470053) = 12 \cdot 1.57427344 - 16$

Step 7.2.1.2

Multiply $12$ by $1.57427344$ .

$f'' (1.25470053) = 18.89128129 - 16$

Step 7.2.2

Subtract $16$ from $18.89128129$ .

$f'' (1.25470053) = 2.89128129$

Step 7.2.3

The final answer is $2.89128129$ .

$2.89128129$

Step 7.3

At $1.25470053$ , the second derivative is $2.89128129$ . Since this is positive, the second derivative is increasing on the interval $(\frac{2 \sqrt{3}}{3}, \infty)$ .

Increasing on $(\frac{2 \sqrt{3}}{3}, \infty)$ since $f'' (x) > 0$

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection points in this case are $(- \frac{2 \sqrt{3}}{3}, - \frac{35}{9}), (\frac{2 \sqrt{3}}{3}, - \frac{35}{9})$ .

$(- \frac{2 \sqrt{3}}{3}, - \frac{35}{9}), (\frac{2 \sqrt{3}}{3}, - \frac{35}{9})$

Step 9