Calculus Examples

$h (x) = 5 x^{3} - 3 x^{5}$

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

By the Sum Rule, the derivative of $5 x^{3} - 3 x^{5}$ with respect to $x$ is $\frac{d}{d x} [5 x^{3}] + \frac{d}{d x} [- 3 x^{5}]$ .

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

Step 1.1.2

Evaluate $\frac{d}{d x} [5 x^{3}]$ .

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Step 1.1.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{3}$ with respect to $x$ is $5 \frac{d}{d x} [x^{3}]$ .

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

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 $n = 3$ .

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

Step 1.1.2.3

Multiply $3$ by $5$ .

$f' (x) = 15 x^{2} + \frac{d}{d x} (- 3 x^{5})$

Step 1.1.3

Evaluate $\frac{d}{d x} [- 3 x^{5}]$ .

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Step 1.1.3.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x^{5}$ with respect to $x$ is $- 3 \frac{d}{d x} [x^{5}]$ .

$f' (x) = 15 x^{2} - 3 \frac{d}{d x} x^{5}$

Step 1.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$f' (x) = 15 x^{2} - 3 (5 x^{4})$

Step 1.1.3.3

Multiply $5$ by $- 3$ .

$f' (x) = 15 x^{2} - 15 x^{4}$

Step 1.1.4

Reorder terms.

$f' (x) = - 15 x^{4} + 15 x^{2}$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $- 15 x^{4} + 15 x^{2}$ with respect to $x$ is $\frac{d}{d x} [- 15 x^{4}] + \frac{d}{d x} [15 x^{2}]$ .

$f'' (x) = \frac{d}{d x} (- 15 x^{4}) + \frac{d}{d x} (15 x^{2})$

Step 1.2.2

Evaluate $\frac{d}{d x} [- 15 x^{4}]$ .

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Step 1.2.2.1

Since $- 15$ is constant with respect to $x$ , the derivative of $- 15 x^{4}$ with respect to $x$ is $- 15 \frac{d}{d x} [x^{4}]$ .

$f'' (x) = - 15 \frac{d}{d x} x^{4} + \frac{d}{d x} (15 x^{2})$

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 $n = 4$ .

$f'' (x) = - 15 (4 x^{3}) + \frac{d}{d x} (15 x^{2})$

Step 1.2.2.3

Multiply $4$ by $- 15$ .

$f'' (x) = - 60 x^{3} + \frac{d}{d x} (15 x^{2})$

Step 1.2.3

Evaluate $\frac{d}{d x} [15 x^{2}]$ .

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Step 1.2.3.1

Since $15$ is constant with respect to $x$ , the derivative of $15 x^{2}$ with respect to $x$ is $15 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = - 60 x^{3} + 15 \frac{d}{d x} (x^{2})$

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 $n = 2$ .

$f'' (x) = - 60 x^{3} + 15 (2 x)$

Step 1.2.3.3

Multiply $2$ by $15$ .

$f'' (x) = - 60 x^{3} + 30 x$

Step 1.3

The second derivative of $h (x)$ with respect to $x$ is $- 60 x^{3} + 30 x$ .

$- 60 x^{3} + 30 x$

Step 2

Set the second derivative equal to $0$ then solve the equation $- 60 x^{3} + 30 x = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$- 60 x^{3} + 30 x = 0$

Step 2.2

Factor $- 30 x$ out of $- 60 x^{3} + 30 x$ .

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Step 2.2.1

Factor $- 30 x$ out of $- 60 x^{3}$ .

$- 30 x (2 x^{2}) + 30 x = 0$

Step 2.2.2

Factor $- 30 x$ out of $30 x$ .

$- 30 x (2 x^{2}) - 30 x \cdot - 1 = 0$

Step 2.2.3

Factor $- 30 x$ out of $- 30 x (2 x^{2}) - 30 x (- 1)$ .

$- 30 x (2 x^{2} - 1) = 0$

Step 2.3

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

$x = 0$

$2 x^{2} - 1 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

Set $2 x^{2} - 1$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $2 x^{2} - 1$ equal to $0$ .

$2 x^{2} - 1 = 0$

Step 2.5.2

Solve $2 x^{2} - 1 = 0$ for $x$ .

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Step 2.5.2.1

Add $1$ to both sides of the equation.

$2 x^{2} = 1$

Step 2.5.2.2

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

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Step 2.5.2.2.1

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

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

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

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

Step 2.5.2.2.2.1.2

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

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

Step 2.5.2.3

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

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

Step 2.5.2.4

Simplify $\pm \sqrt{\frac{1}{2}}$ .

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Step 2.5.2.4.1

Rewrite $\sqrt{\frac{1}{2}}$ as $\frac{\sqrt{1}}{\sqrt{2}}$ .

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

Step 2.5.2.4.2

Any root of $1$ is $1$ .

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

Step 2.5.2.4.3

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

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

Step 2.5.2.4.4

Combine and simplify the denominator.

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Step 2.5.2.4.4.1

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

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

Step 2.5.2.4.4.2

Raise $\sqrt{2}$ to the power of $1$ .

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

Step 2.5.2.4.4.3

Raise $\sqrt{2}$ to the power of $1$ .

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

Step 2.5.2.4.4.4

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

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

Step 2.5.2.4.4.5

Add $1$ and $1$ .

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

Step 2.5.2.4.4.6

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

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Step 2.5.2.4.4.6.1

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

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

Step 2.5.2.4.4.6.2

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

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

Step 2.5.2.4.4.6.3

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

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

Step 2.5.2.4.4.6.4

Cancel the common factor of $2$ .

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Step 2.5.2.4.4.6.4.1

Cancel the common factor.

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

Step 2.5.2.4.4.6.4.2

Rewrite the expression.

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

Step 2.5.2.4.4.6.5

Evaluate the exponent.

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

Step 2.5.2.5

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

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Step 2.5.2.5.1

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

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

Step 2.5.2.5.2

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

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

Step 2.5.2.5.3

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

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

Step 2.6

The final solution is all the values that make $- 30 x (2 x^{2} - 1) = 0$ true.

$x = 0, \frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2}$

Step 3

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

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Step 3.1

Substitute $0$ in $h (x) = 5 x^{3} - 3 x^{5}$ to find the value of $y$ .

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Step 3.1.1

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

$h (0) = 5 {(0)}^{3} - 3 {(0)}^{5}$

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

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

$h (0) = 5 \cdot 0 - 3 {(0)}^{5}$

Step 3.1.2.1.2

Multiply $5$ by $0$ .

$h (0) = 0 - 3 {(0)}^{5}$

Step 3.1.2.1.3

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

$h (0) = 0 - 3 \cdot 0$

Step 3.1.2.1.4

Multiply $- 3$ by $0$ .

$h (0) = 0 + 0$

Step 3.1.2.2

Add $0$ and $0$ .

$h (0) = 0$

Step 3.1.2.3

The final answer is $0$ .

$0$

Step 3.2

The point found by substituting $0$ in $h (x) = 5 x^{3} - 3 x^{5}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Step 3.3

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

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Step 3.3.1

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

$h (\frac{\sqrt{2}}{2}) = 5 {(\frac{\sqrt{2}}{2})}^{3} - 3 {(\frac{\sqrt{2}}{2})}^{5}$

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

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

$h (\frac{\sqrt{2}}{2}) = 5 (\frac{{\sqrt{2}}^{3}}{2^{3}}) - 3 {(\frac{\sqrt{2}}{2})}^{5}$

Step 3.3.2.1.2

Simplify the numerator.

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Step 3.3.2.1.2.1

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

$h (\frac{\sqrt{2}}{2}) = 5 (\frac{\sqrt{2^{3}}}{2^{3}}) - 3 {(\frac{\sqrt{2}}{2})}^{5}$

Step 3.3.2.1.2.2

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

$h (\frac{\sqrt{2}}{2}) = 5 (\frac{\sqrt{8}}{2^{3}}) - 3 {(\frac{\sqrt{2}}{2})}^{5}$

Step 3.3.2.1.2.3

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 3.3.2.1.2.3.1

Factor $4$ out of $8$ .

$h (\frac{\sqrt{2}}{2}) = 5 (\frac{\sqrt{4 (2)}}{2^{3}}) - 3 {(\frac{\sqrt{2}}{2})}^{5}$

Step 3.3.2.1.2.3.2

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

$h (\frac{\sqrt{2}}{2}) = 5 (\frac{\sqrt{2^{2} \cdot 2}}{2^{3}}) - 3 {(\frac{\sqrt{2}}{2})}^{5}$

Step 3.3.2.1.2.4

Pull terms out from under the radical.

$h (\frac{\sqrt{2}}{2}) = 5 (\frac{2 \sqrt{2}}{2^{3}}) - 3 {(\frac{\sqrt{2}}{2})}^{5}$

Step 3.3.2.1.3

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

$h (\frac{\sqrt{2}}{2}) = 5 (\frac{2 \sqrt{2}}{8}) - 3 {(\frac{\sqrt{2}}{2})}^{5}$

Step 3.3.2.1.4

Cancel the common factor of $2$ and $8$ .

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Step 3.3.2.1.4.1

Factor $2$ out of $2 \sqrt{2}$ .

$h (\frac{\sqrt{2}}{2}) = 5 (\frac{2 (\sqrt{2})}{8}) - 3 {(\frac{\sqrt{2}}{2})}^{5}$

Step 3.3.2.1.4.2

Cancel the common factors.

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Step 3.3.2.1.4.2.1

Factor $2$ out of $8$ .

$h (\frac{\sqrt{2}}{2}) = 5 (\frac{2 \sqrt{2}}{2 \cdot 4}) - 3 {(\frac{\sqrt{2}}{2})}^{5}$

Step 3.3.2.1.4.2.2

Cancel the common factor.

$h (\frac{\sqrt{2}}{2}) = 5 (\frac{2 \sqrt{2}}{2 \cdot 4}) - 3 {(\frac{\sqrt{2}}{2})}^{5}$

Step 3.3.2.1.4.2.3

Rewrite the expression.

$h (\frac{\sqrt{2}}{2}) = 5 (\frac{\sqrt{2}}{4}) - 3 {(\frac{\sqrt{2}}{2})}^{5}$

Step 3.3.2.1.5

Combine $5$ and $\frac{\sqrt{2}}{4}$ .

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2}}{4} - 3 {(\frac{\sqrt{2}}{2})}^{5}$

Step 3.3.2.1.6

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

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2}}{4} - 3 \frac{{\sqrt{2}}^{5}}{2^{5}}$

Step 3.3.2.1.7

Simplify the numerator.

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Step 3.3.2.1.7.1

Rewrite ${\sqrt{2}}^{5}$ as $\sqrt{2^{5}}$ .

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2}}{4} - 3 \frac{\sqrt{2^{5}}}{2^{5}}$

Step 3.3.2.1.7.2

Raise $2$ to the power of $5$ .

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2}}{4} - 3 \frac{\sqrt{32}}{2^{5}}$

Step 3.3.2.1.7.3

Rewrite $32$ as $4^{2} \cdot 2$ .

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Step 3.3.2.1.7.3.1

Factor $16$ out of $32$ .

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

Step 3.3.2.1.7.3.2

Rewrite $16$ as $4^{2}$ .

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2}}{4} - 3 \frac{\sqrt{4^{2} \cdot 2}}{2^{5}}$

Step 3.3.2.1.7.4

Pull terms out from under the radical.

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2}}{4} - 3 \frac{4 \sqrt{2}}{2^{5}}$

Step 3.3.2.1.8

Raise $2$ to the power of $5$ .

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2}}{4} - 3 \frac{4 \sqrt{2}}{32}$

Step 3.3.2.1.9

Cancel the common factor of $4$ and $32$ .

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Step 3.3.2.1.9.1

Factor $4$ out of $4 \sqrt{2}$ .

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2}}{4} - 3 \frac{4 (\sqrt{2})}{32}$

Step 3.3.2.1.9.2

Cancel the common factors.

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Step 3.3.2.1.9.2.1

Factor $4$ out of $32$ .

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2}}{4} - 3 \frac{4 \sqrt{2}}{4 \cdot 8}$

Step 3.3.2.1.9.2.2

Cancel the common factor.

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2}}{4} - 3 \frac{4 \sqrt{2}}{4 \cdot 8}$

Step 3.3.2.1.9.2.3

Rewrite the expression.

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2}}{4} - 3 \frac{\sqrt{2}}{8}$

Step 3.3.2.1.10

Combine $- 3$ and $\frac{\sqrt{2}}{8}$ .

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

Step 3.3.2.1.11

Move the negative in front of the fraction.

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2}}{4} - \frac{3 \sqrt{2}}{8}$

Step 3.3.2.2

To write $\frac{5 \sqrt{2}}{4}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2}}{4} \cdot \frac{2}{2} - \frac{3 \sqrt{2}}{8}$

Step 3.3.2.3

Write each expression with a common denominator of $8$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.3.2.3.1

Multiply $\frac{5 \sqrt{2}}{4}$ by $\frac{2}{2}$ .

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2} \cdot 2}{4 \cdot 2} - \frac{3 \sqrt{2}}{8}$

Step 3.3.2.3.2

Multiply $4$ by $2$ .

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2} \cdot 2}{8} - \frac{3 \sqrt{2}}{8}$

Step 3.3.2.4

Combine the numerators over the common denominator.

$h (\frac{\sqrt{2}}{2}) = \frac{5 \sqrt{2} \cdot 2 - 3 \sqrt{2}}{8}$

Step 3.3.2.5

Simplify the numerator.

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Step 3.3.2.5.1

Multiply $2$ by $5$ .

$h (\frac{\sqrt{2}}{2}) = \frac{10 \sqrt{2} - 3 \sqrt{2}}{8}$

Step 3.3.2.5.2

Subtract $3 \sqrt{2}$ from $10 \sqrt{2}$ .

$h (\frac{\sqrt{2}}{2}) = \frac{7 \sqrt{2}}{8}$

Step 3.3.2.6

The final answer is $\frac{7 \sqrt{2}}{8}$ .

$\frac{7 \sqrt{2}}{8}$

Step 3.4

The point found by substituting $\frac{\sqrt{2}}{2}$ in $h (x) = 5 x^{3} - 3 x^{5}$ is $(\frac{\sqrt{2}}{2}, \frac{7 \sqrt{2}}{8})$ . This point can be an inflection point.

$(\frac{\sqrt{2}}{2}, \frac{7 \sqrt{2}}{8})$

Step 3.5

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

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Step 3.5.1

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

$h (- \frac{\sqrt{2}}{2}) = 5 {(- \frac{\sqrt{2}}{2})}^{3} - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2

Simplify the result.

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Step 3.5.2.1

Simplify each term.

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Step 3.5.2.1.1

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

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Step 3.5.2.1.1.1

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

$h (- \frac{\sqrt{2}}{2}) = 5 ({(- 1)}^{3} {(\frac{\sqrt{2}}{2})}^{3}) - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.1.2

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

$h (- \frac{\sqrt{2}}{2}) = 5 ({(- 1)}^{3} (\frac{{\sqrt{2}}^{3}}{2^{3}})) - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.2

Raise $- 1$ to the power of $3$ .

$h (- \frac{\sqrt{2}}{2}) = 5 (- \frac{{\sqrt{2}}^{3}}{2^{3}}) - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.3

Simplify the numerator.

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Step 3.5.2.1.3.1

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

$h (- \frac{\sqrt{2}}{2}) = 5 (- \frac{\sqrt{2^{3}}}{2^{3}}) - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.3.2

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

$h (- \frac{\sqrt{2}}{2}) = 5 (- \frac{\sqrt{8}}{2^{3}}) - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.3.3

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 3.5.2.1.3.3.1

Factor $4$ out of $8$ .

$h (- \frac{\sqrt{2}}{2}) = 5 (- \frac{\sqrt{4 (2)}}{2^{3}}) - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.3.3.2

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

$h (- \frac{\sqrt{2}}{2}) = 5 (- \frac{\sqrt{2^{2} \cdot 2}}{2^{3}}) - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.3.4

Pull terms out from under the radical.

$h (- \frac{\sqrt{2}}{2}) = 5 (- \frac{2 \sqrt{2}}{2^{3}}) - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.4

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

$h (- \frac{\sqrt{2}}{2}) = 5 (- \frac{2 \sqrt{2}}{8}) - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.5

Cancel the common factor of $2$ and $8$ .

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Step 3.5.2.1.5.1

Factor $2$ out of $2 \sqrt{2}$ .

$h (- \frac{\sqrt{2}}{2}) = 5 (- \frac{2 (\sqrt{2})}{8}) - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.5.2

Cancel the common factors.

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Step 3.5.2.1.5.2.1

Factor $2$ out of $8$ .

$h (- \frac{\sqrt{2}}{2}) = 5 (- \frac{2 \sqrt{2}}{2 \cdot 4}) - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.5.2.2

Cancel the common factor.

$h (- \frac{\sqrt{2}}{2}) = 5 (- \frac{2 \sqrt{2}}{2 \cdot 4}) - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.5.2.3

Rewrite the expression.

$h (- \frac{\sqrt{2}}{2}) = 5 (- \frac{\sqrt{2}}{4}) - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.6

Multiply $5 (- \frac{\sqrt{2}}{4})$ .

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Step 3.5.2.1.6.1

Multiply $- 1$ by $5$ .

$h (- \frac{\sqrt{2}}{2}) = - 5 \frac{\sqrt{2}}{4} - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.6.2

Combine $- 5$ and $\frac{\sqrt{2}}{4}$ .

$h (- \frac{\sqrt{2}}{2}) = \frac{- 5 \sqrt{2}}{4} - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.7

Move the negative in front of the fraction.

$h (- \frac{\sqrt{2}}{2}) = - \frac{(5) \sqrt{2}}{4} - 3 {(- \frac{\sqrt{2}}{2})}^{5}$

Step 3.5.2.1.8

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

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Step 3.5.2.1.8.1

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

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

Step 3.5.2.1.8.2

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

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

Step 3.5.2.1.9

Raise $- 1$ to the power of $5$ .

$h (- \frac{\sqrt{2}}{2}) = - \frac{5 \sqrt{2}}{4} - 3 (- \frac{{\sqrt{2}}^{5}}{2^{5}})$

Step 3.5.2.1.10

Simplify the numerator.

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Step 3.5.2.1.10.1

Rewrite ${\sqrt{2}}^{5}$ as $\sqrt{2^{5}}$ .

$h (- \frac{\sqrt{2}}{2}) = - \frac{5 \sqrt{2}}{4} - 3 (- \frac{\sqrt{2^{5}}}{2^{5}})$

Step 3.5.2.1.10.2

Raise $2$ to the power of $5$ .

$h (- \frac{\sqrt{2}}{2}) = - \frac{5 \sqrt{2}}{4} - 3 (- \frac{\sqrt{32}}{2^{5}})$

Step 3.5.2.1.10.3

Rewrite $32$ as $4^{2} \cdot 2$ .

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Step 3.5.2.1.10.3.1

Factor $16$ out of $32$ .

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

Step 3.5.2.1.10.3.2

Rewrite $16$ as $4^{2}$ .

$h (- \frac{\sqrt{2}}{2}) = - \frac{5 \sqrt{2}}{4} - 3 (- \frac{\sqrt{4^{2} \cdot 2}}{2^{5}})$

Step 3.5.2.1.10.4

Pull terms out from under the radical.

$h (- \frac{\sqrt{2}}{2}) = - \frac{5 \sqrt{2}}{4} - 3 (- \frac{4 \sqrt{2}}{2^{5}})$

Step 3.5.2.1.11

Raise $2$ to the power of $5$ .

$h (- \frac{\sqrt{2}}{2}) = - \frac{5 \sqrt{2}}{4} - 3 (- \frac{4 \sqrt{2}}{32})$

Step 3.5.2.1.12

Cancel the common factor of $4$ and $32$ .

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Step 3.5.2.1.12.1

Factor $4$ out of $4 \sqrt{2}$ .

$h (- \frac{\sqrt{2}}{2}) = - \frac{5 \sqrt{2}}{4} - 3 (- \frac{4 (\sqrt{2})}{32})$

Step 3.5.2.1.12.2

Cancel the common factors.

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Step 3.5.2.1.12.2.1

Factor $4$ out of $32$ .

$h (- \frac{\sqrt{2}}{2}) = - \frac{5 \sqrt{2}}{4} - 3 (- \frac{4 \sqrt{2}}{4 \cdot 8})$

Step 3.5.2.1.12.2.2

Cancel the common factor.

$h (- \frac{\sqrt{2}}{2}) = - \frac{5 \sqrt{2}}{4} - 3 (- \frac{4 \sqrt{2}}{4 \cdot 8})$

Step 3.5.2.1.12.2.3

Rewrite the expression.

$h (- \frac{\sqrt{2}}{2}) = - \frac{5 \sqrt{2}}{4} - 3 (- \frac{\sqrt{2}}{8})$

Step 3.5.2.1.13

Multiply $- 3 (- \frac{\sqrt{2}}{8})$ .

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Step 3.5.2.1.13.1

Multiply $- 1$ by $- 3$ .

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

Step 3.5.2.1.13.2

Combine $3$ and $\frac{\sqrt{2}}{8}$ .

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

Step 3.5.2.2

To write $- \frac{5 \sqrt{2}}{4}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.5.2.3

Write each expression with a common denominator of $8$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.5.2.3.1

Multiply $\frac{5 \sqrt{2}}{4}$ by $\frac{2}{2}$ .

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

Step 3.5.2.3.2

Multiply $4$ by $2$ .

$h (- \frac{\sqrt{2}}{2}) = - \frac{5 \sqrt{2} \cdot 2}{8} + \frac{3 \sqrt{2}}{8}$

Step 3.5.2.4

Combine the numerators over the common denominator.

$h (- \frac{\sqrt{2}}{2}) = \frac{- 5 \sqrt{2} \cdot 2 + 3 \sqrt{2}}{8}$

Step 3.5.2.5

Simplify the numerator.

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Step 3.5.2.5.1

Multiply $2$ by $- 5$ .

$h (- \frac{\sqrt{2}}{2}) = \frac{- 10 \sqrt{2} + 3 \sqrt{2}}{8}$

Step 3.5.2.5.2

Add $- 10 \sqrt{2}$ and $3 \sqrt{2}$ .

$h (- \frac{\sqrt{2}}{2}) = \frac{- 7 \sqrt{2}}{8}$

Step 3.5.2.6

Move the negative in front of the fraction.

$h (- \frac{\sqrt{2}}{2}) = - \frac{7 \sqrt{2}}{8}$

Step 3.5.2.7

The final answer is $- \frac{7 \sqrt{2}}{8}$ .

$- \frac{7 \sqrt{2}}{8}$

Step 3.6

The point found by substituting $- \frac{\sqrt{2}}{2}$ in $h (x) = 5 x^{3} - 3 x^{5}$ is $(- \frac{\sqrt{2}}{2}, - \frac{7 \sqrt{2}}{8})$ . This point can be an inflection point.

$(- \frac{\sqrt{2}}{2}, - \frac{7 \sqrt{2}}{8})$

Step 3.7

Determine the points that could be inflection points.

$(0, 0), (\frac{\sqrt{2}}{2}, \frac{7 \sqrt{2}}{8}), (- \frac{\sqrt{2}}{2}, - \frac{7 \sqrt{2}}{8})$

Step 4

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

$(- \infty, - \frac{\sqrt{2}}{2}) \cup (- \frac{\sqrt{2}}{2}, 0) \cup (0, \frac{\sqrt{2}}{2}) \cup (\frac{\sqrt{2}}{2}, \infty)$

Step 5

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

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Step 5.1

Replace the variable $x$ with $- 0.80710678$ in the expression.

$h'' (- 0.80710678) = - 60 {(- 0.80710678)}^{3} + 30 (- 0.80710678)$

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 $- 0.80710678$ to the power of $3$ .

$h'' (- 0.80710678) = - 60 \cdot - 0.52576659 + 30 (- 0.80710678)$

Step 5.2.1.2

Multiply $- 60$ by $- 0.52576659$ .

$h'' (- 0.80710678) = 31.54599564 + 30 (- 0.80710678)$

Step 5.2.1.3

Multiply $30$ by $- 0.80710678$ .

$h'' (- 0.80710678) = 31.54599564 - 24.21320343$

Step 5.2.2

Subtract $24.21320343$ from $31.54599564$ .

$h'' (- 0.80710678) = 7.3327922$

Step 5.2.3

The final answer is $7.3327922$ .

$7.3327922$

Step 5.3

At $- 0.80710678$ , the second derivative is $7.3327922$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, - \frac{\sqrt{2}}{2})$ .

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

Step 6

Substitute a value from the interval $(- \frac{\sqrt{2}}{2}, 0)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 0.35355339$ in the expression.

$h'' (- 0.35355339) = - 60 {(- 0.35355339)}^{3} + 30 (- 0.35355339)$

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

Raise $- 0.35355339$ to the power of $3$ .

$h'' (- 0.35355339) = - 60 \cdot - 0.04419417 + 30 (- 0.35355339)$

Step 6.2.1.2

Multiply $- 60$ by $- 0.04419417$ .

$h'' (- 0.35355339) = 2.65165042 + 30 (- 0.35355339)$

Step 6.2.1.3

Multiply $30$ by $- 0.35355339$ .

$h'' (- 0.35355339) = 2.65165042 - 10.60660171$

Step 6.2.2

Subtract $10.60660171$ from $2.65165042$ .

$h'' (- 0.35355339) = - 7.95495128$

Step 6.2.3

The final answer is $- 7.95495128$ .

$- 7.95495128$

Step 6.3

At $- 0.35355339$ , the second derivative is $- 7.95495128$ . Since this is negative, the second derivative is decreasing on the interval $(- \frac{\sqrt{2}}{2}, 0)$

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

Step 7

Substitute a value from the interval $(0, \frac{\sqrt{2}}{2})$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $0.35355339$ in the expression.

$h'' (0.35355339) = - 60 {(0.35355339)}^{3} + 30 (0.35355339)$

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 $0.35355339$ to the power of $3$ .

$h'' (0.35355339) = - 60 \cdot 0.04419417 + 30 (0.35355339)$

Step 7.2.1.2

Multiply $- 60$ by $0.04419417$ .

$h'' (0.35355339) = - 2.65165042 + 30 (0.35355339)$

Step 7.2.1.3

Multiply $30$ by $0.35355339$ .

$h'' (0.35355339) = - 2.65165042 + 10.60660171$

Step 7.2.2

Add $- 2.65165042$ and $10.60660171$ .

$h'' (0.35355339) = 7.95495128$

Step 7.2.3

The final answer is $7.95495128$ .

$7.95495128$

Step 7.3

At $0.35355339$ , the second derivative is $7.95495128$ . Since this is positive, the second derivative is increasing on the interval $(0, \frac{\sqrt{2}}{2})$ .

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

Step 8

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

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Step 8.1

Replace the variable $x$ with $0.80710678$ in the expression.

$h'' (0.80710678) = - 60 {(0.80710678)}^{3} + 30 (0.80710678)$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

Raise $0.80710678$ to the power of $3$ .

$h'' (0.80710678) = - 60 \cdot 0.52576659 + 30 (0.80710678)$

Step 8.2.1.2

Multiply $- 60$ by $0.52576659$ .

$h'' (0.80710678) = - 31.54599564 + 30 (0.80710678)$

Step 8.2.1.3

Multiply $30$ by $0.80710678$ .

$h'' (0.80710678) = - 31.54599564 + 24.21320343$

Step 8.2.2

Add $- 31.54599564$ and $24.21320343$ .

$h'' (0.80710678) = - 7.3327922$

Step 8.2.3

The final answer is $- 7.3327922$ .

$- 7.3327922$

Step 8.3

At $0.80710678$ , the second derivative is $- 7.3327922$ . Since this is negative, the second derivative is decreasing on the interval $(\frac{\sqrt{2}}{2}, \infty)$

Decreasing on $(\frac{\sqrt{2}}{2}, \infty)$ since $h'' (x) < 0$

Step 9

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 $(- \frac{\sqrt{2}}{2}, - \frac{7 \sqrt{2}}{8}), (0, 0), (\frac{\sqrt{2}}{2}, \frac{7 \sqrt{2}}{8})$ .

$(- \frac{\sqrt{2}}{2}, - \frac{7 \sqrt{2}}{8}), (0, 0), (\frac{\sqrt{2}}{2}, \frac{7 \sqrt{2}}{8})$

Step 10