Calculus Examples

$f (x) = 9 x^{5} - 2 x^{3} - x$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $9 x^{5} - 2 x^{3} - x$ with respect to $x$ is $\frac{d}{d x} [9 x^{5}] + \frac{d}{d x} [- 2 x^{3}] + \frac{d}{d x} [- x]$ .

$\frac{d}{d x} [9 x^{5}] + \frac{d}{d x} [- 2 x^{3}] + \frac{d}{d x} [- x]$

Step 1.2

Evaluate $\frac{d}{d x} [9 x^{5}]$ .

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Step 1.2.1

Since $9$ is constant with respect to $x$ , the derivative of $9 x^{5}$ with respect to $x$ is $9 \frac{d}{d x} [x^{5}]$ .

$9 \frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 2 x^{3}] + \frac{d}{d x} [- x]$

Step 1.2.2

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

$9 (5 x^{4}) + \frac{d}{d x} [- 2 x^{3}] + \frac{d}{d x} [- x]$

Step 1.2.3

Multiply $5$ by $9$ .

$45 x^{4} + \frac{d}{d x} [- 2 x^{3}] + \frac{d}{d x} [- x]$

Step 1.3

Evaluate $\frac{d}{d x} [- 2 x^{3}]$ .

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Step 1.3.1

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

$45 x^{4} - 2 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- x]$

Step 1.3.2

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

$45 x^{4} - 2 (3 x^{2}) + \frac{d}{d x} [- x]$

Step 1.3.3

Multiply $3$ by $- 2$ .

$45 x^{4} - 6 x^{2} + \frac{d}{d x} [- x]$

Step 1.4

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

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Step 1.4.1

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

$45 x^{4} - 6 x^{2} - \frac{d}{d x} [x]$

Step 1.4.2

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

$45 x^{4} - 6 x^{2} - 1 \cdot 1$

Step 1.4.3

Multiply $- 1$ by $1$ .

$45 x^{4} - 6 x^{2} - 1$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $45 x^{4} - 6 x^{2} - 1$ with respect to $x$ is $\frac{d}{d x} [45 x^{4}] + \frac{d}{d x} [- 6 x^{2}] + \frac{d}{d x} [- 1]$ .

$f'' (x) = \frac{d}{d x} (45 x^{4}) + \frac{d}{d x} (- 6 x^{2}) + \frac{d}{d x} (- 1)$

Step 2.2

Evaluate $\frac{d}{d x} [45 x^{4}]$ .

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Step 2.2.1

Since $45$ is constant with respect to $x$ , the derivative of $45 x^{4}$ with respect to $x$ is $45 \frac{d}{d x} [x^{4}]$ .

$f'' (x) = 45 \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 6 x^{2}) + \frac{d}{d x} (- 1)$

Step 2.2.2

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

$f'' (x) = 45 (4 x^{3}) + \frac{d}{d x} (- 6 x^{2}) + \frac{d}{d x} (- 1)$

Step 2.2.3

Multiply $4$ by $45$ .

$f'' (x) = 180 x^{3} + \frac{d}{d x} (- 6 x^{2}) + \frac{d}{d x} (- 1)$

Step 2.3

Evaluate $\frac{d}{d x} [- 6 x^{2}]$ .

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Step 2.3.1

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

$f'' (x) = 180 x^{3} - 6 \frac{d}{d x} x^{2} + \frac{d}{d x} (- 1)$

Step 2.3.2

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

$f'' (x) = 180 x^{3} - 6 (2 x) + \frac{d}{d x} (- 1)$

Step 2.3.3

Multiply $2$ by $- 6$ .

$f'' (x) = 180 x^{3} - 12 x + \frac{d}{d x} (- 1)$

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

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

$f'' (x) = 180 x^{3} - 12 x + 0$

Step 2.4.2

Add $180 x^{3} - 12 x$ and $0$ .

$f'' (x) = 180 x^{3} - 12 x$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$45 x^{4} - 6 x^{2} - 1 = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

By the Sum Rule, the derivative of $9 x^{5} - 2 x^{3} - x$ with respect to $x$ is $\frac{d}{d x} [9 x^{5}] + \frac{d}{d x} [- 2 x^{3}] + \frac{d}{d x} [- x]$ .

$\frac{d}{d x} [9 x^{5}] + \frac{d}{d x} [- 2 x^{3}] + \frac{d}{d x} [- x]$

Step 4.1.2

Evaluate $\frac{d}{d x} [9 x^{5}]$ .

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Step 4.1.2.1

Since $9$ is constant with respect to $x$ , the derivative of $9 x^{5}$ with respect to $x$ is $9 \frac{d}{d x} [x^{5}]$ .

$9 \frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 2 x^{3}] + \frac{d}{d x} [- x]$

Step 4.1.2.2

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

$9 (5 x^{4}) + \frac{d}{d x} [- 2 x^{3}] + \frac{d}{d x} [- x]$

Step 4.1.2.3

Multiply $5$ by $9$ .

$45 x^{4} + \frac{d}{d x} [- 2 x^{3}] + \frac{d}{d x} [- x]$

Step 4.1.3

Evaluate $\frac{d}{d x} [- 2 x^{3}]$ .

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Step 4.1.3.1

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

$45 x^{4} - 2 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- x]$

Step 4.1.3.2

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

$45 x^{4} - 2 (3 x^{2}) + \frac{d}{d x} [- x]$

Step 4.1.3.3

Multiply $3$ by $- 2$ .

$45 x^{4} - 6 x^{2} + \frac{d}{d x} [- x]$

Step 4.1.4

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

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Step 4.1.4.1

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

$45 x^{4} - 6 x^{2} - \frac{d}{d x} [x]$

Step 4.1.4.2

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

$45 x^{4} - 6 x^{2} - 1 \cdot 1$

Step 4.1.4.3

Multiply $- 1$ by $1$ .

$f' (x) = 45 x^{4} - 6 x^{2} - 1$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $45 x^{4} - 6 x^{2} - 1$ .

$45 x^{4} - 6 x^{2} - 1$

Step 5

Set the first derivative equal to $0$ then solve the equation $45 x^{4} - 6 x^{2} - 1 = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$45 x^{4} - 6 x^{2} - 1 = 0$

Step 5.2

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

$45 u^{2} - 6 u - 1 = 0$

$u = x^{2}$

Step 5.3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5.4

Substitute the values $a = 45$ , $b = - 6$ , and $c = - 1$ into the quadratic formula and solve for $u$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (45 \cdot - 1)}}{2 \cdot 45}$

Step 5.5

Simplify.

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Step 5.5.1

Simplify the numerator.

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Step 5.5.1.1

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

$u = \frac{6 \pm \sqrt{36 - 4 \cdot 45 \cdot - 1}}{2 \cdot 45}$

Step 5.5.1.2

Multiply $- 4 \cdot 45 \cdot - 1$ .

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Step 5.5.1.2.1

Multiply $- 4$ by $45$ .

$u = \frac{6 \pm \sqrt{36 - 180 \cdot - 1}}{2 \cdot 45}$

Step 5.5.1.2.2

Multiply $- 180$ by $- 1$ .

$u = \frac{6 \pm \sqrt{36 + 180}}{2 \cdot 45}$

Step 5.5.1.3

Add $36$ and $180$ .

$u = \frac{6 \pm \sqrt{216}}{2 \cdot 45}$

Step 5.5.1.4

Rewrite $216$ as $6^{2} \cdot 6$ .

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Step 5.5.1.4.1

Factor $36$ out of $216$ .

$u = \frac{6 \pm \sqrt{36 (6)}}{2 \cdot 45}$

Step 5.5.1.4.2

Rewrite $36$ as $6^{2}$ .

$u = \frac{6 \pm \sqrt{6^{2} \cdot 6}}{2 \cdot 45}$

Step 5.5.1.5

Pull terms out from under the radical.

$u = \frac{6 \pm 6 \sqrt{6}}{2 \cdot 45}$

Step 5.5.2

Multiply $2$ by $45$ .

$u = \frac{6 \pm 6 \sqrt{6}}{90}$

Step 5.5.3

Simplify $\frac{6 \pm 6 \sqrt{6}}{90}$ .

$u = \frac{1 \pm \sqrt{6}}{15}$

Step 5.6

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 5.6.1

Simplify the numerator.

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Step 5.6.1.1

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

$u = \frac{6 \pm \sqrt{36 - 4 \cdot 45 \cdot - 1}}{2 \cdot 45}$

Step 5.6.1.2

Multiply $- 4 \cdot 45 \cdot - 1$ .

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Step 5.6.1.2.1

Multiply $- 4$ by $45$ .

$u = \frac{6 \pm \sqrt{36 - 180 \cdot - 1}}{2 \cdot 45}$

Step 5.6.1.2.2

Multiply $- 180$ by $- 1$ .

$u = \frac{6 \pm \sqrt{36 + 180}}{2 \cdot 45}$

Step 5.6.1.3

Add $36$ and $180$ .

$u = \frac{6 \pm \sqrt{216}}{2 \cdot 45}$

Step 5.6.1.4

Rewrite $216$ as $6^{2} \cdot 6$ .

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Step 5.6.1.4.1

Factor $36$ out of $216$ .

$u = \frac{6 \pm \sqrt{36 (6)}}{2 \cdot 45}$

Step 5.6.1.4.2

Rewrite $36$ as $6^{2}$ .

$u = \frac{6 \pm \sqrt{6^{2} \cdot 6}}{2 \cdot 45}$

Step 5.6.1.5

Pull terms out from under the radical.

$u = \frac{6 \pm 6 \sqrt{6}}{2 \cdot 45}$

Step 5.6.2

Multiply $2$ by $45$ .

$u = \frac{6 \pm 6 \sqrt{6}}{90}$

Step 5.6.3

Simplify $\frac{6 \pm 6 \sqrt{6}}{90}$ .

$u = \frac{1 \pm \sqrt{6}}{15}$

Step 5.6.4

Change the $\pm$ to $+$ .

$u = \frac{1 + \sqrt{6}}{15}$

Step 5.7

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 5.7.1

Simplify the numerator.

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Step 5.7.1.1

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

$u = \frac{6 \pm \sqrt{36 - 4 \cdot 45 \cdot - 1}}{2 \cdot 45}$

Step 5.7.1.2

Multiply $- 4 \cdot 45 \cdot - 1$ .

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Step 5.7.1.2.1

Multiply $- 4$ by $45$ .

$u = \frac{6 \pm \sqrt{36 - 180 \cdot - 1}}{2 \cdot 45}$

Step 5.7.1.2.2

Multiply $- 180$ by $- 1$ .

$u = \frac{6 \pm \sqrt{36 + 180}}{2 \cdot 45}$

Step 5.7.1.3

Add $36$ and $180$ .

$u = \frac{6 \pm \sqrt{216}}{2 \cdot 45}$

Step 5.7.1.4

Rewrite $216$ as $6^{2} \cdot 6$ .

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Step 5.7.1.4.1

Factor $36$ out of $216$ .

$u = \frac{6 \pm \sqrt{36 (6)}}{2 \cdot 45}$

Step 5.7.1.4.2

Rewrite $36$ as $6^{2}$ .

$u = \frac{6 \pm \sqrt{6^{2} \cdot 6}}{2 \cdot 45}$

Step 5.7.1.5

Pull terms out from under the radical.

$u = \frac{6 \pm 6 \sqrt{6}}{2 \cdot 45}$

Step 5.7.2

Multiply $2$ by $45$ .

$u = \frac{6 \pm 6 \sqrt{6}}{90}$

Step 5.7.3

Simplify $\frac{6 \pm 6 \sqrt{6}}{90}$ .

$u = \frac{1 \pm \sqrt{6}}{15}$

Step 5.7.4

Change the $\pm$ to $-$ .

$u = \frac{1 - \sqrt{6}}{15}$

Step 5.8

The final answer is the combination of both solutions.

$u = \frac{1 + \sqrt{6}}{15}, \frac{1 - \sqrt{6}}{15}$

Step 5.9

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = 0.22996598$

${(x^{2})}^{1} = - 0.09663264$

Step 5.10

Solve the first equation for $x$ .

$x^{2} = 0.22996598$

Step 5.11

Solve the equation for $x$ .

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Step 5.11.1

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

$x = \pm \sqrt{0.22996598}$

Step 5.11.2

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

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Step 5.11.2.1

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

$x = \sqrt{0.22996598}$

Step 5.11.2.2

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

$x = - \sqrt{0.22996598}$

Step 5.11.2.3

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

$x = \sqrt{0.22996598}, - \sqrt{0.22996598}$

Step 5.12

Solve the second equation for $x$ .

${(x^{2})}^{1} = - 0.09663264$

Step 5.13

Solve the equation for $x$ .

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Step 5.13.1

Remove parentheses.

$x^{2} = - 0.09663264$

Step 5.13.2

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

$x = \pm \sqrt{- 0.09663264}$

Step 5.13.3

Simplify $\pm \sqrt{- 0.09663264}$ .

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Step 5.13.3.1

Rewrite $- 0.09663264$ as $- 1 (0.09663264)$ .

$x = \pm \sqrt{- 1 (0.09663264)}$

Step 5.13.3.2

Rewrite $\sqrt{- 1 (0.09663264)}$ as $\sqrt{- 1} \cdot \sqrt{0.09663264}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{0.09663264}$

Step 5.13.3.3

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

$x = \pm i \sqrt{0.09663264}$

Step 5.13.4

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

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Step 5.13.4.1

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

$x = i \sqrt{0.09663264}$

Step 5.13.4.2

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

$x = - i \sqrt{0.09663264}$

Step 5.13.4.3

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

$x = i \sqrt{0.09663264}, - i \sqrt{0.09663264}$

Step 5.14

The solution to $45 x^{4} - 6 x^{2} - 1 = 0$ is $x = \sqrt{0.22996598}, - \sqrt{0.22996598}, i \sqrt{0.09663264}, - i \sqrt{0.09663264}$ .

$x = \sqrt{0.22996598}, - \sqrt{0.22996598}, i \sqrt{0.09663264}, - i \sqrt{0.09663264}$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = \sqrt{0.22996598}, - \sqrt{0.22996598}$

Step 8

Evaluate the second derivative at $x = \sqrt{0.22996598}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$180 {(\sqrt{0.22996598})}^{3} - 12 \sqrt{0.22996598}$

Step 9

Simplify each term.

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Step 9.1

Rewrite ${\sqrt{0.22996598}}^{3}$ as $\sqrt{{0.22996598}^{3}}$ .

$180 \sqrt{{0.22996598}^{3}} - 12 \sqrt{0.22996598}$

Step 9.2

Raise $0.22996598$ to the power of $3$ .

$180 \sqrt{0.0121616} - 12 \sqrt{0.22996598}$

Step 10

$x = \sqrt{0.22996598}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \sqrt{0.22996598}$ is a local minimum

Step 11

Find the y-value when $x = \sqrt{0.22996598}$ .

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Step 11.1

Replace the variable $x$ with $\sqrt{0.22996598}$ in the expression.

$f (\sqrt{0.22996598}) = 9 {(\sqrt{0.22996598})}^{5} - 2 {(\sqrt{0.22996598})}^{3} - (\sqrt{0.22996598})$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Rewrite ${\sqrt{0.22996598}}^{5}$ as $\sqrt{{0.22996598}^{5}}$ .

$f (\sqrt{0.22996598}) = 9 \sqrt{{0.22996598}^{5}} - 2 {(\sqrt{0.22996598})}^{3} - (\sqrt{0.22996598})$

Step 11.2.1.2

Raise $0.22996598$ to the power of $5$ .

$f (\sqrt{0.22996598}) = 9 \sqrt{0.00064315} - 2 {(\sqrt{0.22996598})}^{3} - (\sqrt{0.22996598})$

Step 11.2.1.3

Rewrite ${\sqrt{0.22996598}}^{3}$ as $\sqrt{{0.22996598}^{3}}$ .

$f (\sqrt{0.22996598}) = 9 \sqrt{0.00064315} - 2 \sqrt{{0.22996598}^{3}} - (\sqrt{0.22996598})$

Step 11.2.1.4

Raise $0.22996598$ to the power of $3$ .

$f (\sqrt{0.22996598}) = 9 \sqrt{0.00064315} - 2 \sqrt{0.0121616} - \sqrt{0.22996598}$

Step 11.2.2

The final answer is $9 \sqrt{0.00064315} - 2 \sqrt{0.0121616} - \sqrt{0.22996598}$ .

$y = 9 \sqrt{0.00064315} - 2 \sqrt{0.0121616} - \sqrt{0.22996598}$

Step 12

Evaluate the second derivative at $x = - \sqrt{0.22996598}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$180 {(- \sqrt{0.22996598})}^{3} - 12 (- \sqrt{0.22996598})$

Step 13

Simplify each term.

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Step 13.1

Apply the product rule to $- \sqrt{0.22996598}$ .

$180 ({(- 1)}^{3} {\sqrt{0.22996598}}^{3}) - 12 (- \sqrt{0.22996598})$

Step 13.2

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

$180 (- {\sqrt{0.22996598}}^{3}) - 12 (- \sqrt{0.22996598})$

Step 13.3

Rewrite ${\sqrt{0.22996598}}^{3}$ as $\sqrt{{0.22996598}^{3}}$ .

$180 (- \sqrt{{0.22996598}^{3}}) - 12 (- \sqrt{0.22996598})$

Step 13.4

Raise $0.22996598$ to the power of $3$ .

$180 (- \sqrt{0.0121616}) - 12 (- \sqrt{0.22996598})$

Step 13.5

Multiply $- 1$ by $180$ .

$- 180 \sqrt{0.0121616} - 12 (- \sqrt{0.22996598})$

Step 13.6

Multiply $- 1$ by $- 12$ .

$- 180 \sqrt{0.0121616} + 12 \sqrt{0.22996598}$

Step 14

$x = - \sqrt{0.22996598}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - \sqrt{0.22996598}$ is a local maximum

Step 15

Find the y-value when $x = - \sqrt{0.22996598}$ .

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Step 15.1

Replace the variable $x$ with $- \sqrt{0.22996598}$ in the expression.

$f (- \sqrt{0.22996598}) = 9 {(- \sqrt{0.22996598})}^{5} - 2 {(- \sqrt{0.22996598})}^{3} - (- \sqrt{0.22996598})$

Step 15.2

Simplify the result.

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Step 15.2.1

Simplify each term.

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Step 15.2.1.1

Apply the product rule to $- \sqrt{0.22996598}$ .

$f (- \sqrt{0.22996598}) = 9 ({(- 1)}^{5} {\sqrt{0.22996598}}^{5}) - 2 {(- \sqrt{0.22996598})}^{3} - (- \sqrt{0.22996598})$

Step 15.2.1.2

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

$f (- \sqrt{0.22996598}) = 9 (- {\sqrt{0.22996598}}^{5}) - 2 {(- \sqrt{0.22996598})}^{3} - (- \sqrt{0.22996598})$

Step 15.2.1.3

Rewrite ${\sqrt{0.22996598}}^{5}$ as $\sqrt{{0.22996598}^{5}}$ .

$f (- \sqrt{0.22996598}) = 9 (- \sqrt{{0.22996598}^{5}}) - 2 {(- \sqrt{0.22996598})}^{3} - (- \sqrt{0.22996598})$

Step 15.2.1.4

Raise $0.22996598$ to the power of $5$ .

$f (- \sqrt{0.22996598}) = 9 (- \sqrt{0.00064315}) - 2 {(- \sqrt{0.22996598})}^{3} - (- \sqrt{0.22996598})$

Step 15.2.1.5

Multiply $- 1$ by $9$ .

$f (- \sqrt{0.22996598}) = - 9 \sqrt{0.00064315} - 2 {(- \sqrt{0.22996598})}^{3} - (- \sqrt{0.22996598})$

Step 15.2.1.6

Apply the product rule to $- \sqrt{0.22996598}$ .

$f (- \sqrt{0.22996598}) = - 9 \sqrt{0.00064315} - 2 ({(- 1)}^{3} {\sqrt{0.22996598}}^{3}) - (- \sqrt{0.22996598})$

Step 15.2.1.7

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

$f (- \sqrt{0.22996598}) = - 9 \sqrt{0.00064315} - 2 (- {\sqrt{0.22996598}}^{3}) - (- \sqrt{0.22996598})$

Step 15.2.1.8

Rewrite ${\sqrt{0.22996598}}^{3}$ as $\sqrt{{0.22996598}^{3}}$ .

$f (- \sqrt{0.22996598}) = - 9 \sqrt{0.00064315} - 2 (- \sqrt{{0.22996598}^{3}}) - (- \sqrt{0.22996598})$

Step 15.2.1.9

Raise $0.22996598$ to the power of $3$ .

$f (- \sqrt{0.22996598}) = - 9 \sqrt{0.00064315} - 2 (- \sqrt{0.0121616}) - (- \sqrt{0.22996598})$

Step 15.2.1.10

Multiply $- 1$ by $- 2$ .

$f (- \sqrt{0.22996598}) = - 9 \sqrt{0.00064315} + 2 \sqrt{0.0121616} - (- \sqrt{0.22996598})$

Step 15.2.1.11

Multiply $- (- \sqrt{0.22996598})$ .

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Step 15.2.1.11.1

Multiply $- 1$ by $- 1$ .

$f (- \sqrt{0.22996598}) = - 9 \sqrt{0.00064315} + 2 \sqrt{0.0121616} + 1 \sqrt{0.22996598}$

Step 15.2.1.11.2

Multiply $\sqrt{0.22996598}$ by $1$ .

$f (- \sqrt{0.22996598}) = - 9 \sqrt{0.00064315} + 2 \sqrt{0.0121616} + \sqrt{0.22996598}$

Step 15.2.2

The final answer is $- 9 \sqrt{0.00064315} + 2 \sqrt{0.0121616} + \sqrt{0.22996598}$ .

$y = - 9 \sqrt{0.00064315} + 2 \sqrt{0.0121616} + \sqrt{0.22996598}$

Step 16

These are the local extrema for $f (x) = 9 x^{5} - 2 x^{3} - x$ .

$(\sqrt{0.22996598}, 9 \sqrt{0.00064315} - 2 \sqrt{0.0121616} - \sqrt{0.22996598})$ is a local minima

$(- \sqrt{0.22996598}, - 9 \sqrt{0.00064315} + 2 \sqrt{0.0121616} + \sqrt{0.22996598})$ is a local maxima

Step 17