Calculus Examples

f (x) = \frac{1}{12} x^{4} + 3 x^{3} + 40 x^{2}

$f (x) = \frac{1}{12} x^{4} + 3 x^{3} + 40 x^{2}$

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

\frac{1}{12} x^{4} + 3 x^{3} + 40 x^{2}

$\frac{1}{12} x^{4} + 3 x^{3} + 40 x^{2}$ with respect to

x

$x$ is

\frac{d}{d x} [\frac{1}{12} x^{4}] + \frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [40 x^{2}]

$\frac{d}{d x} [\frac{1}{12} x^{4}] + \frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [40 x^{2}]$ .

$\frac{d}{d x} [\frac{1}{12} x^{4}] + \frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [40 x^{2}]$

Step 1.1.2

Evaluate

$\frac{d}{d x} [\frac{1}{12} x^{4}]$ .

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Step 1.1.2.1

Since

$\frac{1}{12}$ is constant with respect to

$x$ , the derivative of

$\frac{1}{12} x^{4}$ with respect to

$x$ is

$\frac{1}{12} \frac{d}{d x} [x^{4}]$ .

$\frac{1}{12} \frac{d}{d x} [x^{4}] + \frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [40 x^{2}]$

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

$n = 4$ .

$\frac{1}{12} (4 x^{3}) + \frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [40 x^{2}]$

Step 1.1.2.3

Combine

$4$ and

$\frac{1}{12}$ .

$\frac{4}{12} x^{3} + \frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [40 x^{2}]$

Step 1.1.2.4

Combine

$\frac{4}{12}$ and

$x^{3}$ .

$\frac{4 x^{3}}{12} + \frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [40 x^{2}]$

Step 1.1.2.5

Cancel the common factor of

$4$ and

$12$ .

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Step 1.1.2.5.1

Factor

$4$ out of

$4 x^{3}$ .

$\frac{4 (x^{3})}{12} + \frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [40 x^{2}]$

Step 1.1.2.5.2

Cancel the common factors.

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Step 1.1.2.5.2.1

Factor

$4$ out of

$12$ .

$\frac{4 x^{3}}{4 \cdot 3} + \frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [40 x^{2}]$

Step 1.1.2.5.2.2

Cancel the common factor.

$\frac{4 x^{3}}{4 \cdot 3} + \frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [40 x^{2}]$

Step 1.1.2.5.2.3

Rewrite the expression.

$\frac{x^{3}}{3} + \frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [40 x^{2}]$

Step 1.1.3

Evaluate

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

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Step 1.1.3.1

Since

$3$ is constant with respect to

$x$ , the derivative of

$3 x^{3}$ with respect to

$x$ is

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

$\frac{x^{3}}{3} + 3 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [40 x^{2}]$

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

$n = 3$ .

$\frac{x^{3}}{3} + 3 (3 x^{2}) + \frac{d}{d x} [40 x^{2}]$

Step 1.1.3.3

Multiply

$3$ by

$3$ .

$\frac{x^{3}}{3} + 9 x^{2} + \frac{d}{d x} [40 x^{2}]$

Step 1.1.4

Evaluate

$\frac{d}{d x} [40 x^{2}]$ .

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Step 1.1.4.1

Since

$40$ is constant with respect to

$x$ , the derivative of

$40 x^{2}$ with respect to

$x$ is

$40 \frac{d}{d x} [x^{2}]$ .

$\frac{x^{3}}{3} + 9 x^{2} + 40 \frac{d}{d x} [x^{2}]$

Step 1.1.4.2

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 2$ .

$\frac{x^{3}}{3} + 9 x^{2} + 40 (2 x)$

Step 1.1.4.3

Multiply

$2$ by

$40$ .

$f' (x) = \frac{x^{3}}{3} + 9 x^{2} + 80 x$

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of

$\frac{x^{3}}{3} + 9 x^{2} + 80 x$ with respect to

$x$ is

$\frac{d}{d x} [\frac{x^{3}}{3}] + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [80 x]$ .

$\frac{d}{d x} [\frac{x^{3}}{3}] + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [80 x]$

Step 1.2.2

Evaluate

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

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Step 1.2.2.1

Since

$\frac{1}{3}$ is constant with respect to

$x$ , the derivative of

$\frac{x^{3}}{3}$ with respect to

$x$ is

$\frac{1}{3} \frac{d}{d x} [x^{3}]$ .

$\frac{1}{3} \frac{d}{d x} [x^{3}] + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [80 x]$

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

$n = 3$ .

$\frac{1}{3} (3 x^{2}) + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [80 x]$

Step 1.2.2.3

Combine

$3$ and

$\frac{1}{3}$ .

$\frac{3}{3} x^{2} + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [80 x]$

Step 1.2.2.4

Combine

$\frac{3}{3}$ and

$x^{2}$ .

$\frac{3 x^{2}}{3} + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [80 x]$

Step 1.2.2.5

Cancel the common factor of

$3$ .

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Step 1.2.2.5.1

Cancel the common factor.

$\frac{3 x^{2}}{3} + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [80 x]$

Step 1.2.2.5.2

Divide

$x^{2}$ by

$1$ .

$x^{2} + \frac{d}{d x} [9 x^{2}] + \frac{d}{d x} [80 x]$

Step 1.2.3

Evaluate

$\frac{d}{d x} [9 x^{2}]$ .

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Step 1.2.3.1

Since

$9$ is constant with respect to

$x$ , the derivative of

$9 x^{2}$ with respect to

$x$ is

$9 \frac{d}{d x} [x^{2}]$ .

$x^{2} + 9 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [80 x]$

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

$n = 2$ .

$x^{2} + 9 (2 x) + \frac{d}{d x} [80 x]$

Step 1.2.3.3

Multiply

$2$ by

$9$ .

$x^{2} + 18 x + \frac{d}{d x} [80 x]$

Step 1.2.4

Evaluate

$\frac{d}{d x} [80 x]$ .

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Step 1.2.4.1

Since

$80$ is constant with respect to

$x$ , the derivative of

$80 x$ with respect to

$x$ is

$80 \frac{d}{d x} [x]$ .

$x^{2} + 18 x + 80 \frac{d}{d x} [x]$

Step 1.2.4.2

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

$x^{2} + 18 x + 80 \cdot 1$

Step 1.2.4.3

Multiply

$80$ by

$1$ .

$f'' (x) = x^{2} + 18 x + 80$

Step 1.3

The second derivative of

$f (x)$ with respect to

$x$ is

$x^{2} + 18 x + 80$ .

$x^{2} + 18 x + 80$

Step 2

Set the second derivative equal to

$0$ then solve the equation

$x^{2} + 18 x + 80 = 0$ .

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Step 2.1

Set the second derivative equal to

$0$ .

$x^{2} + 18 x + 80 = 0$

Step 2.2

Factor

$x^{2} + 18 x + 80$ using the AC method.

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Step 2.2.1

Consider the form

$x^{2} + b x + c$ . Find a pair of integers whose product is

$c$ and whose sum is

$b$ . In this case, whose product is

$80$ and whose sum is

$18$ .

$8, 10$

Step 2.2.2

Write the factored form using these integers.

$(x + 8) (x + 10) = 0$

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

$0$ .

$x + 8 = 0$

$x + 10 = 0$

Step 2.4

Set

$x + 8$ equal to

$0$ and solve for

$x$ .

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Step 2.4.1

Set

$x + 8$ equal to

$0$ .

$x + 8 = 0$

Step 2.4.2

Subtract

$8$ from both sides of the equation.

$x = - 8$

Step 2.5

Set

$x + 10$ equal to

$0$ and solve for

$x$ .

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Step 2.5.1

Set

$x + 10$ equal to

$0$ .

$x + 10 = 0$

Step 2.5.2

Subtract

$10$ from both sides of the equation.

$x = - 10$

Step 2.6

The final solution is all the values that make

$(x + 8) (x + 10) = 0$ true.

$x = - 8, - 10$

Step 3

Find the points where the second derivative is

$0$ .

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Step 3.1

Substitute

$- 8$ in

$f (x) = \frac{1}{12} x^{4} + 3 x^{3} + 40 x^{2}$ to find the value of

$y$ .

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Step 3.1.1

Replace the variable

$x$ with

$- 8$ in the expression.

$f (- 8) = \frac{1}{12} \cdot {(- 8)}^{4} + 3 {(- 8)}^{3} + 40 {(- 8)}^{2}$

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

Raise

$- 8$ to the power of

$4$ .

$f (- 8) = \frac{1}{12} \cdot 4096 + 3 {(- 8)}^{3} + 40 {(- 8)}^{2}$

Step 3.1.2.1.2

Cancel the common factor of

$4$ .

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Step 3.1.2.1.2.1

Factor

$4$ out of

$12$ .

$f (- 8) = \frac{1}{4 (3)} \cdot 4096 + 3 {(- 8)}^{3} + 40 {(- 8)}^{2}$

Step 3.1.2.1.2.2

Factor

$4$ out of

$4096$ .

$f (- 8) = \frac{1}{4 \cdot 3} \cdot (4 \cdot 1024) + 3 {(- 8)}^{3} + 40 {(- 8)}^{2}$

Step 3.1.2.1.2.3

Cancel the common factor.

$f (- 8) = \frac{1}{4 \cdot 3} \cdot (4 \cdot 1024) + 3 {(- 8)}^{3} + 40 {(- 8)}^{2}$

Step 3.1.2.1.2.4

Rewrite the expression.

$f (- 8) = \frac{1}{3} \cdot 1024 + 3 {(- 8)}^{3} + 40 {(- 8)}^{2}$

Step 3.1.2.1.3

Combine

$\frac{1}{3}$ and

$1024$ .

$f (- 8) = \frac{1024}{3} + 3 {(- 8)}^{3} + 40 {(- 8)}^{2}$

Step 3.1.2.1.4

Raise

$- 8$ to the power of

$3$ .

$f (- 8) = \frac{1024}{3} + 3 \cdot - 512 + 40 {(- 8)}^{2}$

Step 3.1.2.1.5

Multiply

$3$ by

$- 512$ .

$f (- 8) = \frac{1024}{3} - 1536 + 40 {(- 8)}^{2}$

Step 3.1.2.1.6

Raise

$- 8$ to the power of

$2$ .

$f (- 8) = \frac{1024}{3} - 1536 + 40 \cdot 64$

Step 3.1.2.1.7

Multiply

$40$ by

$64$ .

$f (- 8) = \frac{1024}{3} - 1536 + 2560$

Step 3.1.2.2

Find the common denominator.

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Step 3.1.2.2.1

Write

$- 1536$ as a fraction with denominator

$1$ .

$f (- 8) = \frac{1024}{3} + \frac{- 1536}{1} + 2560$

Step 3.1.2.2.2

Multiply

$\frac{- 1536}{1}$ by

$\frac{3}{3}$ .

$f (- 8) = \frac{1024}{3} + \frac{- 1536}{1} \cdot \frac{3}{3} + 2560$

Step 3.1.2.2.3

Multiply

$\frac{- 1536}{1}$ by

$\frac{3}{3}$ .

$f (- 8) = \frac{1024}{3} + \frac{- 1536 \cdot 3}{3} + 2560$

Step 3.1.2.2.4

Write

$2560$ as a fraction with denominator

$1$ .

$f (- 8) = \frac{1024}{3} + \frac{- 1536 \cdot 3}{3} + \frac{2560}{1}$

Step 3.1.2.2.5

Multiply

$\frac{2560}{1}$ by

$\frac{3}{3}$ .

$f (- 8) = \frac{1024}{3} + \frac{- 1536 \cdot 3}{3} + \frac{2560}{1} \cdot \frac{3}{3}$

Step 3.1.2.2.6

Multiply

$\frac{2560}{1}$ by

$\frac{3}{3}$ .

$f (- 8) = \frac{1024}{3} + \frac{- 1536 \cdot 3}{3} + \frac{2560 \cdot 3}{3}$

Step 3.1.2.3

Combine the numerators over the common denominator.

$f (- 8) = \frac{1024 - 1536 \cdot 3 + 2560 \cdot 3}{3}$

Step 3.1.2.4

Simplify each term.

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Step 3.1.2.4.1

Multiply

$- 1536$ by

$3$ .

$f (- 8) = \frac{1024 - 4608 + 2560 \cdot 3}{3}$

Step 3.1.2.4.2

Multiply

$2560$ by

$3$ .

$f (- 8) = \frac{1024 - 4608 + 7680}{3}$

Step 3.1.2.5

Simplify by adding and subtracting.

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Step 3.1.2.5.1

Subtract

$4608$ from

$1024$ .

$f (- 8) = \frac{- 3584 + 7680}{3}$

Step 3.1.2.5.2

Add

$- 3584$ and

$7680$ .

$f (- 8) = \frac{4096}{3}$

Step 3.1.2.6

The final answer is

$\frac{4096}{3}$ .

$\frac{4096}{3}$

Step 3.2

The point found by substituting

$- 8$ in

$f (x) = \frac{1}{12} x^{4} + 3 x^{3} + 40 x^{2}$ is

$(- 8, \frac{4096}{3})$ . This point can be an inflection point.

$(- 8, \frac{4096}{3})$

Step 3.3

Substitute

$- 10$ in

$f (x) = \frac{1}{12} x^{4} + 3 x^{3} + 40 x^{2}$ to find the value of

$y$ .

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Step 3.3.1

Replace the variable

$x$ with

$- 10$ in the expression.

$f (- 10) = \frac{1}{12} \cdot {(- 10)}^{4} + 3 {(- 10)}^{3} + 40 {(- 10)}^{2}$

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

Raise

$- 10$ to the power of

$4$ .

$f (- 10) = \frac{1}{12} \cdot 10000 + 3 {(- 10)}^{3} + 40 {(- 10)}^{2}$

Step 3.3.2.1.2

Cancel the common factor of

$4$ .

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Step 3.3.2.1.2.1

Factor

$4$ out of

$12$ .

$f (- 10) = \frac{1}{4 (3)} \cdot 10000 + 3 {(- 10)}^{3} + 40 {(- 10)}^{2}$

Step 3.3.2.1.2.2

Factor

$4$ out of

$10000$ .

$f (- 10) = \frac{1}{4 \cdot 3} \cdot (4 \cdot 2500) + 3 {(- 10)}^{3} + 40 {(- 10)}^{2}$

Step 3.3.2.1.2.3

Cancel the common factor.

$f (- 10) = \frac{1}{4 \cdot 3} \cdot (4 \cdot 2500) + 3 {(- 10)}^{3} + 40 {(- 10)}^{2}$

Step 3.3.2.1.2.4

Rewrite the expression.

$f (- 10) = \frac{1}{3} \cdot 2500 + 3 {(- 10)}^{3} + 40 {(- 10)}^{2}$

Step 3.3.2.1.3

Combine

$\frac{1}{3}$ and

$2500$ .

$f (- 10) = \frac{2500}{3} + 3 {(- 10)}^{3} + 40 {(- 10)}^{2}$

Step 3.3.2.1.4

Raise

$- 10$ to the power of

$3$ .

$f (- 10) = \frac{2500}{3} + 3 \cdot - 1000 + 40 {(- 10)}^{2}$

Step 3.3.2.1.5

Multiply

$3$ by

$- 1000$ .

$f (- 10) = \frac{2500}{3} - 3000 + 40 {(- 10)}^{2}$

Step 3.3.2.1.6

Raise

$- 10$ to the power of

$2$ .

$f (- 10) = \frac{2500}{3} - 3000 + 40 \cdot 100$

Step 3.3.2.1.7

Multiply

$40$ by

$100$ .

$f (- 10) = \frac{2500}{3} - 3000 + 4000$

Step 3.3.2.2

Find the common denominator.

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Step 3.3.2.2.1

Write

$- 3000$ as a fraction with denominator

$1$ .

$f (- 10) = \frac{2500}{3} + \frac{- 3000}{1} + 4000$

Step 3.3.2.2.2

Multiply

$\frac{- 3000}{1}$ by

$\frac{3}{3}$ .

$f (- 10) = \frac{2500}{3} + \frac{- 3000}{1} \cdot \frac{3}{3} + 4000$

Step 3.3.2.2.3

Multiply

$\frac{- 3000}{1}$ by

$\frac{3}{3}$ .

$f (- 10) = \frac{2500}{3} + \frac{- 3000 \cdot 3}{3} + 4000$

Step 3.3.2.2.4

Write

$4000$ as a fraction with denominator

$1$ .

$f (- 10) = \frac{2500}{3} + \frac{- 3000 \cdot 3}{3} + \frac{4000}{1}$

Step 3.3.2.2.5

Multiply

$\frac{4000}{1}$ by

$\frac{3}{3}$ .

$f (- 10) = \frac{2500}{3} + \frac{- 3000 \cdot 3}{3} + \frac{4000}{1} \cdot \frac{3}{3}$

Step 3.3.2.2.6

Multiply

$\frac{4000}{1}$ by

$\frac{3}{3}$ .

$f (- 10) = \frac{2500}{3} + \frac{- 3000 \cdot 3}{3} + \frac{4000 \cdot 3}{3}$

Step 3.3.2.3

Combine the numerators over the common denominator.

$f (- 10) = \frac{2500 - 3000 \cdot 3 + 4000 \cdot 3}{3}$

Step 3.3.2.4

Simplify each term.

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Step 3.3.2.4.1

Multiply

$- 3000$ by

$3$ .

$f (- 10) = \frac{2500 - 9000 + 4000 \cdot 3}{3}$

Step 3.3.2.4.2

Multiply

$4000$ by

$3$ .

$f (- 10) = \frac{2500 - 9000 + 12000}{3}$

Step 3.3.2.5

Simplify by adding and subtracting.

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Step 3.3.2.5.1

Subtract

$9000$ from

$2500$ .

$f (- 10) = \frac{- 6500 + 12000}{3}$

Step 3.3.2.5.2

Add

$- 6500$ and

$12000$ .

$f (- 10) = \frac{5500}{3}$

Step 3.3.2.6

The final answer is

$\frac{5500}{3}$ .

$\frac{5500}{3}$

Step 3.4

The point found by substituting

$- 10$ in

$f (x) = \frac{1}{12} x^{4} + 3 x^{3} + 40 x^{2}$ is

$(- 10, \frac{5500}{3})$ . This point can be an inflection point.

$(- 10, \frac{5500}{3})$

Step 3.5

Determine the points that could be inflection points.

$(- 8, \frac{4096}{3}), (- 10, \frac{5500}{3})$

Step 4

Split

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

$(- \infty, - 10) \cup (- 10, - 8) \cup (- 8, \infty)$

Step 5

Substitute a value from the interval

$(- \infty, - 10)$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable

$x$ with

$- 10.1$ in the expression.

$f'' (- 10.1) = {(- 10.1)}^{2} + 18 (- 10.1) + 80$

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

$- 10.1$ to the power of

$2$ .

$f'' (- 10.1) = 102.01 + 18 (- 10.1) + 80$

Step 5.2.1.2

Multiply

$18$ by

$- 10.1$ .

$f'' (- 10.1) = 102.01 - 181.8 + 80$

Step 5.2.2

Simplify by adding and subtracting.

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Step 5.2.2.1

Subtract

$181.8$ from

$102.01$ .

$f'' (- 10.1) = - 79.79 + 80$

Step 5.2.2.2

Add

$- 79.79$ and

$80$ .

$f'' (- 10.1) = 0.21$

Step 5.2.3

The final answer is

$0.21$ .

$0.21$

Step 5.3

$- 10.1$ , the second derivative is

$0.21$ . Since this is positive, the second derivative is increasing on the interval

$(- \infty, - 10)$ .

Increasing on

$(- \infty, - 10)$ since

$f'' (x) > 0$

Increasing on

$(- \infty, - 10)$ since

$f'' (x) > 0$

Step 6

Substitute a value from the interval

$(- 10, - 8)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable

$x$ with

$- 9$ in the expression.

$f'' (- 9) = {(- 9)}^{2} + 18 (- 9) + 80$

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

$- 9$ to the power of

$2$ .

$f'' (- 9) = 81 + 18 (- 9) + 80$

Step 6.2.1.2

Multiply

$18$ by

$- 9$ .

$f'' (- 9) = 81 - 162 + 80$

Step 6.2.2

Simplify by adding and subtracting.

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Step 6.2.2.1

Subtract

$162$ from

$81$ .

$f'' (- 9) = - 81 + 80$

Step 6.2.2.2

Add

$- 81$ and

$80$ .

$f'' (- 9) = - 1$

Step 6.2.3

The final answer is

$- 1$ .

$- 1$

Step 6.3

$- 9$ , the second derivative is

$- 1$ . Since this is negative, the second derivative is decreasing on the interval

$(- 10, - 8)$

Decreasing on

$(- 10, - 8)$ since

$f'' (x) < 0$

Decreasing on

$(- 10, - 8)$ since

$f'' (x) < 0$

Step 7

Substitute a value from the interval

$(- 8, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable

$x$ with

$- 7.9$ in the expression.

$f'' (- 7.9) = {(- 7.9)}^{2} + 18 (- 7.9) + 80$

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

$- 7.9$ to the power of

$2$ .

$f'' (- 7.9) = 62.41 + 18 (- 7.9) + 80$

Step 7.2.1.2

Multiply

$18$ by

$- 7.9$ .

$f'' (- 7.9) = 62.41 - 142.2 + 80$

Step 7.2.2

Simplify by adding and subtracting.

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Step 7.2.2.1

Subtract

$142.2$ from

$62.41$ .

$f'' (- 7.9) = - 79.79 + 80$

Step 7.2.2.2

Add

$- 79.79$ and

$80$ .

$f'' (- 7.9) = 0.21$

Step 7.2.3

The final answer is

$0.21$ .

$0.21$

Step 7.3

$- 7.9$ , the second derivative is

$0.21$ . Since this is positive, the second derivative is increasing on the interval

$(- 8, \infty)$ .

Increasing on

$(- 8, \infty)$ since

$f'' (x) > 0$

Increasing on

$(- 8, \infty)$ 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

$(- 10, \frac{5500}{3}), (- 8, \frac{4096}{3})$ .

$(- 10, \frac{5500}{3}), (- 8, \frac{4096}{3})$

Step 9