Calculus Examples

$\frac{1}{10} x^{5} - 3 x^{3}$

Step 1

Write $\frac{1}{10} x^{5} - 3 x^{3}$ as a function.

$f (x) = \frac{1}{10} x^{5} - 3 x^{3}$

Step 2

Find the second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

By the Sum Rule, the derivative of $\frac{1}{10} x^{5} - 3 x^{3}$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{10} x^{5}] + \frac{d}{d x} [- 3 x^{3}]$ .

$\frac{d}{d x} [\frac{1}{10} x^{5}] + \frac{d}{d x} [- 3 x^{3}]$

Step 2.1.2

Evaluate $\frac{d}{d x} [\frac{1}{10} x^{5}]$ .

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Step 2.1.2.1

Since $\frac{1}{10}$ is constant with respect to $x$ , the derivative of $\frac{1}{10} x^{5}$ with respect to $x$ is $\frac{1}{10} \frac{d}{d x} [x^{5}]$ .

$\frac{1}{10} \frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 3 x^{3}]$

Step 2.1.2.2

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

$\frac{1}{10} (5 x^{4}) + \frac{d}{d x} [- 3 x^{3}]$

Step 2.1.2.3

Combine $5$ and $\frac{1}{10}$ .

$\frac{5}{10} x^{4} + \frac{d}{d x} [- 3 x^{3}]$

Step 2.1.2.4

Combine $\frac{5}{10}$ and $x^{4}$ .

$\frac{5 x^{4}}{10} + \frac{d}{d x} [- 3 x^{3}]$

Step 2.1.2.5

Cancel the common factor of $5$ and $10$ .

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Step 2.1.2.5.1

Factor $5$ out of $5 x^{4}$ .

$\frac{5 (x^{4})}{10} + \frac{d}{d x} [- 3 x^{3}]$

Step 2.1.2.5.2

Cancel the common factors.

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Step 2.1.2.5.2.1

Factor $5$ out of $10$ .

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

Step 2.1.2.5.2.2

Cancel the common factor.

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

Step 2.1.2.5.2.3

Rewrite the expression.

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

Step 2.1.3

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

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Step 2.1.3.1

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

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

Step 2.1.3.2

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

$\frac{x^{4}}{2} - 3 (3 x^{2})$

Step 2.1.3.3

Multiply $3$ by $- 3$ .

$f' (x) = \frac{x^{4}}{2} - 9 x^{2}$

Step 2.2

Find the second derivative.

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Step 2.2.1

By the Sum Rule, the derivative of $\frac{x^{4}}{2} - 9 x^{2}$ with respect to $x$ is $\frac{d}{d x} [\frac{x^{4}}{2}] + \frac{d}{d x} [- 9 x^{2}]$ .

$\frac{d}{d x} [\frac{x^{4}}{2}] + \frac{d}{d x} [- 9 x^{2}]$

Step 2.2.2

Evaluate $\frac{d}{d x} [\frac{x^{4}}{2}]$ .

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Step 2.2.2.1

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{x^{4}}{2}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [x^{4}]$ .

$\frac{1}{2} \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 9 x^{2}]$

Step 2.2.2.2

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

$\frac{1}{2} (4 x^{3}) + \frac{d}{d x} [- 9 x^{2}]$

Step 2.2.2.3

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

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

Step 2.2.2.4

Combine $\frac{4}{2}$ and $x^{3}$ .

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

Step 2.2.2.5

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

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Step 2.2.2.5.1

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

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

Step 2.2.2.5.2

Cancel the common factors.

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Step 2.2.2.5.2.1

Factor $2$ out of $2$ .

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

Step 2.2.2.5.2.2

Cancel the common factor.

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

Step 2.2.2.5.2.3

Rewrite the expression.

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

Step 2.2.2.5.2.4

Divide $2 x^{3}$ by $1$ .

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

Step 2.2.3

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

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Step 2.2.3.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x^{2}$ with respect to $x$ is $- 9 \frac{d}{d x} [x^{2}]$ .

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

Step 2.2.3.2

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

$2 x^{3} - 9 (2 x)$

Step 2.2.3.3

Multiply $2$ by $- 9$ .

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

Step 2.3

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

$2 x^{3} - 18 x$

Step 3

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

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Step 3.1

Set the second derivative equal to $0$ .

$2 x^{3} - 18 x = 0$

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

Factor $2 x$ out of $2 x^{3} - 18 x$ .

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Step 3.2.1.1

Factor $2 x$ out of $2 x^{3}$ .

$2 x (x^{2}) - 18 x = 0$

Step 3.2.1.2

Factor $2 x$ out of $- 18 x$ .

$2 x (x^{2}) + 2 x (- 9) = 0$

Step 3.2.1.3

Factor $2 x$ out of $2 x (x^{2}) + 2 x (- 9)$ .

$2 x (x^{2} - 9) = 0$

Step 3.2.2

Rewrite $9$ as $3^{2}$ .

$2 x (x^{2} - 3^{2}) = 0$

Step 3.2.3

Factor.

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Step 3.2.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$2 x ((x + 3) (x - 3)) = 0$

Step 3.2.3.2

Remove unnecessary parentheses.

$2 x (x + 3) (x - 3) = 0$

Step 3.3

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

$x = 0$

$x + 3 = 0$

$x - 3 = 0$

Step 3.4

Set $x$ equal to $0$ .

$x = 0$

Step 3.5

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 3.5.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 3.6

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 3.6.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 3.6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 3.7

The final solution is all the values that make $2 x (x + 3) (x - 3) = 0$ true.

$x = 0, - 3, 3$

Step 4

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

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Step 4.1

Substitute $0$ in $f (x) = \frac{1}{10} x^{5} - 3 x^{3}$ to find the value of $y$ .

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Step 4.1.1

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

$f (0) = \frac{1}{10} \cdot {(0)}^{5} - 3 {(0)}^{3}$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Simplify each term.

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Step 4.1.2.1.1

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

$f (0) = \frac{1}{10} \cdot 0 - 3 {(0)}^{3}$

Step 4.1.2.1.2

Multiply $\frac{1}{10}$ by $0$ .

$f (0) = 0 - 3 {(0)}^{3}$

Step 4.1.2.1.3

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

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

Step 4.1.2.1.4

Multiply $- 3$ by $0$ .

$f (0) = 0 + 0$

Step 4.1.2.2

Add $0$ and $0$ .

$f (0) = 0$

Step 4.1.2.3

The final answer is $0$ .

$0$

Step 4.2

The point found by substituting $0$ in $f (x) = \frac{1}{10} x^{5} - 3 x^{3}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Step 4.3

Substitute $- 3$ in $f (x) = \frac{1}{10} x^{5} - 3 x^{3}$ to find the value of $y$ .

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Step 4.3.1

Replace the variable $x$ with $- 3$ in the expression.

$f (- 3) = \frac{1}{10} \cdot {(- 3)}^{5} - 3 {(- 3)}^{3}$

Step 4.3.2

Simplify the result.

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Step 4.3.2.1

Simplify each term.

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Step 4.3.2.1.1

Raise $- 3$ to the power of $5$ .

$f (- 3) = \frac{1}{10} \cdot - 243 - 3 {(- 3)}^{3}$

Step 4.3.2.1.2

Combine $\frac{1}{10}$ and $- 243$ .

$f (- 3) = \frac{- 243}{10} - 3 {(- 3)}^{3}$

Step 4.3.2.1.3

Move the negative in front of the fraction.

$f (- 3) = - \frac{243}{10} - 3 {(- 3)}^{3}$

Step 4.3.2.1.4

Multiply $- 3$ by ${(- 3)}^{3}$ by adding the exponents.

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Step 4.3.2.1.4.1

Multiply $- 3$ by ${(- 3)}^{3}$ .

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Step 4.3.2.1.4.1.1

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

$f (- 3) = - \frac{243}{10} + (- 3) {(- 3)}^{3}$

Step 4.3.2.1.4.1.2

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

$f (- 3) = - \frac{243}{10} + {(- 3)}^{1 + 3}$

Step 4.3.2.1.4.2

Add $1$ and $3$ .

$f (- 3) = - \frac{243}{10} + {(- 3)}^{4}$

Step 4.3.2.1.5

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

$f (- 3) = - \frac{243}{10} + 81$

Step 4.3.2.2

To write $81$ as a fraction with a common denominator, multiply by $\frac{10}{10}$ .

$f (- 3) = - \frac{243}{10} + 81 \cdot \frac{10}{10}$

Step 4.3.2.3

Combine $81$ and $\frac{10}{10}$ .

$f (- 3) = - \frac{243}{10} + \frac{81 \cdot 10}{10}$

Step 4.3.2.4

Combine the numerators over the common denominator.

$f (- 3) = \frac{- 243 + 81 \cdot 10}{10}$

Step 4.3.2.5

Simplify the numerator.

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Step 4.3.2.5.1

Multiply $81$ by $10$ .

$f (- 3) = \frac{- 243 + 810}{10}$

Step 4.3.2.5.2

Add $- 243$ and $810$ .

$f (- 3) = \frac{567}{10}$

Step 4.3.2.6

The final answer is $\frac{567}{10}$ .

$\frac{567}{10}$

Step 4.4

The point found by substituting $- 3$ in $f (x) = \frac{1}{10} x^{5} - 3 x^{3}$ is $(- 3, \frac{567}{10})$ . This point can be an inflection point.

$(- 3, \frac{567}{10})$

Step 4.5

Substitute $3$ in $f (x) = \frac{1}{10} x^{5} - 3 x^{3}$ to find the value of $y$ .

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Step 4.5.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = \frac{1}{10} \cdot {(3)}^{5} - 3 {(3)}^{3}$

Step 4.5.2

Simplify the result.

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Step 4.5.2.1

Simplify each term.

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Step 4.5.2.1.1

Raise $3$ to the power of $5$ .

$f (3) = \frac{1}{10} \cdot 243 - 3 {(3)}^{3}$

Step 4.5.2.1.2

Combine $\frac{1}{10}$ and $243$ .

$f (3) = \frac{243}{10} - 3 {(3)}^{3}$

Step 4.5.2.1.3

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

$f (3) = \frac{243}{10} - 3 \cdot 27$

Step 4.5.2.1.4

Multiply $- 3$ by $27$ .

$f (3) = \frac{243}{10} - 81$

Step 4.5.2.2

To write $- 81$ as a fraction with a common denominator, multiply by $\frac{10}{10}$ .

$f (3) = \frac{243}{10} - 81 \cdot \frac{10}{10}$

Step 4.5.2.3

Combine $- 81$ and $\frac{10}{10}$ .

$f (3) = \frac{243}{10} + \frac{- 81 \cdot 10}{10}$

Step 4.5.2.4

Combine the numerators over the common denominator.

$f (3) = \frac{243 - 81 \cdot 10}{10}$

Step 4.5.2.5

Simplify the numerator.

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Step 4.5.2.5.1

Multiply $- 81$ by $10$ .

$f (3) = \frac{243 - 810}{10}$

Step 4.5.2.5.2

Subtract $810$ from $243$ .

$f (3) = \frac{- 567}{10}$

Step 4.5.2.6

Move the negative in front of the fraction.

$f (3) = - \frac{567}{10}$

Step 4.5.2.7

The final answer is $- \frac{567}{10}$ .

$- \frac{567}{10}$

Step 4.6

The point found by substituting $3$ in $f (x) = \frac{1}{10} x^{5} - 3 x^{3}$ is $(3, - \frac{567}{10})$ . This point can be an inflection point.

$(3, - \frac{567}{10})$

Step 4.7

Determine the points that could be inflection points.

$(0, 0), (- 3, \frac{567}{10}), (3, - \frac{567}{10})$

Step 5

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

$(- \infty, - 3) \cup (- 3, 0) \cup (0, 3) \cup (3, \infty)$

Step 6

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

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Step 6.1

Replace the variable $x$ with $- 3.1$ in the expression.

$f'' (- 3.1) = 2 {(- 3.1)}^{3} - 18 \cdot - 3.1$

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 $- 3.1$ to the power of $3$ .

$f'' (- 3.1) = 2 \cdot - 29.791 - 18 \cdot - 3.1$

Step 6.2.1.2

Multiply $2$ by $- 29.791$ .

$f'' (- 3.1) = - 59.582 - 18 \cdot - 3.1$

Step 6.2.1.3

Multiply $- 18$ by $- 3.1$ .

$f'' (- 3.1) = - 59.582 + 55.8$

Step 6.2.2

Add $- 59.582$ and $55.8$ .

$f'' (- 3.1) = - 3.782$

Step 6.2.3

The final answer is $- 3.782$ .

$- 3.782$

Step 6.3

At $- 3.1$ , the second derivative is $- 3.782$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, - 3)$

Decreasing on $(- \infty, - 3)$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(- 3, 0)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

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

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

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

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

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Step 7.2.1.1.1

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

$f'' (- \frac{3}{2}) = 2 ({(- 1)}^{3} {(\frac{3}{2})}^{3}) - 18 (- \frac{3}{2})$

Step 7.2.1.1.2

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

$f'' (- \frac{3}{2}) = 2 ({(- 1)}^{3} (\frac{3^{3}}{2^{3}})) - 18 (- \frac{3}{2})$

Step 7.2.1.2

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

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

Step 7.2.1.3

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

$f'' (- \frac{3}{2}) = 2 (- \frac{27}{2^{3}}) - 18 (- \frac{3}{2})$

Step 7.2.1.4

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

$f'' (- \frac{3}{2}) = 2 (- \frac{27}{8}) - 18 (- \frac{3}{2})$

Step 7.2.1.5

Cancel the common factor of $2$ .

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Step 7.2.1.5.1

Move the leading negative in $- \frac{27}{8}$ into the numerator.

$f'' (- \frac{3}{2}) = 2 (\frac{- 27}{8}) - 18 (- \frac{3}{2})$

Step 7.2.1.5.2

Factor $2$ out of $8$ .

$f'' (- \frac{3}{2}) = 2 (\frac{- 27}{2 (4)}) - 18 (- \frac{3}{2})$

Step 7.2.1.5.3

Cancel the common factor.

$f'' (- \frac{3}{2}) = 2 (\frac{- 27}{2 \cdot 4}) - 18 (- \frac{3}{2})$

Step 7.2.1.5.4

Rewrite the expression.

$f'' (- \frac{3}{2}) = \frac{- 27}{4} - 18 (- \frac{3}{2})$

Step 7.2.1.6

Move the negative in front of the fraction.

$f'' (- \frac{3}{2}) = - \frac{27}{4} - 18 (- \frac{3}{2})$

Step 7.2.1.7

Cancel the common factor of $2$ .

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Step 7.2.1.7.1

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$f'' (- \frac{3}{2}) = - \frac{27}{4} - 18 (\frac{- 3}{2})$

Step 7.2.1.7.2

Factor $2$ out of $- 18$ .

$f'' (- \frac{3}{2}) = - \frac{27}{4} + 2 (- 9) (\frac{- 3}{2})$

Step 7.2.1.7.3

Cancel the common factor.

$f'' (- \frac{3}{2}) = - \frac{27}{4} + 2 \cdot (- 9 (\frac{- 3}{2}))$

Step 7.2.1.7.4

Rewrite the expression.

$f'' (- \frac{3}{2}) = - \frac{27}{4} - 9 \cdot - 3$

Step 7.2.1.8

Multiply $- 9$ by $- 3$ .

$f'' (- \frac{3}{2}) = - \frac{27}{4} + 27$

Step 7.2.2

To write $27$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f'' (- \frac{3}{2}) = - \frac{27}{4} + 27 \cdot \frac{4}{4}$

Step 7.2.3

Combine $27$ and $\frac{4}{4}$ .

$f'' (- \frac{3}{2}) = - \frac{27}{4} + \frac{27 \cdot 4}{4}$

Step 7.2.4

Combine the numerators over the common denominator.

$f'' (- \frac{3}{2}) = \frac{- 27 + 27 \cdot 4}{4}$

Step 7.2.5

Simplify the numerator.

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Step 7.2.5.1

Multiply $27$ by $4$ .

$f'' (- \frac{3}{2}) = \frac{- 27 + 108}{4}$

Step 7.2.5.2

Add $- 27$ and $108$ .

$f'' (- \frac{3}{2}) = \frac{81}{4}$

Step 7.2.6

The final answer is $\frac{81}{4}$ .

$\frac{81}{4}$

Step 7.3

At $- \frac{3}{2}$ , the second derivative is $20.25$ . Since this is positive, the second derivative is increasing on the interval $(- 3, 0)$ .

Increasing on $(- 3, 0)$ since $f'' (x) > 0$

Step 8

Substitute a value from the interval $(0, 3)$ into the second derivative to determine if it is increasing or decreasing.

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Step 8.1

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

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

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

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

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

Step 8.2.1.2

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

$f'' (\frac{3}{2}) = 2 (\frac{27}{2^{3}}) - 18 (\frac{3}{2})$

Step 8.2.1.3

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

$f'' (\frac{3}{2}) = 2 (\frac{27}{8}) - 18 (\frac{3}{2})$

Step 8.2.1.4

Cancel the common factor of $2$ .

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Step 8.2.1.4.1

Factor $2$ out of $8$ .

$f'' (\frac{3}{2}) = 2 (\frac{27}{2 (4)}) - 18 (\frac{3}{2})$

Step 8.2.1.4.2

Cancel the common factor.

$f'' (\frac{3}{2}) = 2 (\frac{27}{2 \cdot 4}) - 18 (\frac{3}{2})$

Step 8.2.1.4.3

Rewrite the expression.

$f'' (\frac{3}{2}) = \frac{27}{4} - 18 (\frac{3}{2})$

Step 8.2.1.5

Cancel the common factor of $2$ .

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Step 8.2.1.5.1

Factor $2$ out of $- 18$ .

$f'' (\frac{3}{2}) = \frac{27}{4} + 2 (- 9) (\frac{3}{2})$

Step 8.2.1.5.2

Cancel the common factor.

$f'' (\frac{3}{2}) = \frac{27}{4} + 2 \cdot (- 9 (\frac{3}{2}))$

Step 8.2.1.5.3

Rewrite the expression.

$f'' (\frac{3}{2}) = \frac{27}{4} - 9 \cdot 3$

Step 8.2.1.6

Multiply $- 9$ by $3$ .

$f'' (\frac{3}{2}) = \frac{27}{4} - 27$

Step 8.2.2

To write $- 27$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f'' (\frac{3}{2}) = \frac{27}{4} - 27 \cdot \frac{4}{4}$

Step 8.2.3

Combine $- 27$ and $\frac{4}{4}$ .

$f'' (\frac{3}{2}) = \frac{27}{4} + \frac{- 27 \cdot 4}{4}$

Step 8.2.4

Combine the numerators over the common denominator.

$f'' (\frac{3}{2}) = \frac{27 - 27 \cdot 4}{4}$

Step 8.2.5

Simplify the numerator.

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Step 8.2.5.1

Multiply $- 27$ by $4$ .

$f'' (\frac{3}{2}) = \frac{27 - 108}{4}$

Step 8.2.5.2

Subtract $108$ from $27$ .

$f'' (\frac{3}{2}) = \frac{- 81}{4}$

Step 8.2.6

Move the negative in front of the fraction.

$f'' (\frac{3}{2}) = - \frac{81}{4}$

Step 8.2.7

The final answer is $- \frac{81}{4}$ .

$- \frac{81}{4}$

Step 8.3

At $\frac{3}{2}$ , the second derivative is $- 20.25$ . Since this is negative, the second derivative is decreasing on the interval $(0, 3)$

Decreasing on $(0, 3)$ since $f'' (x) < 0$

Step 9

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

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Step 9.1

Replace the variable $x$ with $3.1$ in the expression.

$f'' (3.1) = 2 {(3.1)}^{3} - 18 \cdot 3.1$

Step 9.2

Simplify the result.

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Step 9.2.1

Simplify each term.

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Step 9.2.1.1

Raise $3.1$ to the power of $3$ .

$f'' (3.1) = 2 \cdot 29.791 - 18 \cdot 3.1$

Step 9.2.1.2

Multiply $2$ by $29.791$ .

$f'' (3.1) = 59.582 - 18 \cdot 3.1$

Step 9.2.1.3

Multiply $- 18$ by $3.1$ .

$f'' (3.1) = 59.582 - 55.8$

Step 9.2.2

Subtract $55.8$ from $59.582$ .

$f'' (3.1) = 3.782$

Step 9.2.3

The final answer is $3.782$ .

$3.782$

Step 9.3

At $3.1$ , the second derivative is $3.782$ . Since this is positive, the second derivative is increasing on the interval $(3, \infty)$ .

Increasing on $(3, \infty)$ since $f'' (x) > 0$

Step 10

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 $(- 3, \frac{567}{10}), (0, 0), (3, - \frac{567}{10})$ .

$(- 3, \frac{567}{10}), (0, 0), (3, - \frac{567}{10})$

Step 11