Calculus Examples

$\frac{3}{10} x^{5} + x^{4} + x^{3}$

Step 1

Write $\frac{3}{10} x^{5} + x^{4} + x^{3}$ as a function.

$f (x) = \frac{3}{10} x^{5} + x^{4} + x^{3}$

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 $\frac{3}{10} x^{5} + x^{4} + x^{3}$ with respect to $x$ is $\frac{d}{d x} [\frac{3}{10} x^{5}] + \frac{d}{d x} [x^{4}] + \frac{d}{d x} [x^{3}]$ .

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

Step 2.1.2

Evaluate $\frac{d}{d x} [\frac{3}{10} x^{5}]$ .

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Step 2.1.2.1

Since $\frac{3}{10}$ is constant with respect to $x$ , the derivative of $\frac{3}{10} x^{5}$ with respect to $x$ is $\frac{3}{10} \frac{d}{d x} [x^{5}]$ .

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

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 $n = 5$ .

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

Step 2.1.2.3

Combine $5$ and $\frac{3}{10}$ .

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

Step 2.1.2.4

Multiply $5$ by $3$ .

$\frac{15}{10} x^{4} + \frac{d}{d x} [x^{4}] + \frac{d}{d x} [x^{3}]$

Step 2.1.2.5

Combine $\frac{15}{10}$ and $x^{4}$ .

$\frac{15 x^{4}}{10} + \frac{d}{d x} [x^{4}] + \frac{d}{d x} [x^{3}]$

Step 2.1.2.6

Cancel the common factor of $15$ and $10$ .

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Step 2.1.2.6.1

Factor $5$ out of $15 x^{4}$ .

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

Step 2.1.2.6.2

Cancel the common factors.

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Step 2.1.2.6.2.1

Factor $5$ out of $10$ .

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

Step 2.1.2.6.2.2

Cancel the common factor.

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

Step 2.1.2.6.2.3

Rewrite the expression.

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

Step 2.1.3

Differentiate using the Power Rule.

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Step 2.1.3.1

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

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

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 $n = 3$ .

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

Step 2.2

Find the second derivative.

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Step 2.2.1

By the Sum Rule, the derivative of $\frac{3 x^{4}}{2} + 4 x^{3} + 3 x^{2}$ with respect to $x$ is $\frac{d}{d x} [\frac{3 x^{4}}{2}] + \frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [3 x^{2}]$ .

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

Step 2.2.2

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

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Step 2.2.2.1

Since $\frac{3}{2}$ is constant with respect to $x$ , the derivative of $\frac{3 x^{4}}{2}$ with respect to $x$ is $\frac{3}{2} \frac{d}{d x} [x^{4}]$ .

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

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 $n = 4$ .

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

Step 2.2.2.3

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

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

Step 2.2.2.4

Multiply $4$ by $3$ .

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

Step 2.2.2.5

Combine $\frac{12}{2}$ and $x^{3}$ .

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

Step 2.2.2.6

Cancel the common factor of $12$ and $2$ .

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Step 2.2.2.6.1

Factor $2$ out of $12 x^{3}$ .

$\frac{2 (6 x^{3})}{2} + \frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [3 x^{2}]$

Step 2.2.2.6.2

Cancel the common factors.

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Step 2.2.2.6.2.1

Factor $2$ out of $2$ .

$\frac{2 (6 x^{3})}{2 (1)} + \frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [3 x^{2}]$

Step 2.2.2.6.2.2

Cancel the common factor.

$\frac{2 (6 x^{3})}{2 \cdot 1} + \frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [3 x^{2}]$

Step 2.2.2.6.2.3

Rewrite the expression.

$\frac{6 x^{3}}{1} + \frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [3 x^{2}]$

Step 2.2.2.6.2.4

Divide $6 x^{3}$ by $1$ .

$6 x^{3} + \frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [3 x^{2}]$

Step 2.2.3

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

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Step 2.2.3.1

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

$6 x^{3} + 4 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 x^{2}]$

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 $n = 3$ .

$6 x^{3} + 4 (3 x^{2}) + \frac{d}{d x} [3 x^{2}]$

Step 2.2.3.3

Multiply $3$ by $4$ .

$6 x^{3} + 12 x^{2} + \frac{d}{d x} [3 x^{2}]$

Step 2.2.4

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

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Step 2.2.4.1

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

$6 x^{3} + 12 x^{2} + 3 \frac{d}{d x} [x^{2}]$

Step 2.2.4.2

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

$6 x^{3} + 12 x^{2} + 3 (2 x)$

Step 2.2.4.3

Multiply $2$ by $3$ .

$f'' (x) = 6 x^{3} + 12 x^{2} + 6 x$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $6 x^{3} + 12 x^{2} + 6 x$ .

$6 x^{3} + 12 x^{2} + 6 x$

Step 3

Set the second derivative equal to $0$ then solve the equation $6 x^{3} + 12 x^{2} + 6 x = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$6 x^{3} + 12 x^{2} + 6 x = 0$

Step 3.2

Factor the left side of the equation.

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Step 3.2.1

Factor $6 x$ out of $6 x^{3} + 12 x^{2} + 6 x$ .

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Step 3.2.1.1

Factor $6 x$ out of $6 x^{3}$ .

$6 x (x^{2}) + 12 x^{2} + 6 x = 0$

Step 3.2.1.2

Factor $6 x$ out of $12 x^{2}$ .

$6 x (x^{2}) + 6 x (2 x) + 6 x = 0$

Step 3.2.1.3

Factor $6 x$ out of $6 x$ .

$6 x (x^{2}) + 6 x (2 x) + 6 x (1) = 0$

Step 3.2.1.4

Factor $6 x$ out of $6 x (x^{2}) + 6 x (2 x)$ .

$6 x (x^{2} + 2 x) + 6 x (1) = 0$

Step 3.2.1.5

Factor $6 x$ out of $6 x (x^{2} + 2 x) + 6 x (1)$ .

$6 x (x^{2} + 2 x + 1) = 0$

Step 3.2.2

Factor using the perfect square rule.

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Step 3.2.2.1

Rewrite $1$ as $1^{2}$ .

$6 x (x^{2} + 2 x + 1^{2}) = 0$

Step 3.2.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 3.2.2.3

Rewrite the polynomial.

$6 x (x^{2} + 2 \cdot x \cdot 1 + 1^{2}) = 0$

Step 3.2.2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

$6 x {(x + 1)}^{2} = 0$

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 $0$ .

$x = 0$

${(x + 1)}^{2} = 0$

Step 3.4

Set $x$ equal to $0$ .

$x = 0$

Step 3.5

Set ${(x + 1)}^{2}$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set ${(x + 1)}^{2}$ equal to $0$ .

${(x + 1)}^{2} = 0$

Step 3.5.2

Solve ${(x + 1)}^{2} = 0$ for $x$ .

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Step 3.5.2.1

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 3.5.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.6

The final solution is all the values that make $6 x {(x + 1)}^{2} = 0$ true.

$x = 0, - 1$

Step 4

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

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Step 4.1

Substitute $0$ in $f (x) = \frac{3}{10} x^{5} + x^{4} + x^{3}$ to find the value of $y$ .

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Step 4.1.1

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

$f (0) = \frac{3}{10} \cdot {(0)}^{5} + {(0)}^{4} + {(0)}^{3}$

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 $0$ .

$f (0) = \frac{3}{10} \cdot 0 + {(0)}^{4} + {(0)}^{3}$

Step 4.1.2.1.2

Multiply $\frac{3}{10}$ by $0$ .

$f (0) = 0 + {(0)}^{4} + {(0)}^{3}$

Step 4.1.2.1.3

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

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

Step 4.1.2.1.4

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

$f (0) = 0 + 0 + 0$

Step 4.1.2.2

Simplify by adding numbers.

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Step 4.1.2.2.1

Add $0$ and $0$ .

$f (0) = 0 + 0$

Step 4.1.2.2.2

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 $f (x) = \frac{3}{10} x^{5} + x^{4} + x^{3}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Step 4.3

Substitute $- 1$ in $f (x) = \frac{3}{10} x^{5} + x^{4} + x^{3}$ to find the value of $y$ .

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Step 4.3.1

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = \frac{3}{10} \cdot {(- 1)}^{5} + {(- 1)}^{4} + {(- 1)}^{3}$

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

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

Step 4.3.2.1.2

Multiply $\frac{3}{10} \cdot - 1$ .

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Step 4.3.2.1.2.1

Combine $\frac{3}{10}$ and $- 1$ .

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

Step 4.3.2.1.2.2

Multiply $3$ by $- 1$ .

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

Step 4.3.2.1.3

Move the negative in front of the fraction.

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

Step 4.3.2.1.4

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

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

Step 4.3.2.1.5

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

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

Step 4.3.2.2

Find the common denominator.

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Step 4.3.2.2.1

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

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

Step 4.3.2.2.2

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

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

Step 4.3.2.2.3

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

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

Step 4.3.2.2.4

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

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

Step 4.3.2.2.5

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

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

Step 4.3.2.2.6

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

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

Step 4.3.2.3

Combine the numerators over the common denominator.

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

Step 4.3.2.4

Simplify the expression.

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Step 4.3.2.4.1

Multiply $- 1$ by $10$ .

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

Step 4.3.2.4.2

Add $- 3$ and $10$ .

$f (- 1) = \frac{7 - 10}{10}$

Step 4.3.2.4.3

Subtract $10$ from $7$ .

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

Step 4.3.2.4.4

Move the negative in front of the fraction.

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

Step 4.3.2.5

The final answer is $- \frac{3}{10}$ .

$- \frac{3}{10}$

Step 4.4

The point found by substituting $- 1$ in $f (x) = \frac{3}{10} x^{5} + x^{4} + x^{3}$ is $(- 1, - \frac{3}{10})$ . This point can be an inflection point.

$(- 1, - \frac{3}{10})$

Step 4.5

Determine the points that could be inflection points.

$(0, 0), (- 1, - \frac{3}{10})$

Step 5

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

$(- \infty, - 1) \cup (- 1, 0) \cup (0, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - 1)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 1.1$ in the expression.

$f'' (- 1.1) = 6 {(- 1.1)}^{3} + 12 {(- 1.1)}^{2} + 6 (- 1.1)$

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 $- 1.1$ to the power of $3$ .

$f'' (- 1.1) = 6 \cdot - 1.331 + 12 {(- 1.1)}^{2} + 6 (- 1.1)$

Step 6.2.1.2

Multiply $6$ by $- 1.331$ .

$f'' (- 1.1) = - 7.986 + 12 {(- 1.1)}^{2} + 6 (- 1.1)$

Step 6.2.1.3

Raise $- 1.1$ to the power of $2$ .

$f'' (- 1.1) = - 7.986 + 12 \cdot 1.21 + 6 (- 1.1)$

Step 6.2.1.4

Multiply $12$ by $1.21$ .

$f'' (- 1.1) = - 7.986 + 14.52 + 6 (- 1.1)$

Step 6.2.1.5

Multiply $6$ by $- 1.1$ .

$f'' (- 1.1) = - 7.986 + 14.52 - 6.6$

Step 6.2.2

Simplify by adding and subtracting.

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Step 6.2.2.1

Add $- 7.986$ and $14.52$ .

$f'' (- 1.1) = 6.534 - 6.6$

Step 6.2.2.2

Subtract $6.6$ from $6.534$ .

$f'' (- 1.1) = - 0.066$

Step 6.2.3

The final answer is $- 0.066$ .

$- 0.066$

Step 6.3

At $- 1.1$ , the second derivative is $- 0.066$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, - 1)$

Decreasing on $(- \infty, - 1)$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(- 1, 0)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $- \frac{1}{2}$ in the expression.

$f'' (- \frac{1}{2}) = 6 {(- \frac{1}{2})}^{3} + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

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 $- \frac{1}{2}$ .

$f'' (- \frac{1}{2}) = 6 ({(- 1)}^{3} {(\frac{1}{2})}^{3}) + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

Step 7.2.1.1.2

Apply the product rule to $\frac{1}{2}$ .

$f'' (- \frac{1}{2}) = 6 ({(- 1)}^{3} (\frac{1^{3}}{2^{3}})) + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

Step 7.2.1.2

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

$f'' (- \frac{1}{2}) = 6 (- \frac{1^{3}}{2^{3}}) + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

Step 7.2.1.3

One to any power is one.

$f'' (- \frac{1}{2}) = 6 (- \frac{1}{2^{3}}) + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

Step 7.2.1.4

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

$f'' (- \frac{1}{2}) = 6 (- \frac{1}{8}) + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

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 $- \frac{1}{8}$ into the numerator.

$f'' (- \frac{1}{2}) = 6 (\frac{- 1}{8}) + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

Step 7.2.1.5.2

Factor $2$ out of $6$ .

$f'' (- \frac{1}{2}) = 2 (3) (\frac{- 1}{8}) + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

Step 7.2.1.5.3

Factor $2$ out of $8$ .

$f'' (- \frac{1}{2}) = 2 \cdot (3 (\frac{- 1}{2 \cdot 4})) + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

Step 7.2.1.5.4

Cancel the common factor.

$f'' (- \frac{1}{2}) = 2 \cdot (3 (\frac{- 1}{2 \cdot 4})) + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

Step 7.2.1.5.5

Rewrite the expression.

$f'' (- \frac{1}{2}) = 3 (\frac{- 1}{4}) + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

Step 7.2.1.6

Combine $3$ and $\frac{- 1}{4}$ .

$f'' (- \frac{1}{2}) = \frac{3 \cdot - 1}{4} + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

Step 7.2.1.7

Multiply $3$ by $- 1$ .

$f'' (- \frac{1}{2}) = \frac{- 3}{4} + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

Step 7.2.1.8

Move the negative in front of the fraction.

$f'' (- \frac{1}{2}) = - \frac{3}{4} + 12 {(- \frac{1}{2})}^{2} + 6 (- \frac{1}{2})$

Step 7.2.1.9

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

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Step 7.2.1.9.1

Apply the product rule to $- \frac{1}{2}$ .

$f'' (- \frac{1}{2}) = - \frac{3}{4} + 12 ({(- 1)}^{2} {(\frac{1}{2})}^{2}) + 6 (- \frac{1}{2})$

Step 7.2.1.9.2

Apply the product rule to $\frac{1}{2}$ .

$f'' (- \frac{1}{2}) = - \frac{3}{4} + 12 ({(- 1)}^{2} (\frac{1^{2}}{2^{2}})) + 6 (- \frac{1}{2})$

Step 7.2.1.10

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

$f'' (- \frac{1}{2}) = - \frac{3}{4} + 12 (1 (\frac{1^{2}}{2^{2}})) + 6 (- \frac{1}{2})$

Step 7.2.1.11

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

$f'' (- \frac{1}{2}) = - \frac{3}{4} + 12 (\frac{1^{2}}{2^{2}}) + 6 (- \frac{1}{2})$

Step 7.2.1.12

One to any power is one.

$f'' (- \frac{1}{2}) = - \frac{3}{4} + 12 (\frac{1}{2^{2}}) + 6 (- \frac{1}{2})$

Step 7.2.1.13

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

$f'' (- \frac{1}{2}) = - \frac{3}{4} + 12 (\frac{1}{4}) + 6 (- \frac{1}{2})$

Step 7.2.1.14

Cancel the common factor of $4$ .

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Step 7.2.1.14.1

Factor $4$ out of $12$ .

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

Step 7.2.1.14.2

Cancel the common factor.

$f'' (- \frac{1}{2}) = - \frac{3}{4} + 4 \cdot (3 (\frac{1}{4})) + 6 (- \frac{1}{2})$

Step 7.2.1.14.3

Rewrite the expression.

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

Step 7.2.1.15

Cancel the common factor of $2$ .

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Step 7.2.1.15.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Step 7.2.1.15.2

Factor $2$ out of $6$ .

$f'' (- \frac{1}{2}) = - \frac{3}{4} + 3 + 2 (3) (\frac{- 1}{2})$

Step 7.2.1.15.3

Cancel the common factor.

$f'' (- \frac{1}{2}) = - \frac{3}{4} + 3 + 2 \cdot (3 (\frac{- 1}{2}))$

Step 7.2.1.15.4

Rewrite the expression.

$f'' (- \frac{1}{2}) = - \frac{3}{4} + 3 + 3 \cdot - 1$

Step 7.2.1.16

Multiply $3$ by $- 1$ .

$f'' (- \frac{1}{2}) = - \frac{3}{4} + 3 - 3$

Step 7.2.2

Find the common denominator.

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Step 7.2.2.1

Write $3$ as a fraction with denominator $1$ .

$f'' (- \frac{1}{2}) = - \frac{3}{4} + \frac{3}{1} - 3$

Step 7.2.2.2

Multiply $\frac{3}{1}$ by $\frac{4}{4}$ .

$f'' (- \frac{1}{2}) = - \frac{3}{4} + \frac{3}{1} \cdot \frac{4}{4} - 3$

Step 7.2.2.3

Multiply $\frac{3}{1}$ by $\frac{4}{4}$ .

$f'' (- \frac{1}{2}) = - \frac{3}{4} + \frac{3 \cdot 4}{4} - 3$

Step 7.2.2.4

Write $- 3$ as a fraction with denominator $1$ .

$f'' (- \frac{1}{2}) = - \frac{3}{4} + \frac{3 \cdot 4}{4} + \frac{- 3}{1}$

Step 7.2.2.5

Multiply $\frac{- 3}{1}$ by $\frac{4}{4}$ .

$f'' (- \frac{1}{2}) = - \frac{3}{4} + \frac{3 \cdot 4}{4} + \frac{- 3}{1} \cdot \frac{4}{4}$

Step 7.2.2.6

Multiply $\frac{- 3}{1}$ by $\frac{4}{4}$ .

$f'' (- \frac{1}{2}) = - \frac{3}{4} + \frac{3 \cdot 4}{4} + \frac{- 3 \cdot 4}{4}$

Step 7.2.3

Combine the numerators over the common denominator.

$f'' (- \frac{1}{2}) = \frac{- 3 + 3 \cdot 4 - 3 \cdot 4}{4}$

Step 7.2.4

Simplify each term.

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Step 7.2.4.1

Multiply $3$ by $4$ .

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

Step 7.2.4.2

Multiply $- 3$ by $4$ .

$f'' (- \frac{1}{2}) = \frac{- 3 + 12 - 12}{4}$

Step 7.2.5

Simplify the expression.

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Step 7.2.5.1

Add $- 3$ and $12$ .

$f'' (- \frac{1}{2}) = \frac{9 - 12}{4}$

Step 7.2.5.2

Subtract $12$ from $9$ .

$f'' (- \frac{1}{2}) = \frac{- 3}{4}$

Step 7.2.5.3

Move the negative in front of the fraction.

$f'' (- \frac{1}{2}) = - \frac{3}{4}$

Step 7.2.6

The final answer is $- \frac{3}{4}$ .

$- \frac{3}{4}$

Step 7.3

At $- \frac{1}{2}$ , the second derivative is $- 0.75$ . Since this is negative, the second derivative is decreasing on the interval $(- 1, 0)$

Decreasing on $(- 1, 0)$ since $f'' (x) < 0$

Step 8

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

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Step 8.1

Replace the variable $x$ with $0.1$ in the expression.

$f'' (0.1) = 6 {(0.1)}^{3} + 12 {(0.1)}^{2} + 6 (0.1)$

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 $0.1$ to the power of $3$ .

$f'' (0.1) = 6 \cdot 0.001 + 12 {(0.1)}^{2} + 6 (0.1)$

Step 8.2.1.2

Multiply $6$ by $0.001$ .

$f'' (0.1) = 0.006 + 12 {(0.1)}^{2} + 6 (0.1)$

Step 8.2.1.3

Raise $0.1$ to the power of $2$ .

$f'' (0.1) = 0.006 + 12 \cdot 0.01 + 6 (0.1)$

Step 8.2.1.4

Multiply $12$ by $0.01$ .

$f'' (0.1) = 0.006 + 0.12 + 6 (0.1)$

Step 8.2.1.5

Multiply $6$ by $0.1$ .

$f'' (0.1) = 0.006 + 0.12 + 0.6$

Step 8.2.2

Simplify by adding numbers.

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Step 8.2.2.1

Add $0.006$ and $0.12$ .

$f'' (0.1) = 0.126 + 0.6$

Step 8.2.2.2

Add $0.126$ and $0.6$ .

$f'' (0.1) = 0.726$

Step 8.2.3

The final answer is $0.726$ .

$0.726$

Step 8.3

At $0.1$ , the second derivative is $0.726$ . Since this is positive, the second derivative is increasing on the interval $(0, \infty)$ .

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

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 point in this case is $(0, 0)$ .

$(0, 0)$

Step 10