Calculus Examples

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

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 $- \frac{1}{6} x^{4} + x^{2}$ with respect to $x$ is $\frac{d}{d x} [- \frac{1}{6} x^{4}] + \frac{d}{d x} [x^{2}]$ .

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

Step 1.1.2

Evaluate $\frac{d}{d x} [- \frac{1}{6} x^{4}]$ .

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Step 1.1.2.1

Since $- \frac{1}{6}$ is constant with respect to $x$ , the derivative of $- \frac{1}{6} x^{4}$ with respect to $x$ is $- \frac{1}{6} \frac{d}{d x} [x^{4}]$ .

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

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 $n = 4$ .

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

Step 1.1.2.3

Multiply $4$ by $- 1$ .

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

Step 1.1.2.4

Combine $- 4$ and $\frac{1}{6}$ .

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

Step 1.1.2.5

Combine $\frac{- 4}{6}$ and $x^{3}$ .

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

Step 1.1.2.6

Cancel the common factor of $- 4$ and $6$ .

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Step 1.1.2.6.1

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

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

Step 1.1.2.6.2

Cancel the common factors.

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Step 1.1.2.6.2.1

Factor $2$ out of $6$ .

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

Step 1.1.2.6.2.2

Cancel the common factor.

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

Step 1.1.2.6.2.3

Rewrite the expression.

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

Step 1.1.2.7

Move the negative in front of the fraction.

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

Step 1.1.3

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

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

Step 1.2

Find the second derivative.

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Step 1.2.1

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

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

Step 1.2.2

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

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Step 1.2.2.1

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

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

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 $n = 3$ .

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

Step 1.2.2.3

Multiply $3$ by $- 1$ .

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

Step 1.2.2.4

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

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

Step 1.2.2.5

Multiply $- 3$ by $2$ .

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

Step 1.2.2.6

Combine $\frac{- 6}{3}$ and $x^{2}$ .

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

Step 1.2.2.7

Cancel the common factor of $- 6$ and $3$ .

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Step 1.2.2.7.1

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

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

Step 1.2.2.7.2

Cancel the common factors.

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Step 1.2.2.7.2.1

Factor $3$ out of $3$ .

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

Step 1.2.2.7.2.2

Cancel the common factor.

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

Step 1.2.2.7.2.3

Rewrite the expression.

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

Step 1.2.2.7.2.4

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

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

Step 1.2.3

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

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Step 1.2.3.1

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

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

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 $n = 1$ .

$- 2 x^{2} + 2 \cdot 1$

Step 1.2.3.3

Multiply $2$ by $1$ .

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

Step 1.3

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

$- 2 x^{2} + 2$

Step 2

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

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Step 2.1

Set the second derivative equal to $0$ .

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

Step 2.2

Subtract $2$ from both sides of the equation.

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $- 2$ .

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Step 2.3.2.1.1

Cancel the common factor.

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

Step 2.3.2.1.2

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

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

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $- 2$ by $- 2$ .

$x^{2} = 1$

Step 2.4

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

$x = \pm \sqrt{1}$

Step 2.5

Any root of $1$ is $1$ .

$x = \pm 1$

Step 2.6

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

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Step 2.6.1

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

$x = 1$

Step 2.6.2

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

$x = - 1$

Step 2.6.3

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

$x = 1, - 1$

Step 3

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

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Step 3.1

Substitute $1$ in $f (x) = - \frac{1}{6} x^{4} + x^{2}$ to find the value of $y$ .

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Step 3.1.1

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

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

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

One to any power is one.

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

Step 3.1.2.1.2

Multiply $- 1$ by $1$ .

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

Step 3.1.2.1.3

One to any power is one.

$f (1) = - \frac{1}{6} + 1$

Step 3.1.2.2

Simplify the expression.

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Step 3.1.2.2.1

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

$f (1) = - \frac{1}{6} + \frac{6}{6}$

Step 3.1.2.2.2

Combine the numerators over the common denominator.

$f (1) = \frac{- 1 + 6}{6}$

Step 3.1.2.2.3

Add $- 1$ and $6$ .

$f (1) = \frac{5}{6}$

Step 3.1.2.3

The final answer is $\frac{5}{6}$ .

$\frac{5}{6}$

Step 3.2

The point found by substituting $1$ in $f (x) = - \frac{1}{6} x^{4} + x^{2}$ is $(1, \frac{5}{6})$ . This point can be an inflection point.

$(1, \frac{5}{6})$

Step 3.3

Substitute $- 1$ in $f (x) = - \frac{1}{6} x^{4} + x^{2}$ to find the value of $y$ .

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Step 3.3.1

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

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

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

Multiply $- 1$ by ${(- 1)}^{4}$ by adding the exponents.

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Step 3.3.2.1.1.1

Move ${(- 1)}^{4}$ .

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

Step 3.3.2.1.1.2

Multiply ${(- 1)}^{4}$ by $- 1$ .

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Step 3.3.2.1.1.2.1

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

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

Step 3.3.2.1.1.2.2

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

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

Step 3.3.2.1.1.3

Add $4$ and $1$ .

$f (- 1) = {(- 1)}^{5} (\frac{1}{6}) + {(- 1)}^{2}$

Step 3.3.2.1.2

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

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

Step 3.3.2.1.3

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

$f (- 1) = - \frac{1}{6} + 1$

Step 3.3.2.2

Simplify the expression.

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Step 3.3.2.2.1

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

$f (- 1) = - \frac{1}{6} + \frac{6}{6}$

Step 3.3.2.2.2

Combine the numerators over the common denominator.

$f (- 1) = \frac{- 1 + 6}{6}$

Step 3.3.2.2.3

Add $- 1$ and $6$ .

$f (- 1) = \frac{5}{6}$

Step 3.3.2.3

The final answer is $\frac{5}{6}$ .

$\frac{5}{6}$

Step 3.4

The point found by substituting $- 1$ in $f (x) = - \frac{1}{6} x^{4} + x^{2}$ is $(- 1, \frac{5}{6})$ . This point can be an inflection point.

$(- 1, \frac{5}{6})$

Step 3.5

Determine the points that could be inflection points.

$(1, \frac{5}{6}), (- 1, \frac{5}{6})$

Step 4

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

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

Step 5

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 5.1

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

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

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

$f'' (- 1.1) = - 2 \cdot 1.21 + 2$

Step 5.2.1.2

Multiply $- 2$ by $1.21$ .

$f'' (- 1.1) = - 2.42 + 2$

Step 5.2.2

Add $- 2.42$ and $2$ .

$f'' (- 1.1) = - 0.42$

Step 5.2.3

The final answer is $- 0.42$ .

$- 0.42$

Step 5.3

At $- 1.1$ , the second derivative is $- 0.42$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, - 1)$

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

Step 6

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

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Step 6.1

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

$f'' (0) = - 2 {(0)}^{2} + 2$

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

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

$f'' (0) = - 2 \cdot 0 + 2$

Step 6.2.1.2

Multiply $- 2$ by $0$ .

$f'' (0) = 0 + 2$

Step 6.2.2

Add $0$ and $2$ .

$f'' (0) = 2$

Step 6.2.3

The final answer is $2$ .

$2$

Step 6.3

At $0$ , the second derivative is $2$ . Since this is positive, the second derivative is increasing on the interval $(- 1, 1)$ .

Increasing on $(- 1, 1)$ since $f'' (x) > 0$

Step 7

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

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Step 7.1

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

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

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

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

$f'' (1.1) = - 2 \cdot 1.21 + 2$

Step 7.2.1.2

Multiply $- 2$ by $1.21$ .

$f'' (1.1) = - 2.42 + 2$

Step 7.2.2

Add $- 2.42$ and $2$ .

$f'' (1.1) = - 0.42$

Step 7.2.3

The final answer is $- 0.42$ .

$- 0.42$

Step 7.3

At $1.1$ , the second derivative is $- 0.42$ . Since this is negative, the second derivative is decreasing on the interval $(1, \infty)$

Decreasing on $(1, \infty)$ 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 $(- 1, \frac{5}{6}), (1, \frac{5}{6})$ .

$(- 1, \frac{5}{6}), (1, \frac{5}{6})$

Step 9