Calculus Examples

$y = x^{4} - 6 x^{2}$

Step 1

Write $y = x^{4} - 6 x^{2}$ as a function.

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

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

Differentiate.

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Step 2.1.1.1

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

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

Step 2.1.1.2

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

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

Step 2.1.2

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

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Step 2.1.2.1

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

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

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 $n = 2$ .

$f' (x) = 4 x^{3} - 6 (2 x)$

Step 2.1.2.3

Multiply $2$ by $- 6$ .

$f' (x) = 4 x^{3} - 12 x$

Step 2.2

Find the second derivative.

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Step 2.2.1

By the Sum Rule, the derivative of $4 x^{3} - 12 x$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 12 x]$ .

$f'' (x) = \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (- 12 x)$

Step 2.2.2

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

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Step 2.2.2.1

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

$f'' (x) = 4 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 12 x)$

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 $n = 3$ .

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

Step 2.2.2.3

Multiply $3$ by $4$ .

$f'' (x) = 12 x^{2} + \frac{d}{d x} (- 12 x)$

Step 2.2.3

Evaluate $\frac{d}{d x} [- 12 x]$ .

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Step 2.2.3.1

Since $- 12$ is constant with respect to $x$ , the derivative of $- 12 x$ with respect to $x$ is $- 12 \frac{d}{d x} [x]$ .

$f'' (x) = 12 x^{2} - 12 \frac{d}{d x} x$

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 $n = 1$ .

$f'' (x) = 12 x^{2} - 12 \cdot 1$

Step 2.2.3.3

Multiply $- 12$ by $1$ .

$f'' (x) = 12 x^{2} - 12$

Step 2.3

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

$12 x^{2} - 12$

Step 3

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

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Step 3.1

Set the second derivative equal to $0$ .

$12 x^{2} - 12 = 0$

Step 3.2

Add $12$ to both sides of the equation.

$12 x^{2} = 12$

Step 3.3

Divide each term in $12 x^{2} = 12$ by $12$ and simplify.

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Step 3.3.1

Divide each term in $12 x^{2} = 12$ by $12$ .

$\frac{12 x^{2}}{12} = \frac{12}{12}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $12$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{12 x^{2}}{12} = \frac{12}{12}$

Step 3.3.2.1.2

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

$x^{2} = \frac{12}{12}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Divide $12$ by $12$ .

$x^{2} = 1$

Step 3.4

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

$x = \pm \sqrt{1}$

Step 3.5

Any root of $1$ is $1$ .

$x = \pm 1$

Step 3.6

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

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Step 3.6.1

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

$x = 1$

Step 3.6.2

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

$x = - 1$

Step 3.6.3

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

$x = 1, - 1$

Step 4

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

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Step 4.1

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

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Step 4.1.1

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

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

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

One to any power is one.

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

Step 4.1.2.1.2

One to any power is one.

$f (1) = 1 - 6 \cdot 1$

Step 4.1.2.1.3

Multiply $- 6$ by $1$ .

$f (1) = 1 - 6$

Step 4.1.2.2

Subtract $6$ from $1$ .

$f (1) = - 5$

Step 4.1.2.3

The final answer is $- 5$ .

$- 5$

Step 4.2

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

$(1, - 5)$

Step 4.3

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

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Step 4.3.1

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

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

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 $4$ .

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

Step 4.3.2.1.2

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

$f (- 1) = 1 - 6 \cdot 1$

Step 4.3.2.1.3

Multiply $- 6$ by $1$ .

$f (- 1) = 1 - 6$

Step 4.3.2.2

Subtract $6$ from $1$ .

$f (- 1) = - 5$

Step 4.3.2.3

The final answer is $- 5$ .

$- 5$

Step 4.4

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

$(- 1, - 5)$

Step 4.5

Determine the points that could be inflection points.

$(1, - 5), (- 1, - 5)$

Step 5

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

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

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.

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

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 $2$ .

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

Step 6.2.1.2

Multiply $12$ by $1.21$ .

$f'' (- 1.1) = 14.52 - 12$

Step 6.2.2

Subtract $12$ from $14.52$ .

$f'' (- 1.1) = 2.52$

Step 6.2.3

The final answer is $2.52$ .

$2.52$

Step 6.3

At $- 1.1$ , the second derivative is $2.52$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, - 1)$ .

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

Step 7

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 7.1

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

$f'' (0) = 12 {(0)}^{2} - 12$

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

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

$f'' (0) = 12 \cdot 0 - 12$

Step 7.2.1.2

Multiply $12$ by $0$ .

$f'' (0) = 0 - 12$

Step 7.2.2

Subtract $12$ from $0$ .

$f'' (0) = - 12$

Step 7.2.3

The final answer is $- 12$ .

$- 12$

Step 7.3

At $0$ , the second derivative is $- 12$ . Since this is negative, the second derivative is decreasing on the interval $(- 1, 1)$

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

Step 8

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 8.1

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

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

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 $1.1$ to the power of $2$ .

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

Step 8.2.1.2

Multiply $12$ by $1.21$ .

$f'' (1.1) = 14.52 - 12$

Step 8.2.2

Subtract $12$ from $14.52$ .

$f'' (1.1) = 2.52$

Step 8.2.3

The final answer is $2.52$ .

$2.52$

Step 8.3

At $1.1$ , the second derivative is $2.52$ . Since this is positive, the second derivative is increasing on the interval $(1, \infty)$ .

Increasing on $(1, \infty)$ 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 points in this case are $(- 1, - 5), (1, - 5)$ .

$(- 1, - 5), (1, - 5)$

Step 10