Calculus Examples

$f (x) = - 0.5 \cdot x^{4} + 3 x^{2}$

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 $- 0.5 \cdot x^{4} + 3 x^{2}$ with respect to $x$ is $\frac{d}{d x} [- 0.5 \cdot x^{4}] + \frac{d}{d x} [3 x^{2}]$ .

$f' (x) = \frac{d}{d x} (- 0.5 \cdot x^{4}) + \frac{d}{d x} (3 x^{2})$

Step 1.1.2

Evaluate $\frac{d}{d x} [- 0.5 \cdot x^{4}]$ .

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Step 1.1.2.1

Since $- 0.5$ is constant with respect to $x$ , the derivative of $- 0.5 x^{4}$ with respect to $x$ is $- 0.5 \frac{d}{d x} [x^{4}]$ .

$f' (x) = - 0.5 \frac{d}{d x} x^{4} + \frac{d}{d x} (3 x^{2})$

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 $n = 4$ .

$f' (x) = - 0.5 (4 x^{3}) + \frac{d}{d x} (3 x^{2})$

Step 1.1.2.3

Multiply $4$ by $- 0.5$ .

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

Step 1.1.3

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

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Step 1.1.3.1

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

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

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 $n = 2$ .

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

Step 1.1.3.3

Multiply $2$ by $3$ .

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

Step 1.2

Find the second derivative.

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Step 1.2.1

By the Sum Rule, the derivative of $- 2 x^{3} + 6 x$ with respect to $x$ is $\frac{d}{d x} [- 2 x^{3}] + \frac{d}{d x} [6 x]$ .

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

Step 1.2.2

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

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Step 1.2.2.1

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

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

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 $n = 3$ .

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

Step 1.2.2.3

Multiply $3$ by $- 2$ .

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

Step 1.2.3

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

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Step 1.2.3.1

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

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

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 $n = 1$ .

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

Step 1.2.3.3

Multiply $6$ by $1$ .

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

Step 1.3

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

$- 6 x^{2} + 6$

Step 2

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

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Step 2.1

Set the second derivative equal to $0$ .

$- 6 x^{2} + 6 = 0$

Step 2.2

Subtract $6$ from both sides of the equation.

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

Step 2.3

Divide each term in $- 6 x^{2} = - 6$ by $- 6$ and simplify.

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Step 2.3.1

Divide each term in $- 6 x^{2} = - 6$ by $- 6$ .

$\frac{- 6 x^{2}}{- 6} = \frac{- 6}{- 6}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $- 6$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{- 6 x^{2}}{- 6} = \frac{- 6}{- 6}$

Step 2.3.2.1.2

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

$x^{2} = \frac{- 6}{- 6}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Divide $- 6$ by $- 6$ .

$x^{2} = 1$

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 $f (x) = - 0.5 \cdot x^{4} + 3 x^{2}$ to find the value of $y$ .

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Step 3.1.1

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

$f (1) = - 0.5 \cdot {(1)}^{4} + 3 {(1)}^{2}$

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.

$f (1) = - 0.5 \cdot 1 + 3 {(1)}^{2}$

Step 3.1.2.1.2

Multiply $- 0.5$ by $1$ .

$f (1) = - 0.5 + 3 {(1)}^{2}$

Step 3.1.2.1.3

One to any power is one.

$f (1) = - 0.5 + 3 \cdot 1$

Step 3.1.2.1.4

Multiply $3$ by $1$ .

$f (1) = - 0.5 + 3$

Step 3.1.2.2

Add $- 0.5$ and $3$ .

$f (1) = 2.5$

Step 3.1.2.3

The final answer is $2.5$ .

$2.5$

Step 3.2

The point found by substituting $1$ in $f (x) = - 0.5 \cdot x^{4} + 3 x^{2}$ is $(1, 2.5)$ . This point can be an inflection point.

$(1, 2.5)$

Step 3.3

Substitute $- 1$ in $f (x) = - 0.5 \cdot x^{4} + 3 x^{2}$ to find the value of $y$ .

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Step 3.3.1

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

$f (- 1) = - 0.5 \cdot {(- 1)}^{4} + 3 {(- 1)}^{2}$

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

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

$f (- 1) = - 0.5 \cdot 1 + 3 {(- 1)}^{2}$

Step 3.3.2.1.2

Multiply $- 0.5$ by $1$ .

$f (- 1) = - 0.5 + 3 {(- 1)}^{2}$

Step 3.3.2.1.3

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

$f (- 1) = - 0.5 + 3 \cdot 1$

Step 3.3.2.1.4

Multiply $3$ by $1$ .

$f (- 1) = - 0.5 + 3$

Step 3.3.2.2

Add $- 0.5$ and $3$ .

$f (- 1) = 2.5$

Step 3.3.2.3

The final answer is $2.5$ .

$2.5$

Step 3.4

The point found by substituting $- 1$ in $f (x) = - 0.5 \cdot x^{4} + 3 x^{2}$ is $(- 1, 2.5)$ . This point can be an inflection point.

$(- 1, 2.5)$

Step 3.5

Determine the points that could be inflection points.

$(1, 2.5), (- 1, 2.5)$

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.

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

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 $2$ .

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

Step 5.2.1.2

Multiply $- 6$ by $1.21$ .

$f'' (- 1.1) = - 7.26 + 6$

Step 5.2.2

Add $- 7.26$ and $6$ .

$f'' (- 1.1) = - 1.26$

Step 5.2.3

The final answer is $- 1.26$ .

$- 1.26$

Step 5.3

At $- 1.1$ , the second derivative is $- 1.26$ . 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.

$f'' (0) = - 6 {(0)}^{2} + 6$

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 $0$ .

$f'' (0) = - 6 \cdot 0 + 6$

Step 6.2.1.2

Multiply $- 6$ by $0$ .

$f'' (0) = 0 + 6$

Step 6.2.2

Add $0$ and $6$ .

$f'' (0) = 6$

Step 6.2.3

The final answer is $6$ .

$6$

Step 6.3

At $0$ , the second derivative is $6$ . 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.

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

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 $2$ .

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

Step 7.2.1.2

Multiply $- 6$ by $1.21$ .

$f'' (1.1) = - 7.26 + 6$

Step 7.2.2

Add $- 7.26$ and $6$ .

$f'' (1.1) = - 1.26$

Step 7.2.3

The final answer is $- 1.26$ .

$- 1.26$

Step 7.3

At $1.1$ , the second derivative is $- 1.26$ . 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 $(- 1, 2.5), (1, 2.5)$ .

$(- 1, 2.5), (1, 2.5)$

Step 9