Calculus Examples

$f (x) = \frac{1}{2} x^{4} + 2 x^{3}$

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 $\frac{1}{2} x^{4} + 2 x^{3}$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{2} x^{4}] + \frac{d}{d x} [2 x^{3}]$ .

$f' (x) = \frac{d}{d x} (\frac{1}{2} x^{4}) + \frac{d}{d x} (2 x^{3})$

Step 1.1.2

Evaluate $\frac{d}{d x} [\frac{1}{2} x^{4}]$ .

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Step 1.1.2.1

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

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

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 $n = 4$ .

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

Step 1.1.2.3

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

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

Step 1.1.2.4

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

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

Step 1.1.2.5

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

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Step 1.1.2.5.1

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

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

Step 1.1.2.5.2

Cancel the common factors.

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Step 1.1.2.5.2.1

Factor $2$ out of $2$ .

$f' (x) = \frac{2 (2 x^{3})}{2 (1)} + \frac{d}{d x} (2 x^{3})$

Step 1.1.2.5.2.2

Cancel the common factor.

$f' (x) = \frac{2 (2 x^{3})}{2 \cdot 1} + \frac{d}{d x} (2 x^{3})$

Step 1.1.2.5.2.3

Rewrite the expression.

$f' (x) = \frac{2 x^{3}}{1} + \frac{d}{d x} (2 x^{3})$

Step 1.1.2.5.2.4

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

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

Step 1.1.3

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

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Step 1.1.3.1

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

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

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 $n = 3$ .

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

Step 1.1.3.3

Multiply $3$ by $2$ .

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

Step 1.2

Find the second derivative.

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Step 1.2.1

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

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

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

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

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 $n = 3$ .

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

Step 1.2.2.3

Multiply $3$ by $2$ .

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

Step 1.2.3

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

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Step 1.2.3.1

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

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

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 $n = 2$ .

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

Step 1.2.3.3

Multiply $2$ by $6$ .

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

Step 1.3

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

$6 x^{2} + 12 x$

Step 2

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

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Step 2.1

Set the second derivative equal to $0$ .

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

Step 2.2

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

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Step 2.2.1

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

$6 x (x) + 12 x = 0$

Step 2.2.2

Factor $6 x$ out of $12 x$ .

$6 x (x) + 6 x (2) = 0$

Step 2.2.3

Factor $6 x$ out of $6 x (x) + 6 x (2)$ .

$6 x (x + 2) = 0$

Step 2.3

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

$x = 0$

$x + 2 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 2.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.6

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

$x = 0, - 2$

Step 3

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

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Step 3.1

Substitute $0$ in $f (x) = \frac{1}{2} x^{4} + 2 x^{3}$ to find the value of $y$ .

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Step 3.1.1

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

$f (0) = \frac{1}{2} \cdot {(0)}^{4} + 2 {(0)}^{3}$

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

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

$f (0) = \frac{1}{2} \cdot 0 + 2 {(0)}^{3}$

Step 3.1.2.1.2

Multiply $\frac{1}{2}$ by $0$ .

$f (0) = 0 + 2 {(0)}^{3}$

Step 3.1.2.1.3

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

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

Step 3.1.2.1.4

Multiply $2$ by $0$ .

$f (0) = 0 + 0$

Step 3.1.2.2

Add $0$ and $0$ .

$f (0) = 0$

Step 3.1.2.3

The final answer is $0$ .

$0$

Step 3.2

The point found by substituting $0$ in $f (x) = \frac{1}{2} x^{4} + 2 x^{3}$ is $(0, 0)$ . This point can be an inflection point.

$(0, 0)$

Step 3.3

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

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Step 3.3.1

Replace the variable $x$ with $- 2$ in the expression.

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

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 $- 2$ to the power of $4$ .

$f (- 2) = \frac{1}{2} \cdot 16 + 2 {(- 2)}^{3}$

Step 3.3.2.1.2

Cancel the common factor of $2$ .

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Step 3.3.2.1.2.1

Factor $2$ out of $16$ .

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

Step 3.3.2.1.2.2

Cancel the common factor.

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

Step 3.3.2.1.2.3

Rewrite the expression.

$f (- 2) = 8 + 2 {(- 2)}^{3}$

Step 3.3.2.1.3

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

$f (- 2) = 8 + 2 \cdot - 8$

Step 3.3.2.1.4

Multiply $2$ by $- 8$ .

$f (- 2) = 8 - 16$

Step 3.3.2.2

Subtract $16$ from $8$ .

$f (- 2) = - 8$

Step 3.3.2.3

The final answer is $- 8$ .

$- 8$

Step 3.4

The point found by substituting $- 2$ in $f (x) = \frac{1}{2} x^{4} + 2 x^{3}$ is $(- 2, - 8)$ . This point can be an inflection point.

$(- 2, - 8)$

Step 3.5

Determine the points that could be inflection points.

$(0, 0), (- 2, - 8)$

Step 4

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

$(- \infty, - 2) \cup (- 2, 0) \cup (0, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 2)$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 2.1$ in the expression.

$f'' (- 2.1) = 6 {(- 2.1)}^{2} + 12 (- 2.1)$

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 $- 2.1$ to the power of $2$ .

$f'' (- 2.1) = 6 \cdot 4.41 + 12 (- 2.1)$

Step 5.2.1.2

Multiply $6$ by $4.41$ .

$f'' (- 2.1) = 26.46 + 12 (- 2.1)$

Step 5.2.1.3

Multiply $12$ by $- 2.1$ .

$f'' (- 2.1) = 26.46 - 25.2$

Step 5.2.2

Subtract $25.2$ from $26.46$ .

$f'' (- 2.1) = 1.26$

Step 5.2.3

The final answer is $1.26$ .

$1.26$

Step 5.3

At $- 2.1$ , the second derivative is $1.26$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, - 2)$ .

Increasing on $(- \infty, - 2)$ since $f'' (x) > 0$

Step 6

Substitute a value from the interval $(- 2, 0)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

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

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

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

$f'' (- 1) = 6 \cdot 1 + 12 (- 1)$

Step 6.2.1.2

Multiply $6$ by $1$ .

$f'' (- 1) = 6 + 12 (- 1)$

Step 6.2.1.3

Multiply $12$ by $- 1$ .

$f'' (- 1) = 6 - 12$

Step 6.2.2

Subtract $12$ from $6$ .

$f'' (- 1) = - 6$

Step 6.2.3

The final answer is $- 6$ .

$- 6$

Step 6.3

At $- 1$ , the second derivative is $- 6$ . Since this is negative, the second derivative is decreasing on the interval $(- 2, 0)$

Decreasing on $(- 2, 0)$ since $f'' (x) < 0$

Step 7

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 7.1

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

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

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 $0.1$ to the power of $2$ .

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

Step 7.2.1.2

Multiply $6$ by $0.01$ .

$f'' (0.1) = 0.06 + 12 (0.1)$

Step 7.2.1.3

Multiply $12$ by $0.1$ .

$f'' (0.1) = 0.06 + 1.2$

Step 7.2.2

Add $0.06$ and $1.2$ .

$f'' (0.1) = 1.26$

Step 7.2.3

The final answer is $1.26$ .

$1.26$

Step 7.3

At $0.1$ , the second derivative is $1.26$ . Since this is positive, the second derivative is increasing on the interval $(0, \infty)$ .

Increasing on $(0, \infty)$ 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 $(- 2, - 8), (0, 0)$ .

$(- 2, - 8), (0, 0)$

Step 9