Calculus Examples

$f (x) = 6 x^{4} + 32 x^{3}$

Step 1

Find the first 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 $6 x^{4} + 32 x^{3}$ with respect to $x$ is $\frac{d}{d x} [6 x^{4}] + \frac{d}{d x} [32 x^{3}]$ .

$f' (x) = \frac{d}{d x} (6 x^{4}) + \frac{d}{d x} (32 x^{3})$

Step 1.1.2

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

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Step 1.1.2.1

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

$f' (x) = 6 \frac{d}{d x} (x^{4}) + \frac{d}{d x} (32 x^{3})$

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 $n = 4$ .

$f' (x) = 6 (4 x^{3}) + \frac{d}{d x} (32 x^{3})$

Step 1.1.2.3

Multiply $4$ by $6$ .

$f' (x) = 24 x^{3} + \frac{d}{d x} (32 x^{3})$

Step 1.1.3

Evaluate $\frac{d}{d x} [32 x^{3}]$ .

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Step 1.1.3.1

Since $32$ is constant with respect to $x$ , the derivative of $32 x^{3}$ with respect to $x$ is $32 \frac{d}{d x} [x^{3}]$ .

$f' (x) = 24 x^{3} + 32 \frac{d}{d x} (x^{3})$

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 $n = 3$ .

$f' (x) = 24 x^{3} + 32 (3 x^{2})$

Step 1.1.3.3

Multiply $3$ by $32$ .

$f' (x) = 24 x^{3} + 96 x^{2}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $24 x^{3} + 96 x^{2}$ .

$24 x^{3} + 96 x^{2}$

Step 2

Set the first derivative equal to $0$ then solve the equation $24 x^{3} + 96 x^{2} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$24 x^{3} + 96 x^{2} = 0$

Step 2.2

Factor $24 x^{2}$ out of $24 x^{3} + 96 x^{2}$ .

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Step 2.2.1

Factor $24 x^{2}$ out of $24 x^{3}$ .

$24 x^{2} (x) + 96 x^{2} = 0$

Step 2.2.2

Factor $24 x^{2}$ out of $96 x^{2}$ .

$24 x^{2} (x) + 24 x^{2} (4) = 0$

Step 2.2.3

Factor $24 x^{2}$ out of $24 x^{2} (x) + 24 x^{2} (4)$ .

$24 x^{2} (x + 4) = 0$

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 $0$ .

$x^{2} = 0$

$x + 4 = 0$

Step 2.4

Set $x^{2}$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Step 2.4.2

Solve $x^{2} = 0$ for $x$ .

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Step 2.4.2.1

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

$x = \pm \sqrt{0}$

Step 2.4.2.2

Simplify $\pm \sqrt{0}$ .

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Step 2.4.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 2.4.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Step 2.4.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 2.5

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 2.5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 2.6

The final solution is all the values that make $24 x^{2} (x + 4) = 0$ true.

$x = 0, - 4$

Step 3

The values which make the derivative equal to $0$ are $0, - 4$ .

$0, - 4$

Step 4

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the derivative $0$ or undefined.

$(- \infty, - 4) \cup (- 4, 0) \cup (0, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 4)$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $- 5$ in the expression.

$f' (- 5) = 24 {(- 5)}^{3} + 96 {(- 5)}^{2}$

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 $- 5$ to the power of $3$ .

$f' (- 5) = 24 \cdot - 125 + 96 {(- 5)}^{2}$

Step 5.2.1.2

Multiply $24$ by $- 125$ .

$f' (- 5) = - 3000 + 96 {(- 5)}^{2}$

Step 5.2.1.3

Raise $- 5$ to the power of $2$ .

$f' (- 5) = - 3000 + 96 \cdot 25$

Step 5.2.1.4

Multiply $96$ by $25$ .

$f' (- 5) = - 3000 + 2400$

Step 5.2.2

Add $- 3000$ and $2400$ .

$f' (- 5) = - 600$

Step 5.2.3

The final answer is $- 600$ .

$- 600$

Step 5.3

At $x = - 5$ the derivative is $- 600$ . Since this is negative, the function is decreasing on $(- \infty, - 4)$ .

Decreasing on $(- \infty, - 4)$ since $f' (x) < 0$

Step 6

Substitute a value from the interval $(- 4, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

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

$f' (- 2) = 24 {(- 2)}^{3} + 96 {(- 2)}^{2}$

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

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

$f' (- 2) = 24 \cdot - 8 + 96 {(- 2)}^{2}$

Step 6.2.1.2

Multiply $24$ by $- 8$ .

$f' (- 2) = - 192 + 96 {(- 2)}^{2}$

Step 6.2.1.3

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

$f' (- 2) = - 192 + 96 \cdot 4$

Step 6.2.1.4

Multiply $96$ by $4$ .

$f' (- 2) = - 192 + 384$

Step 6.2.2

Add $- 192$ and $384$ .

$f' (- 2) = 192$

Step 6.2.3

The final answer is $192$ .

$192$

Step 6.3

At $x = - 2$ the derivative is $192$ . Since this is positive, the function is increasing on $(- 4, 0)$ .

Increasing on $(- 4, 0)$ since $f' (x) > 0$

Step 7

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

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Step 7.1

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

$f' (1) = 24 {(1)}^{3} + 96 {(1)}^{2}$

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

One to any power is one.

$f' (1) = 24 \cdot 1 + 96 {(1)}^{2}$

Step 7.2.1.2

Multiply $24$ by $1$ .

$f' (1) = 24 + 96 {(1)}^{2}$

Step 7.2.1.3

One to any power is one.

$f' (1) = 24 + 96 \cdot 1$

Step 7.2.1.4

Multiply $96$ by $1$ .

$f' (1) = 24 + 96$

Step 7.2.2

Add $24$ and $96$ .

$f' (1) = 120$

Step 7.2.3

The final answer is $120$ .

$120$

Step 7.3

At $x = 1$ the derivative is $120$ . Since this is positive, the function is increasing on $(0, \infty)$ .

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

Step 8

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- 4, 0), (0, \infty)$

Decreasing on: $(- \infty, - 4)$

Step 9