Calculus Examples

$f (x) = 20 x^{3} - 3 x^{5}$

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 $20 x^{3} - 3 x^{5}$ with respect to $x$ is $\frac{d}{d x} [20 x^{3}] + \frac{d}{d x} [- 3 x^{5}]$ .

$f' (x) = \frac{d}{d x} (20 x^{3}) + \frac{d}{d x} (- 3 x^{5})$

Step 1.1.2

Evaluate $\frac{d}{d x} [20 x^{3}]$ .

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Step 1.1.2.1

Since $20$ is constant with respect to $x$ , the derivative of $20 x^{3}$ with respect to $x$ is $20 \frac{d}{d x} [x^{3}]$ .

$f' (x) = 20 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 3 x^{5})$

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 $n = 3$ .

$f' (x) = 20 (3 x^{2}) + \frac{d}{d x} (- 3 x^{5})$

Step 1.1.2.3

Multiply $3$ by $20$ .

$f' (x) = 60 x^{2} + \frac{d}{d x} (- 3 x^{5})$

Step 1.1.3

Evaluate $\frac{d}{d x} [- 3 x^{5}]$ .

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Step 1.1.3.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x^{5}$ with respect to $x$ is $- 3 \frac{d}{d x} [x^{5}]$ .

$f' (x) = 60 x^{2} - 3 \frac{d}{d x} x^{5}$

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 $n = 5$ .

$f' (x) = 60 x^{2} - 3 (5 x^{4})$

Step 1.1.3.3

Multiply $5$ by $- 3$ .

$f' (x) = 60 x^{2} - 15 x^{4}$

Step 1.1.4

Reorder terms.

$f' (x) = - 15 x^{4} + 60 x^{2}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- 15 x^{4} + 60 x^{2}$ .

$- 15 x^{4} + 60 x^{2}$

Step 2

Set the first derivative equal to $0$ then solve the equation $- 15 x^{4} + 60 x^{2} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$- 15 x^{4} + 60 x^{2} = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$- 15 {(x^{2})}^{2} + 60 x^{2} = 0$

Step 2.2.2

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$- 15 u^{2} + 60 u = 0$

Step 2.2.3

Factor $15 u$ out of $- 15 u^{2} + 60 u$ .

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Step 2.2.3.1

Factor $15 u$ out of $- 15 u^{2}$ .

$15 u (- u) + 60 u = 0$

Step 2.2.3.2

Factor $15 u$ out of $60 u$ .

$15 u (- u) + 15 u (4) = 0$

Step 2.2.3.3

Factor $15 u$ out of $15 u (- u) + 15 u (4)$ .

$15 u (- u + 4) = 0$

Step 2.2.4

Replace all occurrences of $u$ with $x^{2}$ .

$15 x^{2} (- x^{2} + 4) = 0$

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 $0$ .

$x^{2} = 0$

$- x^{2} + 4 = 0$

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 $- x^{2} + 4$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $- x^{2} + 4$ equal to $0$ .

$- x^{2} + 4 = 0$

Step 2.5.2

Solve $- x^{2} + 4 = 0$ for $x$ .

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Step 2.5.2.1

Subtract $4$ from both sides of the equation.

$- x^{2} = - 4$

Step 2.5.2.2

Divide each term in $- x^{2} = - 4$ by $- 1$ and simplify.

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Step 2.5.2.2.1

Divide each term in $- x^{2} = - 4$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 4}{- 1}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 4}{- 1}$

Step 2.5.2.2.2.2

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

$x^{2} = \frac{- 4}{- 1}$

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Divide $- 4$ by $- 1$ .

$x^{2} = 4$

Step 2.5.2.3

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

$x = \pm \sqrt{4}$

Step 2.5.2.4

Simplify $\pm \sqrt{4}$ .

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Step 2.5.2.4.1

Rewrite $4$ as $2^{2}$ .

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

Step 2.5.2.4.2

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

$x = \pm 2$

Step 2.5.2.5

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

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Step 2.5.2.5.1

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

$x = 2$

Step 2.5.2.5.2

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

$x = - 2$

Step 2.5.2.5.3

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

$x = 2, - 2$

Step 2.6

The final solution is all the values that make $15 x^{2} (- x^{2} + 4) = 0$ true.

$x = 0, 2, - 2$

Step 3

The values which make the derivative equal to $0$ are $0, 2, - 2$ .

$0, 2, - 2$

Step 4

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

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

Step 5

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

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Step 5.1

Replace the variable $x$ with $- 3$ in the expression.

$f' (- 3) = - 15 {(- 3)}^{4} + 60 {(- 3)}^{2}$

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 $- 3$ to the power of $4$ .

$f' (- 3) = - 15 \cdot 81 + 60 {(- 3)}^{2}$

Step 5.2.1.2

Multiply $- 15$ by $81$ .

$f' (- 3) = - 1215 + 60 {(- 3)}^{2}$

Step 5.2.1.3

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

$f' (- 3) = - 1215 + 60 \cdot 9$

Step 5.2.1.4

Multiply $60$ by $9$ .

$f' (- 3) = - 1215 + 540$

Step 5.2.2

Add $- 1215$ and $540$ .

$f' (- 3) = - 675$

Step 5.2.3

The final answer is $- 675$ .

$- 675$

Step 5.3

At $x = - 3$ the derivative is $- 675$ . Since this is negative, the function is decreasing on $(- \infty, - 2)$ .

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

Step 6

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

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Step 6.1

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

$f' (- 1) = - 15 {(- 1)}^{4} + 60 {(- 1)}^{2}$

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

$f' (- 1) = - 15 \cdot 1 + 60 {(- 1)}^{2}$

Step 6.2.1.2

Multiply $- 15$ by $1$ .

$f' (- 1) = - 15 + 60 {(- 1)}^{2}$

Step 6.2.1.3

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

$f' (- 1) = - 15 + 60 \cdot 1$

Step 6.2.1.4

Multiply $60$ by $1$ .

$f' (- 1) = - 15 + 60$

Step 6.2.2

Add $- 15$ and $60$ .

$f' (- 1) = 45$

Step 6.2.3

The final answer is $45$ .

$45$

Step 6.3

At $x = - 1$ the derivative is $45$ . Since this is positive, the function is increasing on $(- 2, 0)$ .

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

Step 7

Substitute a value from the interval $(0, 2)$ 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.

$f' (1) = - 15 {(1)}^{4} + 60 {(1)}^{2}$

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.

$f' (1) = - 15 \cdot 1 + 60 {(1)}^{2}$

Step 7.2.1.2

Multiply $- 15$ by $1$ .

$f' (1) = - 15 + 60 {(1)}^{2}$

Step 7.2.1.3

One to any power is one.

$f' (1) = - 15 + 60 \cdot 1$

Step 7.2.1.4

Multiply $60$ by $1$ .

$f' (1) = - 15 + 60$

Step 7.2.2

Add $- 15$ and $60$ .

$f' (1) = 45$

Step 7.2.3

The final answer is $45$ .

$45$

Step 7.3

At $x = 1$ the derivative is $45$ . Since this is positive, the function is increasing on $(0, 2)$ .

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

Step 8

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

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Step 8.1

Replace the variable $x$ with $3$ in the expression.

$f' (3) = - 15 {(3)}^{4} + 60 {(3)}^{2}$

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 $3$ to the power of $4$ .

$f' (3) = - 15 \cdot 81 + 60 {(3)}^{2}$

Step 8.2.1.2

Multiply $- 15$ by $81$ .

$f' (3) = - 1215 + 60 {(3)}^{2}$

Step 8.2.1.3

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

$f' (3) = - 1215 + 60 \cdot 9$

Step 8.2.1.4

Multiply $60$ by $9$ .

$f' (3) = - 1215 + 540$

Step 8.2.2

Add $- 1215$ and $540$ .

$f' (3) = - 675$

Step 8.2.3

The final answer is $- 675$ .

$- 675$

Step 8.3

At $x = 3$ the derivative is $- 675$ . Since this is negative, the function is decreasing on $(2, \infty)$ .

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

Step 9

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

Increasing on: $(- 2, 0), (0, 2)$

Decreasing on: $(- \infty, - 2), (2, \infty)$

Step 10