Calculus Examples

$y = 3 x^{4} + 4 x^{3}$

Step 1

Write $y = 3 x^{4} + 4 x^{3}$ as a function.

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

Step 2

Find the first derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

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

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

Step 2.1.2

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

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Step 2.1.2.1

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

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

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 $n = 4$ .

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

Step 2.1.2.3

Multiply $4$ by $3$ .

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

Step 2.1.3

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

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Step 2.1.3.1

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

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

Step 2.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f' (x) = 12 x^{3} + 4 (3 x^{2})$

Step 2.1.3.3

Multiply $3$ by $4$ .

$f' (x) = 12 x^{3} + 12 x^{2}$

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $12 x^{3} + 12 x^{2}$ .

$12 x^{3} + 12 x^{2}$

Step 3

Set the first derivative equal to $0$ then solve the equation $12 x^{3} + 12 x^{2} = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

$12 x^{3} + 12 x^{2} = 0$

Step 3.2

Factor $12 x^{2}$ out of $12 x^{3} + 12 x^{2}$ .

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Step 3.2.1

Factor $12 x^{2}$ out of $12 x^{3}$ .

$12 x^{2} (x) + 12 x^{2} = 0$

Step 3.2.2

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

$12 x^{2} (x) + 12 x^{2} (1) = 0$

Step 3.2.3

Factor $12 x^{2}$ out of $12 x^{2} (x) + 12 x^{2} (1)$ .

$12 x^{2} (x + 1) = 0$

Step 3.3

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

$x^{2} = 0$

$x + 1 = 0$

Step 3.4

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

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Step 3.4.1

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

$x^{2} = 0$

Step 3.4.2

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

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Step 3.4.2.1

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

$x = \pm \sqrt{0}$

Step 3.4.2.2

Simplify $\pm \sqrt{0}$ .

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Step 3.4.2.2.1

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

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

Step 3.4.2.2.2

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

$x = \pm 0$

Step 3.4.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 3.5

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 3.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.6

The final solution is all the values that make $12 x^{2} (x + 1) = 0$ true.

$x = 0, - 1$

Step 4

The values which make the derivative equal to $0$ are $0, - 1$ .

$0, - 1$

Step 5

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

$(- \infty, - 1) \cup (- 1, 0) \cup (0, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - 1)$ 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.

$f' (- 2) = 12 {(- 2)}^{3} + 12 {(- 2)}^{2}$

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 $3$ .

$f' (- 2) = 12 \cdot - 8 + 12 {(- 2)}^{2}$

Step 6.2.1.2

Multiply $12$ by $- 8$ .

$f' (- 2) = - 96 + 12 {(- 2)}^{2}$

Step 6.2.1.3

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

$f' (- 2) = - 96 + 12 \cdot 4$

Step 6.2.1.4

Multiply $12$ by $4$ .

$f' (- 2) = - 96 + 48$

Step 6.2.2

Add $- 96$ and $48$ .

$f' (- 2) = - 48$

Step 6.2.3

The final answer is $- 48$ .

$- 48$

Step 6.3

At $x = - 2$ the derivative is $- 48$ . Since this is negative, the function is decreasing on $(- \infty, - 1)$ .

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

Step 7

Substitute a value from the interval $(- 1, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $- \frac{1}{2}$ in the expression.

$f' (- \frac{1}{2}) = 12 {(- \frac{1}{2})}^{3} + 12 {(- \frac{1}{2})}^{2}$

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

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 7.2.1.1.1

Apply the product rule to $- \frac{1}{2}$ .

$f' (- \frac{1}{2}) = 12 ({(- 1)}^{3} {(\frac{1}{2})}^{3}) + 12 {(- \frac{1}{2})}^{2}$

Step 7.2.1.1.2

Apply the product rule to $\frac{1}{2}$ .

$f' (- \frac{1}{2}) = 12 ({(- 1)}^{3} (\frac{1^{3}}{2^{3}})) + 12 {(- \frac{1}{2})}^{2}$

Step 7.2.1.2

Raise $- 1$ to the power of $3$ .

$f' (- \frac{1}{2}) = 12 (- \frac{1^{3}}{2^{3}}) + 12 {(- \frac{1}{2})}^{2}$

Step 7.2.1.3

One to any power is one.

$f' (- \frac{1}{2}) = 12 (- \frac{1}{2^{3}}) + 12 {(- \frac{1}{2})}^{2}$

Step 7.2.1.4

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

$f' (- \frac{1}{2}) = 12 (- \frac{1}{8}) + 12 {(- \frac{1}{2})}^{2}$

Step 7.2.1.5

Cancel the common factor of $4$ .

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Step 7.2.1.5.1

Move the leading negative in $- \frac{1}{8}$ into the numerator.

$f' (- \frac{1}{2}) = 12 (\frac{- 1}{8}) + 12 {(- \frac{1}{2})}^{2}$

Step 7.2.1.5.2

Factor $4$ out of $12$ .

$f' (- \frac{1}{2}) = 4 (3) (\frac{- 1}{8}) + 12 {(- \frac{1}{2})}^{2}$

Step 7.2.1.5.3

Factor $4$ out of $8$ .

$f' (- \frac{1}{2}) = 4 \cdot (3 (\frac{- 1}{4 \cdot 2})) + 12 {(- \frac{1}{2})}^{2}$

Step 7.2.1.5.4

Cancel the common factor.

$f' (- \frac{1}{2}) = 4 \cdot (3 (\frac{- 1}{4 \cdot 2})) + 12 {(- \frac{1}{2})}^{2}$

Step 7.2.1.5.5

Rewrite the expression.

$f' (- \frac{1}{2}) = 3 (\frac{- 1}{2}) + 12 {(- \frac{1}{2})}^{2}$

Step 7.2.1.6

Combine $3$ and $\frac{- 1}{2}$ .

$f' (- \frac{1}{2}) = \frac{3 \cdot - 1}{2} + 12 {(- \frac{1}{2})}^{2}$

Step 7.2.1.7

Multiply $3$ by $- 1$ .

$f' (- \frac{1}{2}) = \frac{- 3}{2} + 12 {(- \frac{1}{2})}^{2}$

Step 7.2.1.8

Move the negative in front of the fraction.

$f' (- \frac{1}{2}) = - \frac{3}{2} + 12 {(- \frac{1}{2})}^{2}$

Step 7.2.1.9

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 7.2.1.9.1

Apply the product rule to $- \frac{1}{2}$ .

$f' (- \frac{1}{2}) = - \frac{3}{2} + 12 ({(- 1)}^{2} {(\frac{1}{2})}^{2})$

Step 7.2.1.9.2

Apply the product rule to $\frac{1}{2}$ .

$f' (- \frac{1}{2}) = - \frac{3}{2} + 12 ({(- 1)}^{2} (\frac{1^{2}}{2^{2}}))$

Step 7.2.1.10

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

$f' (- \frac{1}{2}) = - \frac{3}{2} + 12 (1 (\frac{1^{2}}{2^{2}}))$

Step 7.2.1.11

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

$f' (- \frac{1}{2}) = - \frac{3}{2} + 12 (\frac{1^{2}}{2^{2}})$

Step 7.2.1.12

One to any power is one.

$f' (- \frac{1}{2}) = - \frac{3}{2} + 12 (\frac{1}{2^{2}})$

Step 7.2.1.13

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

$f' (- \frac{1}{2}) = - \frac{3}{2} + 12 (\frac{1}{4})$

Step 7.2.1.14

Cancel the common factor of $4$ .

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Step 7.2.1.14.1

Factor $4$ out of $12$ .

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

Step 7.2.1.14.2

Cancel the common factor.

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

Step 7.2.1.14.3

Rewrite the expression.

$f' (- \frac{1}{2}) = - \frac{3}{2} + 3$

Step 7.2.2

To write $3$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 7.2.3

Combine $3$ and $\frac{2}{2}$ .

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

Step 7.2.4

Combine the numerators over the common denominator.

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

Step 7.2.5

Simplify the numerator.

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Step 7.2.5.1

Multiply $3$ by $2$ .

$f' (- \frac{1}{2}) = \frac{- 3 + 6}{2}$

Step 7.2.5.2

Add $- 3$ and $6$ .

$f' (- \frac{1}{2}) = \frac{3}{2}$

Step 7.2.6

The final answer is $\frac{3}{2}$ .

$\frac{3}{2}$

Step 7.3

At $x = - \frac{1}{2}$ the derivative is $\frac{3}{2}$ . Since this is positive, the function is increasing on $(- 1, 0)$ .

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

Step 8

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 8.1

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

$f' (1) = 12 {(1)}^{3} + 12 {(1)}^{2}$

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

One to any power is one.

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

Step 8.2.1.2

Multiply $12$ by $1$ .

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

Step 8.2.1.3

One to any power is one.

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

Step 8.2.1.4

Multiply $12$ by $1$ .

$f' (1) = 12 + 12$

Step 8.2.2

Add $12$ and $12$ .

$f' (1) = 24$

Step 8.2.3

The final answer is $24$ .

$24$

Step 8.3

At $x = 1$ the derivative is $24$ . Since this is positive, the function is increasing on $(0, \infty)$ .

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

Step 9

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

Increasing on: $(- 1, 0), (0, \infty)$

Decreasing on: $(- \infty, - 1)$

Step 10