Calculus Examples

$3 {(x - 4)}^{\frac{2}{3}} + 6$

Step 1

Write $3 {(x - 4)}^{\frac{2}{3}} + 6$ as a function.

$f (x) = 3 {(x - 4)}^{\frac{2}{3}} + 6$

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 $3 {(x - 4)}^{\frac{2}{3}} + 6$ with respect to $x$ is $\frac{d}{d x} [3 {(x - 4)}^{\frac{2}{3}}] + \frac{d}{d x} [6]$ .

$f' (x) = \frac{d}{d x} (3 {(x - 4)}^{\frac{2}{3}}) + \frac{d}{d x} (6)$

Step 2.1.2

Evaluate $\frac{d}{d x} [3 {(x - 4)}^{\frac{2}{3}}]$ .

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Step 2.1.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 {(x - 4)}^{\frac{2}{3}}$ with respect to $x$ is $3 \frac{d}{d x} [{(x - 4)}^{\frac{2}{3}}]$ .

$f' (x) = 3 \frac{d}{d x} ({(x - 4)}^{\frac{2}{3}}) + \frac{d}{d x} (6)$

Step 2.1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{2}{3}}$ and $g (x) = x - 4$ .

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Step 2.1.2.2.1

To apply the Chain Rule, set $u$ as $x - 4$ .

$f' (x) = 3 (\frac{d}{d u} (u^{\frac{2}{3}}) \frac{d}{d x} (x - 4)) + \frac{d}{d x} (6)$

Step 2.1.2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{2}{3}$ .

$f' (x) = 3 (\frac{2}{3} u^{\frac{2}{3} - 1} \frac{d}{d x} (x - 4)) + \frac{d}{d x} (6)$

Step 2.1.2.2.3

Replace all occurrences of $u$ with $x - 4$ .

$f' (x) = 3 (\frac{2}{3} \cdot {(x - 4)}^{\frac{2}{3} - 1} \frac{d}{d x} (x - 4)) + \frac{d}{d x} (6)$

Step 2.1.2.3

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

$f' (x) = 3 (\frac{2}{3} \cdot ({(x - 4)}^{\frac{2}{3} - 1} (\frac{d}{d x} (x) + \frac{d}{d x} (- 4)))) + \frac{d}{d x} (6)$

Step 2.1.2.4

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

$f' (x) = 3 (\frac{2}{3} \cdot ({(x - 4)}^{\frac{2}{3} - 1} (1 + \frac{d}{d x} (- 4)))) + \frac{d}{d x} (6)$

Step 2.1.2.5

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$f' (x) = 3 (\frac{2}{3} \cdot ({(x - 4)}^{\frac{2}{3} - 1} (1 + 0))) + \frac{d}{d x} (6)$

Step 2.1.2.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f' (x) = 3 (\frac{2}{3} \cdot ({(x - 4)}^{\frac{2}{3} - 1 \cdot \frac{3}{3}} (1 + 0))) + \frac{d}{d x} (6)$

Step 2.1.2.7

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

$f' (x) = 3 (\frac{2}{3} \cdot ({(x - 4)}^{\frac{2}{3} + \frac{- 1 \cdot 3}{3}} (1 + 0))) + \frac{d}{d x} (6)$

Step 2.1.2.8

Combine the numerators over the common denominator.

$f' (x) = 3 (\frac{2}{3} \cdot ({(x - 4)}^{\frac{2 - 1 \cdot 3}{3}} (1 + 0))) + \frac{d}{d x} (6)$

Step 2.1.2.9

Simplify the numerator.

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Step 2.1.2.9.1

Multiply $- 1$ by $3$ .

$f' (x) = 3 (\frac{2}{3} \cdot ({(x - 4)}^{\frac{2 - 3}{3}} (1 + 0))) + \frac{d}{d x} (6)$

Step 2.1.2.9.2

Subtract $3$ from $2$ .

$f' (x) = 3 (\frac{2}{3} \cdot ({(x - 4)}^{\frac{- 1}{3}} (1 + 0))) + \frac{d}{d x} (6)$

Step 2.1.2.10

Move the negative in front of the fraction.

$f' (x) = 3 (\frac{2}{3} \cdot ({(x - 4)}^{- \frac{1}{3}} (1 + 0))) + \frac{d}{d x} (6)$

Step 2.1.2.11

Add $1$ and $0$ .

$f' (x) = 3 (\frac{2}{3} \cdot {(x - 4)}^{- \frac{1}{3}} \cdot 1) + \frac{d}{d x} (6)$

Step 2.1.2.12

Combine $\frac{2}{3}$ and ${(x - 4)}^{- \frac{1}{3}}$ .

$f' (x) = 3 (\frac{2 {(x - 4)}^{- \frac{1}{3}}}{3} \cdot 1) + \frac{d}{d x} (6)$

Step 2.1.2.13

Multiply $\frac{2 {(x - 4)}^{- \frac{1}{3}}}{3}$ by $1$ .

$f' (x) = 3 (\frac{2 {(x - 4)}^{- \frac{1}{3}}}{3}) + \frac{d}{d x} (6)$

Step 2.1.2.14

Move ${(x - 4)}^{- \frac{1}{3}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (x) = 3 (\frac{2}{3 {(x - 4)}^{\frac{1}{3}}}) + \frac{d}{d x} (6)$

Step 2.1.2.15

Combine $3$ and $\frac{2}{3 {(x - 4)}^{\frac{1}{3}}}$ .

$f' (x) = \frac{3 \cdot 2}{3 {(x - 4)}^{\frac{1}{3}}} + \frac{d}{d x} (6)$

Step 2.1.2.16

Multiply $3$ by $2$ .

$f' (x) = \frac{6}{3 {(x - 4)}^{\frac{1}{3}}} + \frac{d}{d x} (6)$

Step 2.1.2.17

Factor $3$ out of $6$ .

$f' (x) = \frac{3 \cdot 2}{3 {(x - 4)}^{\frac{1}{3}}} + \frac{d}{d x} (6)$

Step 2.1.2.18

Cancel the common factors.

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Step 2.1.2.18.1

Factor $3$ out of $3 {(x - 4)}^{\frac{1}{3}}$ .

$f' (x) = \frac{3 \cdot 2}{3 ({(x - 4)}^{\frac{1}{3}})} + \frac{d}{d x} (6)$

Step 2.1.2.18.2

Cancel the common factor.

$f' (x) = \frac{3 \cdot 2}{3 {(x - 4)}^{\frac{1}{3}}} + \frac{d}{d x} (6)$

Step 2.1.2.18.3

Rewrite the expression.

$f' (x) = \frac{2}{{(x - 4)}^{\frac{1}{3}}} + \frac{d}{d x} (6)$

Step 2.1.3

Differentiate using the Constant Rule.

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Step 2.1.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

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

Step 2.1.3.2

Add $\frac{2}{{(x - 4)}^{\frac{1}{3}}}$ and $0$ .

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

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{2}{{(x - 4)}^{\frac{1}{3}}}$ .

$\frac{2}{{(x - 4)}^{\frac{1}{3}}}$

Step 3

Set the first derivative equal to $0$ then solve the equation $\frac{2}{{(x - 4)}^{\frac{1}{3}}} = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

$\frac{2}{{(x - 4)}^{\frac{1}{3}}} = 0$

Step 3.2

Set the numerator equal to zero.

$2 = 0$

Step 3.3

Since $2 \neq 0$ , there are no solutions.

No solution

Step 4

There are no values of $x$ in the domain of the original problem where the derivative is $0$ or undefined.

No critical points found

Step 5

Find where the derivative is undefined.

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Step 5.1

Convert expressions with fractional exponents to radicals.

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Step 5.1.1

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\frac{2}{\sqrt[3]{{(x - 4)}^{1}}}$

Step 5.1.2

Anything raised to $1$ is the base itself.

$\frac{2}{\sqrt[3]{x - 4}}$

Step 5.2

Set the denominator in $\frac{2}{\sqrt[3]{x - 4}}$ equal to $0$ to find where the expression is undefined.

$\sqrt[3]{x - 4} = 0$

Step 5.3

Solve for $x$ .

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Step 5.3.1

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{x - 4}}^{3} = 0^{3}$

Step 5.3.2

Simplify each side of the equation.

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Step 5.3.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x - 4}$ as ${(x - 4)}^{\frac{1}{3}}$ .

${({(x - 4)}^{\frac{1}{3}})}^{3} = 0^{3}$

Step 5.3.2.2

Simplify the left side.

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Step 5.3.2.2.1

Simplify ${({(x - 4)}^{\frac{1}{3}})}^{3}$ .

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Step 5.3.2.2.1.1

Multiply the exponents in ${({(x - 4)}^{\frac{1}{3}})}^{3}$ .

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Step 5.3.2.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x - 4)}^{\frac{1}{3} \cdot 3} = 0^{3}$

Step 5.3.2.2.1.1.2

Cancel the common factor of $3$ .

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Step 5.3.2.2.1.1.2.1

Cancel the common factor.

${(x - 4)}^{\frac{1}{3} \cdot 3} = 0^{3}$

Step 5.3.2.2.1.1.2.2

Rewrite the expression.

${(x - 4)}^{1} = 0^{3}$

Step 5.3.2.2.1.2

Simplify.

$x - 4 = 0^{3}$

Step 5.3.2.3

Simplify the right side.

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Step 5.3.2.3.1

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

$x - 4 = 0$

Step 5.3.3

Add $4$ to both sides of the equation.

$x = 4$

Step 6

After finding the point that makes the derivative $f' (x) = \frac{2}{{(x - 4)}^{\frac{1}{3}}}$ equal to $0$ or undefined, the interval to check where $f (x) = 3 {(x - 4)}^{\frac{2}{3}} + 6$ is increasing and where it is decreasing is $(- \infty, 4) \cup (4, \infty)$ .

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

Step 7

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 7.1

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

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

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify the denominator.

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Step 7.2.1.1

Subtract $4$ from $3$ .

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

Step 7.2.1.2

Rewrite $- 1$ as ${(- 1)}^{3}$ .

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

Step 7.2.1.3

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 7.2.1.4

Cancel the common factor of $3$ .

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Step 7.2.1.4.1

Cancel the common factor.

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

Step 7.2.1.4.2

Rewrite the expression.

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

Step 7.2.1.5

Evaluate the exponent.

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

Step 7.2.2

Divide $2$ by $- 1$ .

$f' (3) = - 2$

Step 7.2.3

The final answer is $- 2$ .

$- 2$

Step 7.3

At $x = 3$ the derivative is $- 2$ . Since this is negative, the function is decreasing on $(- \infty, 4)$ .

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

Step 8

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

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Step 8.1

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

$f' (5) = \frac{2}{{((5) - 4)}^{\frac{1}{3}}}$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify the denominator.

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Step 8.2.1.1

Subtract $4$ from $5$ .

$f' (5) = \frac{2}{1^{\frac{1}{3}}}$

Step 8.2.1.2

One to any power is one.

$f' (5) = \frac{2}{1}$

Step 8.2.2

Divide $2$ by $1$ .

$f' (5) = 2$

Step 8.2.3

The final answer is $2$ .

$2$

Step 8.3

At $x = 5$ the derivative is $2$ . Since this is positive, the function is increasing on $(4, \infty)$ .

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

Step 9

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

Increasing on: $(4, \infty)$

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

Step 10