Calculus Examples

$f (x) = {(x + 4)}^{\frac{6}{7}}$

Step 1

Find the $x$ values where the second derivative is equal to $0$ .

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Step 1.1

Find the second derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{6}{7}}$ and $g (x) = x + 4$ .

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Step 1.1.1.1.1

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

$f' (x) = \frac{d}{d u} (u^{\frac{6}{7}}) \frac{d}{d x} (x + 4)$

Step 1.1.1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{6}{7}$ .

$f' (x) = \frac{6}{7} u^{\frac{6}{7} - 1} \frac{d}{d x} (x + 4)$

Step 1.1.1.1.3

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

$f' (x) = \frac{6}{7} \cdot {(x + 4)}^{\frac{6}{7} - 1} \frac{d}{d x} (x + 4)$

Step 1.1.1.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{7}{7}$ .

$f' (x) = \frac{6}{7} \cdot {(x + 4)}^{\frac{6}{7} - 1 \cdot \frac{7}{7}} \frac{d}{d x} (x + 4)$

Step 1.1.1.3

Combine $- 1$ and $\frac{7}{7}$ .

$f' (x) = \frac{6}{7} \cdot {(x + 4)}^{\frac{6}{7} + \frac{- 1 \cdot 7}{7}} \frac{d}{d x} (x + 4)$

Step 1.1.1.4

Combine the numerators over the common denominator.

$f' (x) = \frac{6}{7} \cdot {(x + 4)}^{\frac{6 - 1 \cdot 7}{7}} \frac{d}{d x} (x + 4)$

Step 1.1.1.5

Simplify the numerator.

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Step 1.1.1.5.1

Multiply $- 1$ by $7$ .

$f' (x) = \frac{6}{7} \cdot {(x + 4)}^{\frac{6 - 7}{7}} \frac{d}{d x} (x + 4)$

Step 1.1.1.5.2

Subtract $7$ from $6$ .

$f' (x) = \frac{6}{7} \cdot {(x + 4)}^{\frac{- 1}{7}} \frac{d}{d x} (x + 4)$

Step 1.1.1.6

Combine fractions.

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Step 1.1.1.6.1

Move the negative in front of the fraction.

$f' (x) = \frac{6}{7} \cdot {(x + 4)}^{- \frac{1}{7}} \frac{d}{d x} (x + 4)$

Step 1.1.1.6.2

Combine $\frac{6}{7}$ and ${(x + 4)}^{- \frac{1}{7}}$ .

$f' (x) = \frac{6 {(x + 4)}^{- \frac{1}{7}}}{7} \cdot \frac{d}{d x} (x + 4)$

Step 1.1.1.6.3

Move ${(x + 4)}^{- \frac{1}{7}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (x) = \frac{6}{7 {(x + 4)}^{\frac{1}{7}}} \cdot \frac{d}{d x} (x + 4)$

Step 1.1.1.7

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

$f' (x) = \frac{6}{7 {(x + 4)}^{\frac{1}{7}}} \cdot (\frac{d}{d x} (x) + \frac{d}{d x} (4))$

Step 1.1.1.8

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

$f' (x) = \frac{6}{7 {(x + 4)}^{\frac{1}{7}}} \cdot (1 + \frac{d}{d x} (4))$

Step 1.1.1.9

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

$f' (x) = \frac{6}{7 {(x + 4)}^{\frac{1}{7}}} \cdot (1 + 0)$

Step 1.1.1.10

Simplify the expression.

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Step 1.1.1.10.1

Add $1$ and $0$ .

$f' (x) = \frac{6}{7 {(x + 4)}^{\frac{1}{7}}} \cdot 1$

Step 1.1.1.10.2

Multiply $\frac{6}{7 {(x + 4)}^{\frac{1}{7}}}$ by $1$ .

$f' (x) = \frac{6}{7 {(x + 4)}^{\frac{1}{7}}}$

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

Differentiate using the Constant Multiple Rule.

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Step 1.1.2.1.1

Since $\frac{6}{7}$ is constant with respect to $x$ , the derivative of $\frac{6}{7 {(x + 4)}^{\frac{1}{7}}}$ with respect to $x$ is $\frac{6}{7} \frac{d}{d x} [\frac{1}{{(x + 4)}^{\frac{1}{7}}}]$ .

$f'' (x) = \frac{6}{7} \cdot \frac{d}{d x} (\frac{1}{{(x + 4)}^{\frac{1}{7}}})$

Step 1.1.2.1.2

Apply basic rules of exponents.

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Step 1.1.2.1.2.1

Rewrite $\frac{1}{{(x + 4)}^{\frac{1}{7}}}$ as ${({(x + 4)}^{\frac{1}{7}})}^{- 1}$ .

$f'' (x) = \frac{6}{7} \cdot \frac{d}{d x} ({({(x + 4)}^{\frac{1}{7}})}^{- 1})$

Step 1.1.2.1.2.2

Multiply the exponents in ${({(x + 4)}^{\frac{1}{7}})}^{- 1}$ .

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Step 1.1.2.1.2.2.1

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

$f'' (x) = \frac{6}{7} \cdot \frac{d}{d x} ({(x + 4)}^{\frac{1}{7} \cdot - 1})$

Step 1.1.2.1.2.2.2

Combine $\frac{1}{7}$ and $- 1$ .

$f'' (x) = \frac{6}{7} \cdot \frac{d}{d x} ({(x + 4)}^{\frac{- 1}{7}})$

Step 1.1.2.1.2.2.3

Move the negative in front of the fraction.

$f'' (x) = \frac{6}{7} \cdot \frac{d}{d x} ({(x + 4)}^{- \frac{1}{7}})$

Step 1.1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- \frac{1}{7}}$ and $g (x) = x + 4$ .

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Step 1.1.2.2.1

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

$f'' (x) = \frac{6}{7} \cdot (\frac{d}{d u} (u^{- \frac{1}{7}}) \frac{d}{d x} (x + 4))$

Step 1.1.2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - \frac{1}{7}$ .

$f'' (x) = \frac{6}{7} \cdot (- \frac{1}{7} u^{- \frac{1}{7} - 1} \frac{d}{d x} (x + 4))$

Step 1.1.2.2.3

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

$f'' (x) = \frac{6}{7} \cdot (- \frac{1}{7} \cdot {(x + 4)}^{- \frac{1}{7} - 1} \frac{d}{d x} (x + 4))$

Step 1.1.2.3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{7}{7}$ .

$f'' (x) = \frac{6}{7} \cdot (- \frac{1}{7} \cdot {(x + 4)}^{- \frac{1}{7} - 1 \cdot \frac{7}{7}} \frac{d}{d x} (x + 4))$

Step 1.1.2.4

Combine $- 1$ and $\frac{7}{7}$ .

$f'' (x) = \frac{6}{7} \cdot (- \frac{1}{7} \cdot {(x + 4)}^{- \frac{1}{7} + \frac{- 1 \cdot 7}{7}} \frac{d}{d x} (x + 4))$

Step 1.1.2.5

Combine the numerators over the common denominator.

$f'' (x) = \frac{6}{7} \cdot (- \frac{1}{7} \cdot {(x + 4)}^{\frac{- 1 - 1 \cdot 7}{7}} \frac{d}{d x} (x + 4))$

Step 1.1.2.6

Simplify the numerator.

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Step 1.1.2.6.1

Multiply $- 1$ by $7$ .

$f'' (x) = \frac{6}{7} \cdot (- \frac{1}{7} \cdot {(x + 4)}^{\frac{- 1 - 7}{7}} \frac{d}{d x} (x + 4))$

Step 1.1.2.6.2

Subtract $7$ from $- 1$ .

$f'' (x) = \frac{6}{7} \cdot (- \frac{1}{7} \cdot {(x + 4)}^{\frac{- 8}{7}} \frac{d}{d x} (x + 4))$

Step 1.1.2.7

Combine fractions.

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Step 1.1.2.7.1

Move the negative in front of the fraction.

$f'' (x) = \frac{6}{7} \cdot (- \frac{1}{7} \cdot {(x + 4)}^{- \frac{8}{7}} \frac{d}{d x} (x + 4))$

Step 1.1.2.7.2

Combine ${(x + 4)}^{- \frac{8}{7}}$ and $\frac{1}{7}$ .

$f'' (x) = \frac{6}{7} \cdot (- \frac{{(x + 4)}^{- \frac{8}{7}}}{7} \cdot \frac{d}{d x} (x + 4))$

Step 1.1.2.7.3

Move ${(x + 4)}^{- \frac{8}{7}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = \frac{6}{7} \cdot (- \frac{1}{7 {(x + 4)}^{\frac{8}{7}}} \cdot \frac{d}{d x} (x + 4))$

Step 1.1.2.7.4

Multiply $\frac{6}{7}$ by $\frac{1}{7 {(x + 4)}^{\frac{8}{7}}}$ .

$f'' (x) = - \frac{6}{7 (7 {(x + 4)}^{\frac{8}{7}})} \cdot \frac{d}{d x} (x + 4)$

Step 1.1.2.7.5

Multiply $7$ by $7$ .

$f'' (x) = - \frac{6}{49 {(x + 4)}^{\frac{8}{7}}} \cdot \frac{d}{d x} (x + 4)$

Step 1.1.2.8

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

$f'' (x) = - \frac{6}{49 {(x + 4)}^{\frac{8}{7}}} \cdot (\frac{d}{d x} (x) + \frac{d}{d x} (4))$

Step 1.1.2.9

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

$f'' (x) = - \frac{6}{49 {(x + 4)}^{\frac{8}{7}}} \cdot (1 + \frac{d}{d x} (4))$

Step 1.1.2.10

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

$f'' (x) = - \frac{6}{49 {(x + 4)}^{\frac{8}{7}}} \cdot (1 + 0)$

Step 1.1.2.11

Simplify the expression.

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Step 1.1.2.11.1

Add $1$ and $0$ .

$f'' (x) = - \frac{6}{49 {(x + 4)}^{\frac{8}{7}}} \cdot 1$

Step 1.1.2.11.2

Multiply $- 1$ by $1$ .

$f'' (x) = - \frac{6}{49 {(x + 4)}^{\frac{8}{7}}}$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $- \frac{6}{49 {(x + 4)}^{\frac{8}{7}}}$ .

$- \frac{6}{49 {(x + 4)}^{\frac{8}{7}}}$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $- \frac{6}{49 {(x + 4)}^{\frac{8}{7}}} = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$- \frac{6}{49 {(x + 4)}^{\frac{8}{7}}} = 0$

Step 1.2.2

Set the numerator equal to zero.

$6 = 0$

Step 1.2.3

Since $6 \neq 0$ , there are no solutions.

No solution

Step 2

Find the domain of $f (x) = {(x + 4)}^{\frac{6}{7}}$ .

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Step 2.1

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

$\sqrt[7]{{(x + 4)}^{6}}$

Step 2.2

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 3

The graph is concave down because the second derivative is negative.

The graph is concave down

Step 4