# Find the critical values for f(x) = \frac{\sqrt[3]{x+1}}{3-4x}

## Question:

Find the critical values for {eq}f(x) = \frac{\sqrt[3]{x+1}}{3-4x} {/eq}

## Critical Values:

A part of finding the extreme values of a function is determining its critical values.

Values of {eq}x {/eq} fulfilling any of the following conditions are considered as critical values of {eq}f(x) {/eq}:

1) {eq}x {/eq} is in the domain of {eq}f(x) {/eq} such that {eq}f'(x) =0 {/eq}.

2) {eq}x {/eq} is in the domain of {eq}f(x) {/eq} but is not in the domain of {eq}f'(x) {/eq}.

## Answer and Explanation:

The derivative of {eq}\displaystyle f(x) = \frac{\sqrt[3]{x+1}}{3-4x} {/eq}, which we obtain via the application of the quotient rule, is:

{eq}\begin{align*} \displaystyle f(x)& = \frac{\sqrt[3]{x+1}}{3-4x}\\ f'(x)& =\frac{(3-4x) \frac{\mathrm{d}}{\mathrm{d}x}(\sqrt[3]{x+1})-(\sqrt[3]{x+1})\frac{\mathrm{d}}{\mathrm{d}x}(3-4x)}{(3-4x)^2}\\ f'(x)& =\frac{(3-4x) \left(\frac{1}{3}(x+1)^{-\frac{2}{3}}\right)-(\sqrt[3]{x+1})(-4)}{(3-4x)^2}\\ f'(x)& =\frac{8x+15}{3(3-4x)^2(x+1)^{\frac{2}{3}}}\\ \end{align*} {/eq}

If the denominator of {eq}f'(x) = \displaystyle \frac{8x+15}{3(3-4x)^2(x+1)^{\frac{2}{3}}} {/eq} is {eq}0 {/eq}, then {eq}f'(x) {/eq} does not exist.

Its denominator is zero if {eq}3-4x=0 \implies x = \displaystyle \frac{3}{4} {/eq} and {eq}x+1=0 \implies x=-1 {/eq}.

Of these two points, only {eq}x=-1 {/eq} is in the domain of {eq}f(x) {/eq}.

Thus, {eq}x=-1 {/eq} is one of the critical values of {eq}\displaystyle f(x) = \frac{\sqrt[3]{x+1}}{3-4x} {/eq}.

Attaining the other critical numbers is performed by setting {eq}f'(x) =0 {/eq}:

{eq}\begin{align*} \displaystyle f'(x)& =0 \\ \frac{8x+15}{3(3-4x)^2(x+1)^{\frac{2}{3}}}& = 0\\ 8x+15& =0 \\ x & =- \frac{15}{8}\\ \end{align*} {/eq}.

Hence, the critical values of {eq}\displaystyle f(x) = \frac{\sqrt[3]{x+1}}{3-4x} {/eq} are {eq}x=-1 {/eq} and {eq}\displaystyle x =- \frac{15}{8} {/eq}.

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from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9