Find the critical values for f(x) = \frac{13xe^{-5x/6}}{6x+5}

Question:

Find the critical values for {eq}f(x) = \frac{13xe^{-5x/6}}{6x+5} {/eq}

Critical Values:

The critical values of a function correspond to {eq}x {/eq}-values such that the first derivative of {eq}f(x) {/eq} defined at these values is {eq}0 {/eq} or those values in the domain of {eq}f(x) {/eq} wherein the first derivative does not exist.

Such values are crucial as they can be utilized to determine the intervals of increase and decrease of a function and the extreme points of a function, among others.

Answer and Explanation:

First, we'll take the derivative of {eq}\displaystyle f(x) = \frac{13xe^{-\frac{5x}{6}}}{6x+5} {/eq} by utilizing the quotient rule given by the formula {eq}\displaystyle \left(\frac{b(x)}{c(x)}\right)'=\frac{c(x)b'(x)-b(x)c'(x)}{(c(x))^2} {/eq}:

{eq}\begin{align*} \displaystyle f(x) &= \frac{13xe^{-\frac{5x}{6}}}{6x+5}\\ f'(x) &= \frac{(6x+5)\frac{\mathrm{d}}{\mathrm{d}x}(13xe^{-\frac{5x}{6}})-(13xe^{-\frac{5x}{6}})\frac{\mathrm{d}}{\mathrm{d}x}(6x+5)}{(6x+5)^2} && \left[\mathrm{Quotient \ Rule }\right]\\ f'(x) &= \frac{(6x+5) \left(13e^{-\frac{5x}{6}}- \frac{65xe^{-\frac{5x}{6}}}{6}\right)-(13xe^{-\frac{5x}{6}}) (6 )}{(6x+5)^2}&& \left[\mathrm{ Differentiate \ 13xe^{-\frac{5x}{6}} \ by \ means \ of \ the \ product \ rule }\right]\\ f'(x) &= \frac{-390x^2e^{-\frac{5x}{6}}-325xe^{-\frac{5x}{6}}+390e^{-\frac{5x}{6}}}{6(6x+5)^2}\\ \end{align*} {/eq}

The derivative is not defined at {eq}6x+5=0 \implies x= -\displaystyle \frac{5}{6} {/eq} because the derivative will have a zero denominator at this value, which is not allowed.

But {eq}\displaystyle x= -\displaystyle \frac{5}{6} {/eq} is not in the domain of {eq}\displaystyle f(x) = \frac{13xe^{-\frac{5x}{6}}}{6x+5} {/eq} as it will have a zero denominator at this value.

Thus, {eq}\displaystyle x= -\displaystyle \frac{5}{6} {/eq} is not a critical value for the function {eq}\displaystyle f(x) = \frac{13xe^{-\frac{5x}{6}}}{6x+5} {/eq}.

We then set {eq}f'(x) {/eq} equivalent to {eq}0 {/eq} to find the critical values:

{eq}\begin{align*} \displaystyle f'(x)& =0 \\ \frac{-390x^2e^{-\frac{5x}{6}}-325xe^{-\frac{5x}{6}}+390e^{-\frac{5x}{6}}}{6(6x+5)^2}& =0 \\ -6e^{-\frac{5x}{6}}x^2-5e^{-\frac{5x}{6}}x+6e^{-\frac{5x}{6}}& =0 \\ -e^{-\frac{5x}{6}}(6x^2+5x-6) & = 0\\ e^{-\frac{5x}{6}}(3x-2)(2x+3) & = 0\\ \end{align*} {/eq}

Via the zero product property, we have that {eq}e^{-\frac{5x}{6}}=0 {/eq}, {eq}3x-2=0 {/eq} and {eq}2x+3=0 {/eq}.

The equation {eq}e^{-\frac{5x}{6}}=0 {/eq} has no solution as {eq}e^{-\frac{5x}{6}}>0 {/eq} for all {eq}x {/eq}.

Meanwhile, the solutions of {eq}3x-2=0 {/eq} and {eq}2x+3=0 {/eq} are {eq}x= \displaystyle \frac{2}{3} {/eq} and {eq}x= - \displaystyle \frac{3}{2} {/eq}, respectively.

Hence, the critical values for {eq}\displaystyle f(x) = \frac{13xe^{-\frac{5x}{6}}}{6x+5} {/eq} are {eq}x= \displaystyle \frac{2}{3} {/eq} and {eq}x= - \displaystyle \frac{3}{2} {/eq}.


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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
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