Copyright

Find the critical points of the function f(x) = (x-9)x^{1/3} and classify them.

Question:

Find the critical points of the function {eq}f(x) = (x-9)x^{1/3} {/eq} and classify them.

Critical Points

A critical point for a function {eq}\displaystyle y=f(x), {/eq} is the point {eq}\displaystyle c {/eq} from the domain of the function, such that {eq}\displaystyle f'(c)=0 \text{ or } f'(c) - DNE \text{ (does not exists)}. {/eq}

A critical point {eq}\displaystyle c {/eq} is maximum point if {eq}\displaystyle f'(x), {/eq} changes from positive to negative at {eq}\displaystyle c, {/eq} and is a minimum point if {eq}\displaystyle f'(x) {/eq} changes from negative to positive at {eq}\displaystyle c. {/eq}

The absolute extreme values are the maximum and minimum values of the function on the entire domain, if they exist.

Answer and Explanation:

To find the critical points of {eq}\displaystyle f(x)=(9-x)x^{1/3}, {/eq} we need first the domain of the fucntion and its derivative.

The domain is the entire real axis: {eq}\displaystyle (-\infty, \infty). {/eq}

The derivative function is calculated as

{eq}\displaystyle \begin{align*} f'(x)&=\frac{d}{dx}\left((x-9)x^{1/3}\right)\\ & =x^{1/3}\frac{d}{dx}(x-9)+(x-9)\frac{d}{dx}\left(x^{1/3}\right), \ \ \ \text{using Product Rule}\\ & =x^{1/3}+(x-9)\frac{1}{3}\cdot x^{-2/3}\\ & =\sqrt[3]{x}+(x-9)\frac{1}{3\sqrt[3]{x^2}}\\ & =\frac{3x+x-9}{3\sqrt[3]{x^2}}\\ & =\frac{4x-9}{3\sqrt[3]{x^2}}. \end{align*} {/eq}

Therefore, the critical points are given by

{eq}\displaystyle \begin{align*} & f'(x)=0 \text{ or } f'(x) - DNE\\ & 4x-9=0 \text{ or } x=0 \\ & x=\frac{9}{4} \text{ or } x=0 \\ &\boxed{\text{ the critical points are } x =0, x= \frac{9}{4}}. \end{align*} {/eq}

To classify the critical points, we look at the sign of the derivative, near them.

{eq}\displaystyle \begin{array}{c|ccccc} \text{critical points} & & 0 & & \frac{9}{4} &\\ \hline x & -1 & 0 & 1 & \frac{9}{4} & 3\\ \hline \text{ sings of } f'(x)& - & DNE & - & 0 & + \end{array} {/eq}

Looking at the row with the signs of {eq}\displaystyle f'(x), {/eq} the derivative changes signs, from negative to positive only at {eq}\displaystyle x=\frac{9}{4}\\ \implies \boxed{x=\frac{9}{4} \text{ is a local /absolute minimum}}, {/eq} while {eq}\displaystyle \boxed{x=0, \text{ is only a critical point}}. {/eq}


Learn more about this topic:

Loading...
Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
163K

Related to this Question

Explore our homework questions and answers library