# Find the critical Numbers a. h (x) = \frac{1}{3}x^3 - 4x b. g (x) = 5 x^2 - 20 x c. f (x) =...

## Question:

Find the critical Numbers

a. {eq}h (x) = \frac{1}{3}x^3 - 4x {/eq}

b. {eq}g (x) = 5 x^2 - 20 x {/eq}

c. {eq}f (x) = (x^2 - 9)^{2/3} {/eq}

d. {eq}y = x^2 -6x^2 {/eq}

e. {eq}V(x) = x^4 + \frac{1}{x^2} {/eq}

f. {eq}W(x) = x^3 - 3x^2 + 3x{/eq}

## Critical Numbers of a Function:

The critical numbers of a function are nothing but the the x-coordinates of the critical points which are obtained by equating

the derivative of the function to zero.Thus, if {eq}y=f(x) {/eq} be a function then the critical numbers are determined by solving {eq}f'(x)=0 {/eq} for {eq}x. {/eq}

## Answer and Explanation:

a. The given function:

{eq}h (x) = \frac{1}{3}x^{3} - 4x {/eq}

The derivative of {eq}h(x) {/eq} is:

{eq}\\\\\begin{align*} h'(x) & =...

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a. The given function:

{eq}h (x) = \frac{1}{3}x^{3} - 4x {/eq}

The derivative of {eq}h(x) {/eq} is:

{eq}\\\\\begin{align*} h'(x) & = \frac{1}{3}(3x^{2}) - 4 \ \ \ \ \texttt{(Using the power rule)} \\\\& = x^{2} - 4 \\\end{align*} {/eq}

For the critical number we put {eq}h'(x)=0 {/eq} to get:

{eq}\\\\ x^{2} - 4 = 0 \\\\\Rightarrow x^{2} = 4 \\\\\Rightarrow x = \pm 2 {/eq}

Hence, the critical numbers are {eq}x=-2 {/eq} and {eq}x = 2. {/eq}

b. The given function:

{eq}g (x) = 5 x^{2} - 20 x {/eq}

The derivative of {eq}g(x) {/eq} is:

{eq}g'(x) = 10x-20 \ \ \ \ \texttt{(Using the power rule)} {/eq}

For the critical number we put {eq}g'(x)=0 {/eq} to get:

{eq}\\\\10x-20 = 0 \\\\\Rightarrow 10x = 20 \\\\\Rightarrow x = 2 {/eq}

Hence, the critical numbers is {eq}x=2. {/eq}

c. The given function:

{eq}f (x) = (x^{2} - 9)^{2/3} {/eq}

The derivative of {eq}f(x) {/eq} is:

{eq}\\\\\begin{align*} f'(x) & = \frac{2}{3}(x^{2} - 9)^{-1/3}\ \frac{d}{dx}(x^{2} - 9) \ \ \ \ \texttt{(Using the power rule)} \\\\& = \frac{2}{3}(x^{2} - 9)^{-1/3}(2x) \\\\& = \frac{4x}{3}(x^{2} - 9)^{-1/3} \\\\& = \frac{4x}{3(x^{2} - 9)^{1/3} } \\\end{align*} {/eq}

For the critical number we put {eq}f'(x)=0 {/eq} to get:

{eq}\\\\ \frac{4x}{3(x^{2} - 9)^{1/3} }= 0 \\\\\Rightarrow x = 0 {/eq}

Hence, the critical numbers is {eq}x=0. {/eq}

d. The given function:

{eq}y = x^{2} -6 x^{2} {/eq}

The derivative of {eq}y {/eq} is:

{eq}\\\\\begin{align*} y'(x) & =2x-12x \ \ \ \ \texttt{(Using the power rule)} \\\\& = -10x \\\\\end{align*} {/eq}

For the critical number we put {eq}y'(x)=0 {/eq} to get:

{eq}\\\\ -10x = 0 \\\\\Rightarrow x = 0 {/eq}

Hence, the critical numbers is {eq}x=0. {/eq}

e. The given function:

{eq}V(x) = x^{4} + \frac{1}{x^{2}} {/eq}

The derivative of {eq}V(x) {/eq} is:

{eq}V'(x) = 4x^{3} - \frac{2}{x^{3}} \ \ \ \ \texttt{(Using the power rule)} {/eq}

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from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9