What are the critical numbers of the function: f(x) = frac{x^2}{x^2 - 4}

Question:

What are the critical numbers of the function: {eq}f(x) = \frac{x^2}{x^2 - 4} {/eq}

Critical Points:

The points in the domain of the function where the function is defined but the gradient of the function is either zero or undefined.


Suppose the point {eq}m {/eq} is a critical point of any function {eq}f(x) {/eq} if and only if any of the following is satisfied,

1.{eq}f^{\prime}(m)=0 {/eq}.

Or

2. {eq}f^{\prime}(m) {/eq} doesn't exist.

Answer and Explanation:

The given function is

{eq}\displaystyle f(x) = \frac{x^2}{x^2 - 4} {/eq}

and we want to find the critical points of the function.


Here we will use the derivative test to find the critical points of the function.

{eq}\displaystyle \begin{align} &\therefore f^{\prime}(x) = 0\\[0.3cm] &\Rightarrow \frac{d}{dx}f(x) = 0\\[0.3cm] &\Rightarrow \frac{d}{dx}\left(\frac{x^2}{x^2-4}\right)=0\\[0.3cm] &\Rightarrow \frac{\frac{d}{dx}\left(x^2\right)\left(x^2-4\right)-\frac{d}{dx}\left(x^2-4\right)x^2}{\left(x^2-4\right)^2}=0\\[0.3cm] &\Rightarrow \frac{2x\left(x^2-4\right)-2xx^2}{\left(x^2-4\right)^2}=0\\[0.3cm] &\Rightarrow -\frac{8x}{\left(x^2-4\right)^2}=0 & (1)\\[0.3cm] &\Rightarrow 8x=0\\[0.3cm] &\Rightarrow x=0\\[0.3cm] &\text{for undefined points we take the denominator of (1) equal to zero}\\[0.3cm] &\Rightarrow \left(x^2-4\right)^2=0\\[0.3cm] &\Rightarrow x^2-4=0\\[0.3cm] &\Rightarrow x^2=4\\[0.3cm] &\Rightarrow x=\pm2.\\[0.3cm] \end{align} {/eq}


Therefore the critical points are {eq}\displaystyle \boxed{x=-2,~x=0,~x=2} {/eq}.


Learn more about this topic:

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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
163K

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