For the function/interval pair f(x)= \frac{x^2 - 1}{x - 2}, ~[3,10], find the absolute extrema...


For the function/interval pair {eq}f(x)= \frac{x^2 - 1}{x - 2}, {/eq} [3,10], find the absolute extrema using the Extreme Value Theorem or the Critical Point Theorem.

Critical Points:

Critical points are some important points in the domain of the function where the derivative of the function either vanishes or zero.


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

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


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

Answer and Explanation:

The given function is

{eq}\displaystyle f(x)= \frac{x^2 - 1}{x - 2} {/eq}

and we are asked to find the absolute extreme values of the function on the interval {eq}\displaystyle [3,10] {/eq}.

First, we will find the critical points of the function using the derivative test.

{eq}\displaystyle \begin{align} &\therefore f^{\prime}(x) = 0\\ &\Rightarrow \frac{d}{dx}f(x)=0\\ &\Rightarrow \frac{d}{dx}\left(\frac{x^2-1}{x-2}\right)=0\\ &\Rightarrow \frac{\frac{d}{dx}\left(x^2-1\right)\left(x-2\right)-\frac{d}{dx}\left(x-2\right)\left(x^2-1\right)}{\left(x-2\right)^2}=0\\ &\Rightarrow \frac{2x\left(x-2\right)-1\cdot \left(x^2-1\right)}{\left(x-2\right)^2}=0\\ &\Rightarrow \frac{x^2-4x+1}{\left(x-2\right)^2}=0\\ &\Rightarrow x^2-4x+1=0\\ &\Rightarrow x_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:1\cdot \:1}}{2\cdot \:1} & \left(\text{using quadractic root formula }x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \right)\\ &\Rightarrow x_{1,\:2}=\frac{4\pm\sqrt{12}}{2}\\ &\Rightarrow x_{1,\:2}=\frac{4\pm2\sqrt{3}}{2}\\ &\Rightarrow x_{1,\:2}=\frac{2\left(2\pm\sqrt{3}\right)}{2}\\ &\Rightarrow x_{1,\:2}=2\pm\sqrt{3}\\ &\text{for undefined points}\\ &\Rightarrow \left(x-2\right)^2=0\\ &\Rightarrow \left(x-2\right)=0\\ &\Rightarrow x=2.\\ \end{align} {/eq}

Therefore the critical points are {eq}\displaystyle x=2-\sqrt{3},~x=2,~x=2+\sqrt{3} {/eq}

but the point {eq}\displaystyle x=2-\sqrt{3},~x=2 \notin [3,10] {/eq}

therefore the only critical points is {eq}\displaystyle x=2+ \sqrt{3} {/eq}.

Now for the absolute maximum and minimum values, we will evaluate the value of the function

at the critical point as well as at the endpoints of the given interval

{eq}\displaystyle \begin{align} &f(3)=\frac{(3)^2 - 1}{3 - 2} = \frac{8}{1} = 8\\[0.3cm] &f\left(2+ \sqrt{3}\right) = \frac{\left(2+ \sqrt{3}\right)^2 - 1}{\left(2+ \sqrt{3}\right) - 2} = \frac{\left( 2^2+2\cdot \:2\sqrt{3}+\left(\sqrt{3}\right)^2 \right) - 1}{\left( \sqrt{3}\right)} =\frac{6+4\sqrt{3}}{\sqrt{3}} = 7.4641\\[0.3cm] &f(10) = \frac{(10)^2 - 1}{10 - 2} = \frac{99}{8} = 12.375\\[0.3cm] \end{align} {/eq}

From the above calculations, we can say that

the given function has an absolute minimum at the point {eq}\displaystyle \boxed{x=2+ \sqrt{3}} {/eq}

and the given function has an absolute maximum at the point {eq}\displaystyle \boxed{x=10} {/eq}.

Learn more about this topic:

Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9

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