# Given F(x)= \frac {3x-4}{x^{2}+1}, on the interval [-2,2], find the critical values.

## Question:

Given {eq}F(x)= \frac {3x-4}{x^{2}+1}, {/eq} on the interval {eq}[-2,2], {/eq} find the critical values.

## Critical Points:

All those points where the derivative of the function is zero or those points where derivative of the function does not exist but the given function exists are known as critical points.

To evaluate the critical points of the function, first, we evaluate the derivative of the function and equate the derivative equal to zero.

Given

{eq}F(x)= \dfrac {3x-4}{x^{2}+1}, {/eq} on the interval {eq}[-2,2] {/eq}

We have to find the critical points of the function.

To calculate the critical points of the function, first, we differentiate the function and then set the derivative equal to zero.

Differentiate the given function with respect to {eq}x {/eq}

{eq}\begin{align} f'(x) &=\dfrac{\mathrm{d} }{\mathrm{d} x}\left (\dfrac {3x-4}{x^{2}+1} \right )\\ &=\dfrac{(x^2+1)\dfrac{\mathrm{d} }{\mathrm{d} x}(3x-4)-(3x-4)\dfrac{\mathrm{d} }{\mathrm{d} x}(x^2+1)}{(x^2+1)^2} \ & \left [ \dfrac{\mathrm{d} }{\mathrm{d} x}\left ( \dfrac{u}{v} \right )=\dfrac{vu'-uv'}{v^2} \right ]\\ &=\dfrac{(x^2+1)(3(1)-0)-(3x-4)(2x+0)}{(x^2+1)^2} \ & \left [ \dfrac{\mathrm{d} }{\mathrm{d} x}(x^n)=nx^{n-1}, \dfrac{\mathrm{d} }{\mathrm{d} x}(constant)=0 \right ]\\ &=\dfrac{3x^2+3-6x^2+8x}{(x^2+1)^2}\\ &=\dfrac{-3x^2+8x+3}{(x^2+1)^2} \end{align} {/eq}

For critical points:

{eq}\begin{align} f'(x) &=0\\ \dfrac{-3x^2+8x+3}{(x^2+1)^2} &=0 \\ -3x^2+8x+3 &=0 & \left [ \dfrac{a}{b}=0 \rightarrow a=0, b\neq 0 \right ]\\ -(3x^2-8x-3) &=0\\ 3x^2-8x-3 &=0\\ 3x^2-9x+x-3 &=0\\ 3x(x-3)+1(x-3) &=0\\ (3x+1)(x-3) &=0\\ x &=\dfrac{-1}{3}, 3\\ \end{align} {/eq}

The only solution {eq}x=-\dfrac{1}{3} {/eq} belongs to the interval {eq}[-2,2] {/eq}.

So, the function has only one critical point at {eq}x=-\dfrac{1}{3} {/eq} on the interval {eq}[-2,2] {/eq}. 