# Find the absolute minimum and absolute maximum values of f on the given interval:...

## Question:

Find the absolute minimum and absolute maximum values of {eq}f {/eq} on the given interval: {eq}f(x)=4x^3-12x^2-96x+3, \ [-3,5] {/eq}

## Finding Minima & Maxima:

The maxima and the minima of the function in the given interval or in the restricted domain can be used to find either the absolute extreme values of the local extreme values.

In order to get the absolute minimum and absolute maximum values of {eq}f(x)=4x^3-12x^2-96x+3 {/eq} on the given interval: {eq}[-3,5] {/eq}

we will first find the derivative of the function and the critical points lying on this interval.

Thus the derivative and the critical points are:

{eq}f'(x)=0\\ \Rightarrow \frac{d}{dx}\left(4x^3-12x^2-96x+3\right)=0\\ \Rightarrow \frac{d}{dx}\left(4x^3\right)-\frac{d}{dx}\left(12x^2\right)-\frac{d}{dx}\left(96x\right)+\frac{d}{dx}\left(3\right)=0\\ \Rightarrow 12x^2-24x-96=0\\ \Rightarrow x=\frac{-\left(-24\right)\pm \sqrt{\left(-24\right)^2-4\cdot \:12\left(-96\right)}}{2\cdot \:12} \\ \Rightarrow x=4,\:x=-2\\ {/eq}

So these are the critical points.

Now we have the find the second derivative of the function first as follows:

{eq}f''(x)=\frac{d}{dx}\left(12x^2-24x-96\right)=24x-24\\ {/eq}

So the value of the second derivative at these points are:

{eq}f''(-2)=24(-2)-24=-72<0 \left [ \therefore absolute ~maxima \right ]~~~~~~~~~~\\ f''(4)=24(4)-24=72>0 \left [ \therefore absolute ~minima \right ]~~~~~~~~~~\\ {/eq}