# Find the absolute minimum m and the absolute maximum value M of the function f(x) = 2x^3 - 3x^2 -...

## Question:

Find the absolute minimum {eq}m {/eq} and the absolute maximum value {eq}M {/eq} of the function {eq}f(x) = 2x^3 - 3x^2 - 72x + 2 {/eq} on the closed interval {eq}[-6, 7]. {/eq}

## Finding Minima & Maxima:

In some intervals, the maxima or the minima is found with the help of the graphs and another way is to use the first level derivative and the second level derivative methods. The critical points are generally the maxima or the minima points.

It is required here to find the absolute minimum {eq}m {/eq} and the absolute maximum value {eq}M {/eq} of the function {eq}f(x) = 2x^3 - 3x^2 - 72x + 2 {/eq} on the closed interval {eq}[-6, 7]. {/eq}

So for that, we find the first-level derivative and then critical points by the condition:

{eq}f'(x)=0\\ \Rightarrow \frac{d}{dx}\left(2x^3-3x^2-72x+2\right)=0\\ \Rightarrow 6x^2-6x-72=0\\ \Rightarrow x=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:6\left(-72\right)}}{2\cdot \:6}\\ \Rightarrow x=4,\:x=-3\\ {/eq}

Now these points are with the given restricted domain.

Now we will find the second level derivative to check the maxima and the minima.

{eq}f''(x)=\frac{d}{dx}\left(6x^2-6x-72\right)\\ =12x-6\\ {/eq}

Now the value of the second derivative at the critical points are:

{eq}f''(-3)=12(-3)-6=-42<0~~~Maxima\\ f''(4)=12(4)-6=42>0~~~Minima\\ {/eq}

So the minimum value is:

{eq}m=f(4) = 2(4)^3 - 3(4)^2 - 72(4) + 2=-206\\ {/eq}

and the maximum value is:

{eq}M=f(-3) = 2(-3)^3 - 3(-3)^2 - 72(-3) + 2=137\\ {/eq}