Question

Find the relative extreme values of the function f (x) = x3 − x2 − 6x + 5.

Answer 1

Given

f(x)=x³-x²-6x+5

We will find extreme

First of find 1st derivative of f(x)

f'(x)=3x²-2x-6

Put f'(x) equal to zero for extremize.

f'(x)=0

3x²-2x-6=0

Now find value of x.

$x=\frac{[2± \sqrt{4-4.3(-6)}]}{2.3}\\ x=\frac{(1±\sqrt{\smash[b]{19}}) }{3}$

this values of x give us extreme value of f(x)

Now find second derivative of f(x) for find maxima or minima.

f''(x)= 6x-2

Now check 

$x=\frac{(1+\sqrt{\smash[b]{19}}) }{3}$

is give maxima or minima.

So put this value in f''(x)

$=6(\frac{(1+\sqrt{\smash[b]{19}}) }{3})-2$

=8.717

It is positive value .it mean this value of x give us minima of f(x).

Put this value of x in f(x).

$f(x)=(\frac{1+\sqrt{\smash[b]{19}})}{3})^3-(\frac{1+\sqrt{\smash[b]{19}})}{3})^2-6(\frac{1+\sqrt{\smash[b]{19}})}{3})+5$

=9.59802825019

This is minima value.

Now check

$x=\frac{(1-\sqrt{\smash[b]{19}}) }{3}$

 

is give maxima or minima.

So put this value in f''(x)

$=6(\frac{(1-\sqrt{\smash[b]{19}}) }{3})-2$

=-8.71779788708

It is negative value .it mean this value of x give us maxima of f(x).

Put this value of x in f(x).

$f(x)=(\frac{1-\sqrt{\smash[b]{19}})}{3})^3-(\frac{1-\sqrt{\smash[b]{19}})}{3})^2-6(\frac{1-\sqrt{\smash[b]{19}})}{3})+5$

=11.7984052203

This is maxima value.