Pl. note correction in absolute minimum.

**Function has absolute minimum=-76 at x=-3**

Given: `f(x)=x^3-6x^2+5,[-3,5].`

Find the critical value(s) by setting the first derivative equal to zero and solving for the x value(s).

`f'(x)=3x^2-12x=0`

`3x(x-4)=0`

`x=0,x=4`

The critical numbers are x=0 and x=4 . Plug in the critical numbers and the endpoints of the interval [-3, 5] into the original f(x) function.

f(-3)=-76

f(0)=5

f(4)=-27

f(5)=-20

Examine the f(x) values to determine the absolute maximum and absolute minimum values.

The **absolute maximum** occurs at the point **(0, 5)**.

The **absolute minimum** occurs at the point **(-3, -76)**.

`f(x)=x^3-6x^2+5`

differentiating.

`f'(x)=3x^2-12x`

`f'(x)=3x(x-4)`

Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.

Now to find the critical numbers, solve for x for f'(x)=0.

`3x(x-4)=0`

`x=0 , x=4`

Now evaluate the function at the critical numbers and at the end points of the interval.

`f(0)=0^3-x(0)^2+5=5`

`f(4)=4^3-6(4)^2+5=-27`

`f(-3)=(-3)^3-6(-3)^2+5=-76`

`f(5)=5^3-6(5)^2+5=-20`

So the function has **absolute maximum=5 at x=0 **

**There in no absolute minimum which is clear from the graph.**