Question

What are the absolute extrema of `f(x)=(x+1)(x-8)^2+9 in[0,16]`?

Answer 1

No absolute maxima or minima, we have a maxima at `x=16` and a minima at `x=0`

Explanation:

The maxima will appear where `f'(x)=0` and `f''(x)<0`

for `f(x)=(x+1)(x-8)^2+9`

`f'(x)=(x-8)^2+2(x+1)(x-8)`

= `(x-8)(x-8+2x+2)=(x-8)(3x-6)=3(x-8)(x-2)`

It is apparent that when `x=2` and `x=8`, we have extrema

but `f''(x)=3(x-2)+3(x-8)=6x-30`

and at `x=2`, `f''(x)=-18` and at `x=8`, `f''(x)=18`

Hence when `x in[0,16]`

we have a local maxima at `x=2` and a local minima at `x=8`

not an absolute maxima or minima.

In the interval `[0,16]`, we have a maxima at `x=16` and a minima at `x=0`

(Graph below not drawn to scale)
graph{(x+1)(x-8)^2+9 [-2, 18, 0, 130]}