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

1 Answer
Jul 22, 2018

Answer:

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]}