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

Question

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

William Abdalla · Accepted Answer

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

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]}

Ezekiel Alexander · Answer

undefined To find the absolute extrema of $ f(x) = (x+1)(x-8)^2 + 9 $ on the interval $[0,16]$, we first find critical points by taking the derivative and setting it equal to zero. Then we evaluate $ f(x) $ at these critical points as well as at the endpoints of the interval $[0,16]$, and compare the values to determine the absolute extrema.

First, find the derivative of $ f(x) $:

$ f'(x) = 3(x-8)(x+1) + (x-8)^2 $

Setting $ f'(x) $ equal to zero and solving for $ x $ gives the critical points.

$ 0 = 3(x-8)(x+1) + (x-8)^2 $

Solving this equation yields $ x = 8 $ and $ x = 5 $ as critical points.

Next, evaluate $ f(x) $ at these critical points and at the endpoints of the interval $[0,16]$:

$ f(0) = (0+1)(0-8)^2 + 9 = 145 $

$ f(5) = (5+1)(5-8)^2 + 9 = 41 $

$ f(8) = (8+1)(8-8)^2 + 9 = 9 $

$ f(16) = (16+1)(16-8)^2 + 9 = 145 $

Comparing these values, we see that the absolute maximum is $ f(0) = 145 $ and the absolute minimum is $ f(8) = 9 $.