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

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Answer 1 · 2018-07-22T12:32:02

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

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]}

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