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

Jul 22, 2018

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

#### Explanation:

The maxima will appear where $f ' \left(x\right) = 0$ and $f ' ' \left(x\right) < 0$

for $f \left(x\right) = \left(x + 1\right) {\left(x - 8\right)}^{2} + 9$

$f ' \left(x\right) = {\left(x - 8\right)}^{2} + 2 \left(x + 1\right) \left(x - 8\right)$

= $\left(x - 8\right) \left(x - 8 + 2 x + 2\right) = \left(x - 8\right) \left(3 x - 6\right) = 3 \left(x - 8\right) \left(x - 2\right)$

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

but $f ' ' \left(x\right) = 3 \left(x - 2\right) + 3 \left(x - 8\right) = 6 x - 30$

and at $x = 2$, $f ' ' \left(x\right) = - 18$ and at $x = 8$, $f ' ' \left(x\right) = 18$

Hence when $x \in \left[0 , 16\right]$

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

not an absolute maxima or minima.

In the interval $\left[0 , 16\right]$, 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]}