villane0

2021-12-03

Find the absolute maximum value and the absolute minimum value of the function.

Mary Darby

Step 1
Given,

Step 2
Now,
$\therefore h\left(t\right)={t}^{3}-9{t}^{2}$
Differentiating w.r.t t, we have
$⇒{h}^{\prime }\left(t\right)=\frac{d}{dt}\left({t}^{3}-9{t}^{2}\right)$
$⇒{h}^{\prime }\left(t\right)=\frac{d}{dt}\left({t}^{3}\right)-9\frac{d}{dt}\left({t}^{2}\right)$
$⇒{h}^{\prime }\left(t\right)=3{t}^{2}-18t$
For critical points: h'(t)=0
$⇒3{t}^{2}-18t=0$
$⇒3t\left(t-6\right)=0$
either

Step 3
At t=0; h(0)=0
At $t=4;h\left(4\right)={4}^{3}-9{\left(4\right)}^{3}$
$⇒h\left(4\right)=-512$
At $t=6;h\left(6\right)={6}^{3}-9{\left(6\right)}^{3}$
$⇒h\left(6\right)=-1728$
At $t=8;h\left(8\right)={8}^{3}-9{\left(8\right)}^{3}$
$⇒h\left(8\right)=-4096$
$\therefore$ The absolute maximum is at x=0, and the value is 0.
$\therefore$ The absolute minimum is at x=8, and the value is -4096.

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