Find the absolute maximum value and the absolute minimum value of the function. h(t)=t3−9t2 on

Step 1
Given,
${\displaystyle h\left(t\right)=t^3-9t^{2}on[4,8]}$
Step 2
Now,
${\displaystyle \therefore h\left(t\right)=t^3-9t^2}$
Differentiating w.r.t t, we have
${\displaystyle \Rightarrow h^{\prime }\left(t\right)=\frac{d}{dt}(t^3-9t^2)}$
${\displaystyle \Rightarrow h^{\prime }\left(t\right)=\frac{d}{dt}\left(t^3\right)-9\frac{d}{dt}\left(t^2\right)}$
$\Rightarrow h^{\prime }(t)=3t^2-18t$
For critical points: h'(t)=0
${\displaystyle \Rightarrow 3t^2-18t=0}$
${\displaystyle \Rightarrow 3t(t-6)=0}$
either
${\displaystyle \Rightarrow t=0\text{or}\Rightarrow t=6}$
Step 3
At t=0; h(0)=0
At ${\displaystyle t=4;h\left(4\right)=4^3-9{\left(4\right)}^3}$
${\displaystyle \Rightarrow h\left(4\right)=-512}$
At ${\displaystyle t=6;h\left(6\right)=6^3-9{\left(6\right)}^3}$
${\displaystyle \Rightarrow h\left(6\right)=-1728}$
At ${\displaystyle t=8;h\left(8\right)=8^3-9{\left(8\right)}^3}$
${\displaystyle \Rightarrow h\left(8\right)=-4096}$
${\displaystyle \therefore }$ The absolute maximum is at x=0, and the value is 0.
${\displaystyle \therefore }$ The absolute minimum is at x=8, and the value is -4096.