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Find the critical numbers $: f(x)=2 x^{3}-3 x^{2}-36 x, f(\theta)=2 \cos \theta+\sin ^{2} \theta$

$f(x)=2 x^{3}-3 x^{2}-36 x \rightarrow f^{\prime}(x)=6 x^{2}-6 x-36$

$f^{\prime}(x)=0 \rightarrow 6 x^{2}-6 x-36=0 \quad \div 6$

$x^{2}-x-6=0$

$(x-3)(x+2)=0$

$x=3 \quad, \quad x=-2$

Critical numbers are $\rightarrow x=3, x=-2$

$f(\theta)=2 \cos (\theta)+\sin ^{2} \theta$

$f^{\prime}(\theta)=-2 \sin \theta+2 \sin \theta \cdot \cos \theta$

$f^{\prime}(\theta)=2 \sin \theta(-1+\cos \theta)$

$f^{\prime}(\theta)=0 \longrightarrow 2 \sin \theta(-1+\cos \theta)=0$

$2 \sin \theta=0 \div 2 \quad or \quad-1+\cos \theta=0$

$\sin \theta=0 \quad \cos \theta=+1$

$\theta=n \pi \quad \theta=2 n \pi$

Critical numbers are : $0, \pi, 2 \pi, 3 \pi, 4 \pi, \dots$

Find the absolute maximum and absolute minimum values of $f(x)=2 x^{3}-3 x^{2}-12 x+1$ on $[-2,3]$

$f(x)=2 x^{3}-3 x^{2}-12 x+1 \quad, \quad[-2,3]$

$F^{\prime}(x)=6 x^{2}-6 x-12 \quad f^{\prime}(x)=0$

$6 x^{2}-6 x-12=0 \div 6 \quad \rightarrow x^{2}-x-2=0$

$(x-2)(x+1)=0$

$x=2 \quad$ or $\quad x=-1 \quad \in[-2,3]$

$F(2)=2(2)^{3}-3(2)^{2}-12(2)+1=-19$
$f(-1)=2(-1)^{3}-3(-1)^{2}-12(-1)+1=8$
$F(-2)=2(-2)^{3}-3(-2)^{2}-12(-2)+1=-3$
$f(3)=2(3)^{3}-3(3)^{2}-12(3)+1=-62$

F has Global Min at
$x=3, f(3)=-62$ in $[-2 , 3]$
F has Globod Max at
$x=-1, f(-1)=8$ in $[-2 , 3]$