Calculus AExtreme Values Of A Function Problem

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 $$[-2 , 3]$$