Min and Max of $f(x) = 2 \sin(x) + \cos^2 (x)$ on $[0, 2\pi]$

Question

Find the absolute minimum and maximum values of,

$$f(x) = 2 \sin(x) + \cos^2 (x) \text{ on } [0, 2\pi]$$

What I did so far is

$$f'(x) = 2\cos(x) -2 \cos(x) \sin(x)$$

Could someone please help me get started?

Answer 1

Method $1$:

Continue from what you have so far,

$$\cos(x)((1-\sin(x))=0$$

Find the stationary point, evaluate the function values at the stationary point as well as the boundaries and conclude the minimal and maximal point.

Method $2$:

\begin{align} f(x)&=2\sin(x)+\cos^2(x)\\ &=2 \sin(x)+1-\sin^2(x)\\ &=-\sin^2(x)+2\sin(x)+1 \\ &=-(\sin(x)-1)^2+2 \end{align}

$$-(-1-1)^2+2\le-(\sin(x)-1)^2+2 \le -(1-1)^2+2$$

Answer 2

Without differentiation: $$f(x)=-(1-\sin x)^2+2; \\ f\left(\frac{3\pi}{2}\right)=-2 (\text{min}); \\ f\left(\frac{\pi}{2}\right)=2 (\text{max}).$$

Answer 3

hint: $\cos^2x = 1 -\sin^2x \implies f(x) = 2-\left(1-\sin x\right)^2$. Then use the fact that $-1 \le \sin x \le 1$ to obtain the min and max.