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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?

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  • $\begingroup$ So you're looking for where $f'$ vanishes. Note $f'(x)=2\cos x(1-\sin x)$. Also, don't forget about the endpoints of the interval. $\endgroup$ – MPW Mar 22 '18 at 1:50
  • $\begingroup$ This may be helpful: mathsisfun.com/calculus/maxima-minima.html $\endgroup$ – Nash J. Mar 22 '18 at 1:51
  • $\begingroup$ $f'(x) = 2(\cos x)(1-\sin x).$ This is $0$ if $\cos x=0$ or if $1-\sin x=0. \qquad$ $\endgroup$ – Michael Hardy Mar 22 '18 at 2:00
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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}

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

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

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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.

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