If f(x) = 1 + 2 sin x + 3 cos^2 x for 0

Question

(A) Given $f(x) = 1 + 2 \sin x + 3 \cos^2 x = 1 + 2 \sin x + 3(1 - \sin^2 x) = 4 + 2 \sin x - 3 \sin^2 x$.Let $u = \sin x$. Since $0 < x < 2\pi / 3$,t

Vedclass · Accepted Answer

Minimum at $x = \pi / 2$

Vedclass · Answer

(A) Given $f(x) = 1 + 2 \sin x + 3 \cos^2 x = 1 + 2 \sin x + 3(1 - \sin^2 x) = 4 + 2 \sin x - 3 \sin^2 x$.Let $u = \sin x$. Since $0 Then $g(u) = 4 + 2u - 3u^2$.To find critical points,$g'(u) = 2 - 6u = 0 \implies u = 1/3$.Since $g''(u) = -6 Thus,$f(x)$ has a maximum at $\sin x = 1/3$,i.e.,$x = \sin^{-1}(1/3)$.At the boundaries of the interval $(0, 2\pi / 3)$:As $x 	o 0^+$,$f(x) 	o 1 + 0 + 3(1) = 4$.At $x = 2\pi / 3$,$f(2\pi / 3) = 1 + 2(\sqrt{3}/2) + 3(1/4) = 1 + \sqrt{3} + 0.75 = 1.75 + \sqrt{3} \approx 3.48$.Since $g(1/3) = 4 + 2/3 - 3(1/9) = 4 + 2/3 - 1/3 = 4.33$,the maximum is at $x = \sin^{-1}(1/3)$.Checking the options,the question asks for the behavior. Based on the derivative $f'(x) = 2 \cos x - 6 \cos x \sin x = 2 \cos x (1 - 3 \sin x)$,$f'(x) = 0$ at $x = \pi / 2$ and $x = \sin^{-1}(1/3)$.At $x = \pi / 2$,$f(\pi / 2) = 1 + 2(1) + 3(0) = 3$. Since $3 < 4$ and $3 < 4.33$,$x = \pi / 2$ is a local minimum.

If $f(x) = 1 + 2 \sin x + 3 \cos^2 x$ for $0 < x < 2\pi / 3$,then:

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