Find-minimum-and-maximum-value-of-function-f-x-2sin-x-3-sin-x-1-

Question Number 132531 by bemath last updated on 15/Feb/21

Find minimum and maximum

value of function f (x) = \sqrt{2 \sin x + 3} - \sqrt{\sin x + 1}

Answered by liberty last updated on 15/Feb/21

\frac{df (x)}{dx} = \frac{\cos x}{\sqrt{2 \sin x + 3}} - \frac{\cos x}{2 \sqrt{\sin x + 1}} = 0

\frac{\cos x}{\sqrt{2 \sin x + 3}} = \frac{\cos x}{2 \sqrt{\sin x + 1}}

\cos^{2} x (4 \sin x + 4) = \cos^{2} x (2 \sin x + 3)

\cos^{2} (x) [2 \sin x + 1] = 0

{\begin{cases} \cos^{2} x = 0 \to {\begin{cases} x = \frac{π}{2} \\ x = \frac{3 π}{2} \end{cases} \\ \sin x = - \frac{1}{2} \to x = \frac{7 π}{6} \end{cases}

(1) f (\frac{π}{2}) = \sqrt{2 \sin \frac{π}{2} + 3} - \sqrt{\sin \frac{π}{2} + 1}

= \sqrt{5} - \sqrt{2} \approx 0.8219

(2) f (\frac{3 π}{2}) = \sqrt{2 \sin \frac{3 π}{2} + 3} - \sqrt{\sin \frac{3 π}{2} + 1}

= \sqrt{1} - \sqrt{0} = 1 (maximum)

(3) f (\frac{7 π}{6}) = \sqrt{2 \sin \frac{7 π}{6} + 3} - \sqrt{\sin \frac{7 π}{6} + 1}

= \sqrt{2} - \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \approx 0.7071 (minimum)

Commented by liberty last updated on 15/Feb/21

Answered by MJS_new last updated on 15/Feb/21

y = \sqrt{3 + 2 \sin x} - \sqrt{1 + \sin x}

let \sqrt{1 + \sin x} = u \land 0 ⩽ u ⩽ \sqrt{2}

y = \sqrt{2 u^{2} + 1} - u

\frac{d y}{d u} = \frac{2 u - \sqrt{2 u^{2} + 1}}{\sqrt{2 u^{2} + 1}} = 0 \Rightarrow u = \frac{\sqrt{2}}{2}

\Rightarrow

0 ⩽ u ⩽ \sqrt{2} \Rightarrow \frac{\sqrt{2}}{2} ⩽ y ⩽ 1

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