If A = {x in R : frac{pi}{4} leq x leq frac{p | vedclass

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Question

(A) Given $f(x) = \sin x - x$. To find the range $f(A)$,we check the derivative $f'(x) = \cos x - 1$. Since $\cos x < 1$ for all $x \in [\frac{\pi}{4}

Vedclass · Accepted Answer

$\left[\frac{\sqrt{3}}{2} - \frac{\pi}{3}, \frac{1}{\sqrt{2}} - \frac{\pi}{4}\right]$

Vedclass · Answer

(A) Given $f(x) = \sin x - x$. To find the range $f(A)$,we check the derivative $f'(x) = \cos x - 1$. Since $\cos x Therefore,the minimum value is at the upper bound $x = \frac{\pi}{3}$ and the maximum value is at the lower bound $x = \frac{\pi}{4}$. $f(\frac{\pi}{3}) = \sin(\frac{\pi}{3}) - \frac{\pi}{3} = \frac{\sqrt{3}}{2} - \frac{\pi}{3}$. $f(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) - \frac{\pi}{4} = \frac{1}{\sqrt{2}} - \frac{\pi}{4}$. Thus,$f(A) = [\frac{\sqrt{3}}{2} - \frac{\pi}{3}, \frac{1}{\sqrt{2}} - \frac{\pi}{4}]$.

If $A = \{x \in R : \frac{\pi}{4} \leq x \leq \frac{\pi}{3}\}$ and $f(x) = \sin x - x$,then $f(A)$ is equal to

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