For what interval is the function f(x) = sin | vedclass

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(C) Given $f(x) = \sin x - \cos x$.To find the interval where the function is strictly increasing,we find the derivative $f'(x)$.$f'(x) = \cos x + \si

Vedclass · Accepted Answer

$x \in (-\pi /4, 3\pi /4)$

Vedclass · Answer

(C) Given $f(x) = \sin x - \cos x$.To find the interval where the function is strictly increasing,we find the derivative $f'(x)$.$f'(x) = \cos x + \sin x$.We can rewrite this as $f'(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x ight) = \sqrt{2} \sin(x + \pi /4)$.The function is strictly increasing when $f'(x) > 0$.$\sqrt{2} \sin(x + \pi /4) > 0 \implies \sin(x + \pi /4) > 0$.Since $\sin 	heta > 0$ for $	heta \in (0, \pi)$,we have $0 Subtracting $\pi /4$ from all parts,we get $-\pi /4 Thus,the function is strictly increasing for $x \in (-\pi /4, 3\pi /4)$.

For what interval is the function $f(x) = \sin x - \cos x$ strictly increasing?

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