Determine the interval, where f(x)=sin x - cos x, 0 x 2pie is strictly increasing or decreasing - Maths - Application of Derivatives

Question

Determine the interval, where f(x)=sin x - cos x, 0<x<2pie
is strictly increasing or decreasing?

Santosh Kumar · Accepted Answer

$f (x) = \sin x - \cos x \frac{d}{d x} [f (x)] = f' (x) = \cos x + \sin x f' (x) = \sqrt{2} (\frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x) \Rightarrow f' (x) = \sqrt{2} (\sin \frac{π}{4} \cdot \cos x + \cos \frac{π}{4} \cdot \sin x) \Rightarrow f' (x) = \sqrt{2} \sin (x + \frac{π}{4})$

For f (x) to be increasing f'(x) > 0

$\sqrt{2} \sin (x + \frac{π}{4}) > 0 \Rightarrow \sin (x + \frac{π}{4}) > 0 \Rightarrow 0 < x + \frac{π}{4} < π (∵ \sin θ > 0 for 0 < θ < π) \Rightarrow - \frac{π}{4} < x < \frac{3 π}{4} \Rightarrow \frac{- π}{4} < x < 0 or 0 < x < \frac{3 π}{4} \Rightarrow x \in (- \frac{π}{4}, 0) or (0, \frac{3 π}{4}) \Rightarrow 2 π - \frac{π}{4} < x < 2 π + 0 or 0 < x < \frac{3 π}{4} \Rightarrow \frac{7 π}{4} < x < 2 π or 0 < x < \frac{3 π}{4} \Rightarrow x \in (0, \frac{3 π}{4}) \cup (\frac{7 π}{4}, 2 π)$

∴ f (x) increases in the interval $(0, \frac{3 π}{4}) \cup (\frac{7 π}{4}, 2 π)$

For f (x) to be decreasing, f'(x) < 0

$\sin (x + \frac{π}{4}) < 0 π < x + \frac{π}{4} < 2 π (∵ \sin θ < 0 for π < θ < 2 π) \Rightarrow π - \frac{π}{4} < x < 2 π - \frac{π}{4} \Rightarrow \frac{3 π}{4} < x < \frac{7 π}{4}$

∴ f (x)decreases in the interval $(\frac{3 π}{4}, \frac{7 π}{4})$

∴ f (x) increases in the interval $(0, \frac{3 π}{4}) \cup (\frac{7 π}{4}, 2 π)$ and decreases in the interval $(\frac{3 π}{4}, \frac{7 π}{4})$

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Determine the interval, where f(x)=sin x - cos x, 0<x<2pie is strictly increasing or decreasing?