Question

Find the interval of increasing and decreasing for,

$f(x) = sinx + cosx \: on \: (0, 2\pi)$


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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = sinx + cosx$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) =\dfrac{d}{dx}( sinx + cosx)$

$\rm :\longmapsto\:f'(x) = cosx - sinx$

For critical points,

$\rm :\longmapsto\:f'(x) = 0$

$\rm :\longmapsto\:cosx - sinx = 0$

$\rm :\longmapsto\:cosx = sinx$

$\rm :\longmapsto\:\dfrac{sinx}{cosx} = 1$

$\rm :\longmapsto\:tanx = 1$

$\bf\implies \:x = \dfrac{\pi}{4} \: \: \: or \: \: \: \dfrac{5\pi}{4}$

Now, Let we check the sign in intervals.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf \: Interval & \bf Sign \: of \: f'(x)\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \bigg(0,\dfrac{\pi}{4}\bigg) & \sf + ve \\ \\ \sf \bigg(\dfrac{\pi}{4},\dfrac{5\pi}{4}\bigg) & \sf - ve \\ \\ \sf \bigg(\dfrac{\pi}{4},2\pi\bigg) & \sf + ve \end{array}} \\ \end{gathered}$

So, For increasing,

$\rm :\longmapsto\:f'(x) > 0$

$\bf\implies \:\boxed{ \tt{ \: x \: \in \: \bigg(0,\dfrac{\pi}{4}\bigg) \cup \: \bigg(\dfrac{\pi}{4},2\pi\bigg)}}$

Now, For decreasing,

$\rm :\longmapsto\:f'(x) < 0$

$\bf\implies \:\boxed{ \tt{ \: x \: \in \: \bigg(\dfrac{\pi}{4},\dfrac{5\pi}{4}\bigg) \: \: }}$

More to know,

$\purple{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$