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Question Number 167098 by mr W last updated on 06/Mar/22

f i n d t h e m a x i m u m o f

f (x) = \sin x + \cos x + \sin x \cos x

Answered by cortano1 last updated on 06/Mar/22

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

f (x) = \sqrt{1 + \sin 2 x} + \frac{1}{2} \sin 2 x

f^{'} (x) = \frac{\cos 2 x}{\sqrt{1 + \sin 2 x}} + \cos 2 x = 0

\Rightarrow \cos 2 x + \cos 2 x \sqrt{1 + \sin 2 x} = 0

\Rightarrow \cos 2 x (1 + \sqrt{1 + \sin 2 x}) = 0

\Rightarrow \cos 2 x = 0 \Rightarrow x = \frac{π}{4}

\max f (x) = \sqrt{2} + \frac{1}{2}

Answered by cortano1 last updated on 06/Mar/22

for minimum

f (x) = \sin x + \cos x + \sin x \cos x

f (x) = \sin x + \cos x + \frac{1}{2} \sin 2 x

f^{'} (x) = \cos x - \sin x + \cos 2 x = 0

\cos x - \sin x + \cos^{2} x - \sin^{2} x = 0

{(\cos x + \frac{1}{2})}^{2} - {(\sin x + \frac{1}{2})}^{2} = 0

\Rightarrow (\cos x + \sin x + 1) (\cos x - \sin x) = 0

\Rightarrow {\begin{cases} \cos x + \sin x = - 1 \Rightarrow 1 + \sin 2 x = 1 \\ \cos x - \sin x = 0 \end{cases}

\Rightarrow {\begin{cases} \max = \sqrt{2} + \frac{1}{2} \\ \min = - 1 \end{cases} \Rightarrow - 1 ⩽ f (x) ⩽ \sqrt{2} + \frac{1}{2}

Answered by mahdipoor last updated on 06/Mar/22

s i n x + c o s x = u = \sqrt{2} s i n (45 + x)

s i n x . c o s x = 0.5 (u^{2} - 1)

f (x) = f (u) = 0.5 (u^{2} + 2 u - 1) = 0.5 {(u + 1)}^{2} - 1

\Rightarrow m a n o f f = f (m a x o f u) = f (\sqrt{2}) =

0.5 {(\sqrt{2} + 1)}^{2} - 1 = \sqrt{2} + 0.5

Commented by mr W last updated on 06/Mar/22

t h a n k s s i r s!

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