Find-the-interval-for-which-the-function-f-x-sinx-cosx-for-x-0-2pi-is-strictly-inceasing-and-srictly-decreasing-

Question Number 75368 by vishalbhardwaj last updated on 10/Dec/19

Find the interval for which the function

f (x) = s i n x + c o s x, for x \in [0, 2 π]

is strictly inceasing and srictly decreasing ?

Commented by vishalbhardwaj last updated on 10/Dec/19

please sir, write complete explanation

of intervals

Answered by MJS last updated on 10/Dec/19

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

f^{'} (x) = 0

\sin x = \cos x

x = \frac{π}{4} \lor x = \frac{5 π}{4}

f^{″} (x) = - \sin x - \cos x

f^{″} (\frac{π}{4}) < 0 \Rightarrow maximum

f^{″} (\frac{5 π}{4}) > 0 \Rightarrow minimum

[0; \frac{π}{4} [f (x) strictly increasing

] \frac{π}{4}; \frac{5 π}{4} [f (x) strictly decreasing

] \frac{5 π}{4}; 2 π] f (x) strictly increasing

Commented by vishalbhardwaj last updated on 10/Dec/19

sir please solve by first derivative rule

and please explane by taking any value in

between the intervals

Commented by MJS last updated on 10/Dec/19

I {don}^{'} t know this rule . the proper way I used

is to find the zeros of the first derivate to get

extremes and then check the second derivate

to find if these are maxima or minima

f^{'} (p) = 0 \land f^{″} (p) < 0 \Leftrightarrow \max at p

f^{'} (p) = 0 \land f^{″} (p) > 0 \Leftrightarrow \min at p

Answered by Kunal12588 last updated on 10/Dec/19

f^{'} (x) = c o s x - s i n x = \sqrt{2} c o s (\frac{π}{4} + x)

f^{'} (x) = 0

\Rightarrow c o s (\frac{π}{4} + x) = 0

\Rightarrow x = 2 n π \pm \frac{π}{2} - \frac{π}{4} = 2 n π + \frac{π}{4}, 2 n π - \frac{3 π}{4}

w h i c h d i v i d e s [0, 2 π] i n t o 3 i n t e r v a l s

[0, \frac{π}{4}), (\frac{π}{4}, \frac{5 π}{4}), (\frac{5 π}{4}, 2 π]

i n t e r v a l ∣ s i g n o f f^{'} (x) ∣ n a t u r e o f f (x)

[0, \frac{π}{4}) ∣ (+) ∣ s t r i c t l y i n c r e a s i n g

(\frac{π}{4}, \frac{5 π}{4}) ∣ (-) ∣ s t r i c t l y d e c r e a s i n g

(\frac{5 π}{4}, 2 π] ∣ (+) ∣ s t i c t l y i n c r e a s i n g

∴ f (x) i s i n c r e a s i n g i n [0, \frac{π}{4}) \cup (\frac{5 π}{4}, 2 π]

f (x) i s d e c r e a s i n g i n (\frac{π}{4}, \frac{5 π}{4})

[N o t e : t h i s a n s w e r i s n o t v a l i d f o r e v e r y

v a l u e o f x, o n l y f o r t h e d o m a i n [0, 2 π]]