Question

Find the intervals in which the function f given by
f(x)=sinx + cosx ,o≤x≤2pi is strictly increasing or strictly decreasing

Answer 1

\[f(x)=sinx+cosx\]\[f'(x)=cosx-sinx\]

Answer 2

Can you solve \(\cos(x)-\sin(x)=0\), as we just did?

Answer 3

sin(x) = -cos(x)
sin(x)/-cos(x) = 1
tan(x) = -1
which gives you the solution 3pi/4, 5pi/4

Answer 4

Take the inequalities at these two points

Answer 5

\[Cos(x)=Sin(x) \]
\[x= \frac{\pi}{4}, \frac{5\pi}{4}\]

Answer 6

f'(x)=cos(x)-sin(x)
set =0
cos(x)-sin(x)=0
=>
cos(x)=sin(x)
this happens at x=pi/4, pi/4+pi=pi/4,5pi/4

| ------|-------|-------|
0 pi/4 5pi/4 2pi
choose test points in each of these intervals to plug into f' to see if you get negative output or positive output
a positive output=> it is increasing on that interval
a negative output=> it is decreasing on that interval

Answer 7

gj diyadiya

Answer 8

Thanks :)
Now?

Answer 9

Nice job!

Answer 10

Amazing!
Well, myin had everything explained, there's nothing more to say! :)

Answer 11

so you should get 3pi/4 + 2pi(k) and 5pi/4 + 2pi(k)
K being an element of all intergers
so you would set it up like this

o≤ (3pi/4) +2pi(k) ≤ 2pi
you should be left with k and two numbers it fits inside
figure out what integers are present in the inequality then sub them into the original domain
(3pi/4) +2pi(k)

Do with the other one and you will find all possible solutions

Answer 12

OK letme go through it!

Answer 13

@ cuddlypony
it will be pi/4 and 5pi/4
we are solving
sinx=cosx
this time

Answer 14

oh sorry I miss read

Answer 15

Can you show the last part ?

Answer 16

no it wouldn't turing test
sin(x) + cos(x) = f(x) or did I mess up

Answer 17

So what integers did you get for k

Answer 18

we want to find where f is decreasing and increasing
what function will tell us where f is increasing and decreasing@ cuddle?

Answer 19

\[(0,\pi/4 ) (\pi/4,5\pi/4)(5\pi/4,2\pi)\]

Answer 20

f tells us where
f' tells us the increasing/decreasing part
f'' tells us about concavity

Answer 21

we need f' here
so f'(x)=cos(x)-sin(x)

Answer 22

sorry i read the question wrong disregard my input :l

Answer 23

set this puppy =0

Answer 24

good Diya, so pick a point in each interval and see if f'(x) is
>0
or
<0

Answer 25

Okay! In general a function is decreasing when its derivative is negative and it's increasing when the derivative is positive. In our case, we have found three intervals of signs for the derivative and we need to test each of them. Following so far?

Answer 26

Yup!

Answer 27

So i should plug in the values of each interval to "cos(x)-sin(x) right ?

Answer 28

It's very easy to test the sign of each given interval: choose any value for x inside the interval (not in the boundaries) and substitute into f'(x) and see what you get.

Answer 29

Yes Diya.

Answer 30

Ok 1 min

Answer 31

cos(pi/6)-sin(pi/6) = 1/2 -sqrt3/2 is it like this ? o.O

Answer 32

I think you switched the values :D

Answer 33

oh yeah

Answer 34

sqrt3/2-1/2

Answer 35

now is that >0 or <0 ???

Answer 36

>0

Answer 37

\(\cos{\frac{\pi}{6}}-\sin{\frac{\pi}{6}}=\frac{\sqrt{3}}{2}-\frac{1}{2}>0\), since \(\sqrt{3}>1\).

Answer 38

So, you conclude that..?

Answer 39

Strictly Increasing

Answer 40

on..?

Answer 41

(0,pii/4)

Answer 42

Good! when you say Strictly use open intervals.

Answer 43

okay 1min brb sorryy

Answer 44

Alright! Apply the same idea to the other two intervals.

Answer 45

ThankYouu Myininaya ,Mr.Math & turingtest :D

Answer 46

Glad to help!

Answer 47

anytime :)