Question

Find the interval in which the function f(x) = sin x + cos x increases, where 0 ≤ x ≤pi/2

Answer 1

You an take the first derivative of f(x) and see where that is positive:

f '(x) = cos x - sin x

In the interval 0 <= x <= pi/2 , you will have cos x >= sin x only in the subinterval:

0 <= x <= pi/4

Answer 2

I thought I was doing Algebra I not Calculus?? I never knew we had to do derivatives in Algebra I?

Answer 3

Well, you're doing whatever you say you are doing! Who's to know except you? lol

Answer 4

I was doing Algebra and this is one of the questions @tcarroll010
Isn't there another way to slove the problem?

Answer 5

Yes, I just thought of a way. It'll take a little time to write it out.

Answer 6

@tcarroll010 Okay, Thank you very much!

Answer 7

f(x) = sin x + cos x -> [f(x)]^2 = (sin x + cos x )^2

[f(x)]^2 = sin^2 x + 2[(sin x)(cos x)] + cos^2 x

[f(x)]^2 = (sin^2 x + cos^2 x) + 2[(sin x)(cos x)]

[f(x)]^2 = 1 + sin 2x

f(x) = sqrt(1 + sin 2x)

We took the positive square root because in the original equation, both sin x and cos x are positive the indicated interval and this is maximized at 2x = pi/2 -> x = pi/4

Now sin (0) + cos(0) = 0 + 1 = 1 and sin (pi/4) + cos (pi/4) = sqrt(2) > 1

So, we increase from 0 <= x <= pi/4 and then decrease in pi/4 <= x <= pi/2

Answer 8

Here's a graph to show your original function. Each horizontal spacing is about pi/8 which is about 0.39

You will see a max at the second horizontal spacing which is about pi/4:

Answer 9

All good now, @Tehsh ?

Answer 10

@tcarroll010 Gotcha! All good brother. Thanks!

Answer 11

uw! Good luck to you in all of your studies and thx for the recognition! @Tehsh

Answer 12

@tcarroll010 You're very welcome and keep in touch!