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I take it the difficulty is in finding where f ' and f " are zero?

f ' (x) = cos(x) - sin(x)

f " (x) = -sin(x) -cos(x).

So, you have to determine where

cos(x) - sin(x) = 0 and where sin(x) + cos(x) = 0.

You can use that the first requires cos(x) = sin(x) and the second cos(x) = -sin(x) and just ask yourself when are the sine and cosine the same---or the same value with opposite signs. It may be easier to divide through by the cosine. Then the equations become:

tan(x) = 1 (this is where f ' = 0) and

tan(x) = -1 (this is where f " = 0)

Recalling that tan(pi/4) = 1, the tangent will be +1 or -1 whenever the reference angle is pi/4. Tangent is positive in quadrants I and III, and negative in quadrants II and IV. So

f ' (x) = 0 when x = pi/4, 5pi/4

f " (x) = 0 when x = 3pi/4, 7pi/4

Now you can take test points and find where its increasing/decreasing concave up and down and answer all the questions.

I hope this was what you were after.

You say you understand how to do this in general, so I'm going to assume you already have the equation

cos x = sin x

for finding local minima and maxima.

you can rewrite that as tan x = 1 and look up what values of x make that true.

or you can answer it in degrees (0 ≤ x ≤ 360) and convert back to radians.

you should find two extrema, at pi/4 and at 5/4 pi.

the inflection points will be where cos x = - sin x, or tan x = -1.