Find Max and Min values of f(x)=cosx-sinx, given interval of -pi

Question

Find Max and Min values of f(x)=cosx-sinx, given interval of -pi<x<pi    (I keep getting stuck after taking the derivatives)

anonymous · Answer

What do you get when you take the derivative?

anonymous · Answer

-sinx-cosx

anonymous · Answer

set the derivative equal 0, and find all the possibilities

anonymous · Answer

For a min or a max to occur, that means that the slope of f(x) equals 0. So find where f'(x)=0.

anonymous · Answer

I did that and I'm stuck at sinx=-cosx

anonymous · Answer

Check your unit circle. Where does the sine of x equal the opposite of the cosine?

anonymous · Answer

at pi/4 and 3pi/4?

anonymous · Answer

it's one of the special angle

anonymous · Answer

you're right, dont forget it's from -pi to pi

anonymous · Answer

so, there are 4 possibilities

anonymous · Answer

does that mean I forgot one?

anonymous · Answer

ohh, so 5pi/4 and 7pi/4 too. Thanks

anonymous · Answer

no, it's -3pi/4, -pi/4, pi/4 and 3pi/4, because the interval is from -pi to pi, 5pi/4 and 7pi/4 is off the interval

anonymous · Answer

ok