let f(x)=e^-x*cos(x)
find the absolute maximum and absolute minimum of f on the interval [-pi/2,pi/2]

Question

let f(x)=e^-x*cos(x)
find the absolute maximum and absolute minimum of f on the interval [-pi/2,pi/2]

anonymous · Answer

Is it just this question or are you having difficulty in other max.min problems?

anonymous · Answer

Or rather, what step of this question is giving you trouble?

anonymous · Answer

i'm not sure what to do to start the problem?

anonymous · Answer

so the steps of how to find the answer would be helpful thanks

anonymous · Answer

When trying to find max or min, you have to:
1.  Take the derivative of the function
2.  Set that function to zero and solve for values of x (which make the derivative zero)

anonymous · Answer

so you are given:
$$\huge f(x)$$
and must solve:$$\huge f'(x)=0$$

anonymous · Answer

so for my derivative i got 
f'(x)=-e^-x(cos(x)+sin(x))
is this right?
how would i set this to 0 if right?

anonymous · Answer

this is right.  because you're setting f'(x) to zero, you just replace f'(x) with zero:
$$\huge -e^{-x}(cosx+sinx) = 0$$

anonymous · Answer

you can now think to yourself "when will -e^(-x) equal zero...here's a rough graph of that function
|dw:1353448776030:dw|