# What is f(0),f(pi), f'(pi/2)? x<pif=xcosx-sinx,x>0

*print*Print*list*Cite

### 1 Answer

Plugging `x = 0` in the equation of the function yields:

`f(0) = 0*cos 0- sin 0 =gt f(0) = 0*1 - 0 =gtf(0) = 0`

Plugging `x =pi` in the equation of the function yields:

`f(pi) = pi*cos pi - sinpi =gt f(pi) = pi*(-1) - 0 =gtf(pi) = -pi`

You need to find the derivative of the function f(x).

`f'(x) = (xcos x - sin x)'`

Use the product rule when differentiate the term xcos x:

`f'(x) = (x'*cos x + x*(cos x)') - cos x`

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

Plugging `x =pi/2` in the equation of the derivative of function yields:

`f'(pi/2) = (pi/2)*sin(pi/2) =gt f'(pi/2) = (pi/2)*1 =gt f'(pi/2) = (pi/2)`

**Evaluating the values of `f(0),f(pi)` and f '(pi/2) yields: `f(0) = 0; f(pi) = -pi` and `f'(pi/2) = (pi/2).` **