Question

How do you find the critical points for `f(x)=sinx cosx` and the local max and min in #0 ≤ x < 2pi?

Answer 1

`x = pi/4`: Local max
`x = (3pi)/4`: Local Min

Explanation:

Differentiate:

`f'(x) = cosx(cosx) + sinx(-sinx)`

`f'(x) = cos^2x - sin^2x`

`f'(x) = cos2x`

Critical points will occur when the derivative equals `0` or is undefined.

There will be no undefined points on the function, since it's continuous.

`0 = cos2x`

`2x = pi/2 and (3pi)/2`

`x= pi/4 and (3pi)/4`

Now check the signs to the left and right of each of these points.

`f'(0) = cos(2(0)) = cos(0) = 1`

`f'(pi/2) = cos(2(pi/2)) = cos(pi) = -1`

So `x = pi/4` is a local maximum (the derivative is increasing, then decreasing).

`f'((2pi)/3) = cos(2((2pi)/3)) = cos((4pi)/3) = -1/2`

`f'((5pi)/6) = cos(2((5pi)/6)) = cos((5pi)/3) = 1/2`

So, `x = (3pi)/4` is a local minimum.

Hopefully this helps!