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

1 Answer
Jan 24, 2017

Answer:

#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!