Optimizing Trigonometric Function on Interval

[Hockeystar]

Homework Statement

Find the max/min of the function for [0,pi/2]

f(x)= kcos²xsin(x)

Homework Equations

Product Rule and chain rule

The Attempt at a Solution

f^'(x)= k(2cosx(-sinx)(sinx)+cos²xcosx)

f^'(x)= kcosx(-2sin²x+cos²x)

Set eqaution to 0

(2sin²x+cos(x)²) = 0

2(1-cos²x)+cos(x)² = 0

cos²x = 2
?
Now I'm stuck. There should be a max or min.

 

[ystael]

f^'(x)= kcosx(-2sin²x+cos²x)

Set eqaution to 0

(2sin²x+cos(x)²) = 0

You appear to have made two errors here. You neglected to account for the other factor, that perhaps $$\cos x = 0$$. You also seem to have dropped a minus sign; if the second factor is zero, you should have $$-2 \sin^2 x + \cos^2 x = 0$$.

You may have an easier time with this second factor if you rewrite it so that only one trig function of $$x$$ appears.

 

[Hockeystar]

Man I hate when I forget a negative sign. I also scrwed up the interval is supposed to be (0,pi/2) instead of [0,pi/2]. Thank you for pointing that out.