How do you find the critical numbers of #f(x)=sinxcosx#?

Question

Cameron Alexander · Accepted Answer

Critical numbers of #f(x)# occur when

# x \ \ = pi/4 + (npi)/2 \ \ \ \ n in ZZ#
eg #x = +-pi/4, +-3pi/4, +-5pi/4, ... #

We have:

# f(x) = sinx cos x #

Differentiating wrt #x# using the product rule:

# f'(x) = (sinx) (-sinx) + (cosx) (cosx) #
# " "= cos^2x -sin^2x #
# " "= cos(2x) #

At a critical point #f'(x)=0#

# :. cos (2x) = 0 #
# :. 2x = pi/2 + npi #
# :. x \ \ = pi/4 + (npi)/2 \ \ \ \ n in ZZ#

Hence critical numbers of #f(x)# occur when

# x \ \ = pi/4 + (npi)/2 \ \ \ \ n in ZZ#
eg #x = +-pi/4, +-3pi/4, +-5pi/4, ... #

We can see these values of #x# correspond to max/min of the graph #y=sinx cos x #:

Samantha Adkison · Answer

undefined To find the critical numbers of $ f(x) = \sin(x)\cos(x) $, you need to find the values of $ x $ where the derivative $ f'(x) $ is equal to zero or undefined.

1. Calculate the derivative $ f'(x) $:
$$ f'(x) = \cos(x)\cos(x) - \sin(x)\sin(x) $$
$$ f'(x) = \cos^2(x) - \sin^2(x) $$

2. Use the trigonometric identity $ \cos^2(x) - \sin^2(x) = \cos(2x) $:
$$ f'(x) = \cos(2x) $$

3. Set $ f'(x) $ to zero and solve for $ x $:
$$ \cos(2x) = 0 $$

4. Solve for $ x $:
$$ 2x = \frac{\pi}{2} + n\pi $$ where $ n $ is an integer.

$$ x = \frac{\pi}{4} + \frac{n\pi}{2} $$