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

1 Answer
Feb 1, 2017

Answer:

Critical numbers of #f(x)# occur when

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

Explanation:

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 #:

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