What are the critical values, if any, of # f(x)= sin|x|#?

1 Answer
Dec 20, 2015

Answer:

#{pi/2 + npi|n in ZZ} uu {0}#

Explanation:

A critical value #c# of a function #f(x)# is a value where #f'(c) = 0# or #f'(c)# is undefined.

The challenge here is to find #f'(x)# as we do not have a simple rule to handle #|x|# in differentiation. We could break it into a piecewise function split at #x = 0#, but let's use another trick in this case.

For all #x in RR# we have #|x| = sqrt(x^2)#. Thus we can rewrite the function as

#f(x) = sin(sqrt(x^2))#

Then, applying the chain rule, we have

#f'(x) = d/dx sin(sqrt(x^2))#

#= cos(sqrt(x^2))(d/dxsqrt(x^2))#

#= cos(sqrt(x^2))(1/(2sqrt(x^2)))(d/dxx^2)#

#= cos(sqrt(x^2))(1/(2sqrt(x^2)))(2x)#

#= (xcos(|x|))/|x|#

Then, #f'(x)# is undefined when #x=0#, and

#f'(x) = 0 <=> cos(|x|) = 0#

#<=> x = pi/2 + npi# where #n in ZZ#

Thus the critical values of #sin|x|# are

#{pi/2 + npi|n in ZZ} uu {0}#