How do you find values of k for which there are no critical points if #h(x)=e^(-x)+kx# where k is any constant?

1 Answer
Mar 10, 2015

First of all, a critical point of a function #h# is defined as a point in which the derivative #h'# equals zero. In your case, the derivative is
#-e^{-x}+k#
in fact, applying the chain rule, the derivative of #e^{-x}# is #e^{-x}# times the derivative of #-x#, which is #-1#; while the derivative of #kx# is of course #k#
The question is: for which values of #x# we have
#-e^{-x}+k=0#
easy manipulations bring us to the equivalent request
#e^{-x}=k#
Since #e^{-x}# is always (strictly) positive, this equation has solutions only if #k# is positive.

So, the final answer is: if #k\leq 0#, the derivative has no zeros, and so the function has no critical points.