Let #h(x) = e^(-x) + kx#, where #k# is any constant. For what value(s) of #k# does #h# have critical points?

1 Answer

It has critical points only for #k>0#

First, let's compute the first derivative of #h(x)#.

#h^(prime) (x) = d/(dx)[e^(-x)+kx] = d/(dx)[e^(-x)] + d/(dx)[kx] = - e^(-x) +k#

Now, for #x_0# to be a critical point of #h#, it must obey the condition #h^(prime)(x_0) = 0#, or:

#h^(prime) (x_0) = -e^(-x_0)+k = 0 <=> e^(-x_0) = k <=> -x_0 = ln(k) <=>#
#<=> x_0 = -ln(k)#

Now, the natural logarithm of #k# is only defined for #k>0#, so, #h(x)# only has critical points for values of #k>0#.