What are the critical points of #f (x) = e^x + e^-(6x)#?

1 Answer
Dec 23, 2015

Answer:

#c=(ln6)/7#

Explanation:

A critical point #c# is when #f'(c)=0# or does not exist.

To find #f'(x)#, know that according to the chain rule #d/dx[e^u]=e^u*(du)/dx#.

#f'(x)=e^x-6e^(-6x)#

Now, we have to know for what values of #x# this equals #0# or is undefined.

An easy way to do this is to turn it into a fraction—if a fraction, #f'(x)=0# when the numerator #=0# and will be undefined when the denominator #=0#.

#f'(x)=e^(-6x)(e^(7x)-6)#

#f'(x)=(e^(7x)-6)/e^(6x)#

#f'(x)=0# when #e^(7x)-6=0#. Solve this.

#e^(7x)=6#
#7x=ln6#
#x=(ln6)/7#

This is a critical value.

#f'(x)# isn't defined when #e^(6x)=0#. However, #e^(6x)# never can equal #0# so the function is never undefined.

Thus, the only critical point is #c=(ln6)/7~~0.256#.

graph{e^x+3^(-6x) [-10.04, 9.96, -5.04, 4.96]}