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

Question

Emily Ammerman · Accepted Answer

#c=(ln6)/7#

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]}

Evelyn Agard · Answer

undefined To find the critical points of $ f(x) = e^x + e^{-6x} $, you need to first find its derivative and then solve for where the derivative equals zero.

The derivative of $ f(x) $ with respect to $ x $ is:

$$ f'(x) = e^x - 6e^{-6x} $$

To find critical points, set $ f'(x) $ equal to zero and solve for $ x $:

$$ e^x - 6e^{-6x} = 0 $$

$$ e^x = 6e^{-6x} $$

$$ e^x = \frac{6}{e^{6x}} $$

$$ e^{7x} = 6 $$

$$ x = \frac{1}{7} \ln(6) $$

Therefore, the critical point of $ f(x) $ is $ x = \frac{1}{7} \ln(6) $.