How do you find the inflection points for the function #f(x)=e^(3x)+e^(-2x)#?

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Answer 1 · 2018-03-16T15:38:24

There are no points of inflections.

First, calculate the first and the second derivatives

#f(x)=e^(3x)+e^(-2x)#

The domain of #f(x)# is #x in RR#

#f'(x)=3e^(3x)-2e^(-2x)#

#f''(x)=9e^(3x)+4e^(-2x)#

The points of inflections are when #f''(x)=0#

#f''(x)=9e^(3x)+4e^(-2x)=9e^(3x)+4/e^(2x)#

#=(9e^(5x)+4)/e^(2x)#

#AA x in RR, f''(x)>0#

So there are no points of inflections

graph{e^(3x)+e^(-2x) [-6.52, 4.58, 0.71, 6.26]}