What are the critical points of #f(x) =2e^(3x)-3xe^(2x)#?

1 Answer
Nov 24, 2015

Answer:

#3e^{2x}(2e^x - 1 - 2x)#

Explanation:

To compute critical points of a function, you need its first derivative.

Derivative of #2e^{3x}#:

We need two rules: first of all, the derivative doesn't care about multiplicative factors, so

#d/dx (2e^{3x}) = 2\ d/dx e^{3x}# .

To compute #d/dx e^{3x}# we need the chain rule, since #e^{3x}# is a composite function #f(g(x))#, where #f(x)=e^x#, and #g(x)=3x#.

The chain rule states that

#(f(g(x)))' = f'(g(x)) * g'(x)#

Which in your case becomes

#e^{3x} * 3 = 3e^{3x}#

So, the whole derivative is #6e^{3x}#

Derivative of #3xe^{2x}#:

In addition to what we saw before, we need to add the rule for deriving a product of two functions, which is

#(f(x)*g(x))' = f'(x)*g(x) + f(x)*g'(x)#

So, we have that

#d/dx 3xe^{2x} = 3e^{2x} + 3x*2*e^{2x} = 3e^{2x} + 6x*e^{2x}#

Finally, sum up the two pieces: we have

#6e^{3x}-3e^{2x} - 6x*e^{2x}#

And we can factor #3e^{2x}#:

#3e^{2x}(2e^x - 1 - 2x)#