Question

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

Answer 1

`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)`