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

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Answer 1 · 2015-11-24T11:54:48

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

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 $3e^{2x}$ :

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

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