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

Nov 24, 2015

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

Explanation:

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

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

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

$\frac{d}{\mathrm{dx}} \left(2 {e}^{3 x}\right) = 2 \setminus \frac{d}{\mathrm{dx}} {e}^{3 x}$ .

To compute $\frac{d}{\mathrm{dx}} {e}^{3 x}$ we need the chain rule, since ${e}^{3 x}$ is a composite function $f \left(g \left(x\right)\right)$, where $f \left(x\right) = {e}^{x}$, and $g \left(x\right) = 3 x$.

The chain rule states that

$\left(f \left(g \left(x\right)\right)\right) ' = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

${e}^{3 x} \cdot 3 = 3 {e}^{3 x}$

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

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

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

$\left(f \left(x\right) \cdot g \left(x\right)\right) ' = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

So, we have that

$\frac{d}{\mathrm{dx}} 3 x {e}^{2 x} = 3 {e}^{2 x} + 3 x \cdot 2 \cdot {e}^{2 x} = 3 {e}^{2 x} + 6 x \cdot {e}^{2 x}$

Finally, sum up the two pieces: we have

$6 {e}^{3 x} - 3 {e}^{2 x} - 6 x \cdot {e}^{2 x}$

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

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