3.3.291 AP Calculus Exam Problem solve for k

[karush]

Let $g(x)$ be the function given by $g(x) = x^2e^{kx}$ , where k is a constant. For what value of k does g have a critical point at $x=\dfrac{2}{3}$?
$$(A)\quad {-3}
\quad (B)\quad -\dfrac{3}{2}
\quad (C)\quad -\dfrac{3}{2}
\quad (D)\quad {0}
\quad (E)\text{ There is no such k}$$

ok I really did not know how you could isolate k to solve this.

if you plug in $x=\dfrac{2}{3}$ then g(2/3)

if you plug in $x=\dfrac{2}{3}$ then $g(2/3)=\dfrac{4}{9}e^\dfrac{2k}{3}$

 

[skeeter]

a critical point is a value(s) of $x$ where the derivative equals zero or is undefined

$g(x) = x^2 \cdot e^{kx}$

$g'(x) = x^2 \cdot ke^{kx} + 2x \cdot e^{kx}$

$x^2 \cdot ke^{kx} + 2x \cdot e^{kx} = 0 \implies xe^{kx}(kx + 2) = 0$

at $x = \dfrac{2}{3}$, the first factor is always > 0

the second factor ...

$\left(k \cdot \dfrac{2}{3} + 2 \right) = 0 \implies k = ?$