Question

Let g be the function given by g(x)=x^2*e^(kx) , where k is a constant. For what value of k does g have a critical point at x=2/3?

Answer 1

A critical point occurs for points at which the derivative will be zero (such as at the vertex of a parabola) or when the derivative does not exist (like at the "vertex" of the absolute value function).

Differentiate the given function and set equal to 0:
\[g(x)=x^2e^{kx}~~\Rightarrow~~g'(x)=2xe^{kx}+kx^2e^{kx}\]
\[\begin{align*}2xe^{kx}+kx^2e^{kx}&=0\\
xe^{kx}\left(2+kx\right)&=0\\
x\left(2+kx\right)&=0&e^{kx}\not=0 \text{ for any } x
\end{align*}\]
This gives you \(x=0\) and \(x=-\dfrac{2}{k}\) as critical points. So, \(g\) will have a critical point at \(x=\dfrac{2}{3}\) when \(k=-3\).