Question

let g 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=2/3

Answer 1

Answer:

K= -3

Step-by-step explanation:

If g(x) be the function given by $g(x)=x^{2},e^{kx}$

where k is a constant.

We have to find the value of k when the function has a critical point at x = $\frac{2}{3}$

Since g'(x) = $x^{2} e^{kx}$

Now for critical point we will find derivative of g(x) and equate the derivative to zero.

$g'(x)=\frac{d}{dx}[(x^{2})(e^{kx})]$

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

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

Now for $x=\frac{2}{3}$

$g'(\frac{2}{3})=0$

$[(2(\frac{2}{3})+k(\frac{2}{3})^{2}](e^{(\frac{2}{3}k)})=0$

Sin $e^{x}$ ≠ 0

therefore, $(\frac{4}{3}+\frac{4}{9}k)=0$

$\frac{4k}{9}=-\frac{4}{3}$

$k=\frac{-4}{3}(\frac{9}{4})=-3$

Therefore, k = -3 is the answer

Answer 2

Hello here is a solution