Answer:
K= -3
Step-by-step explanation:
If g(x) be the function given by [tex]g(x)=x^{2},e^{kx}[/tex]
where k is a constant.
We have to find the value of k when the function has a critical point at x = [tex]\frac{2}{3}[/tex]
Since g'(x) = [tex]x^{2} e^{kx}[/tex]
Now for critical point we will find derivative of g(x) and equate the derivative to zero.
[tex]g'(x)=\frac{d}{dx}[(x^{2})(e^{kx})][/tex]
g'(x) = [tex](2x)(e^{kx})+(kx^{2})(e^{kx})[/tex]
[tex]g'(x)=(2x+kx^{2})(e^{kx})[/tex]
Now for [tex]x=\frac{2}{3}[/tex]
[tex]g'(\frac{2}{3})=0[/tex]
[tex][(2(\frac{2}{3})+k(\frac{2}{3})^{2}](e^{(\frac{2}{3}k)})=0[/tex]
Sin [tex]e^{x}[/tex] ≠ 0
therefore, [tex](\frac{4}{3}+\frac{4}{9}k)=0[/tex]
[tex]\frac{4k}{9}=-\frac{4}{3}[/tex]
[tex]k=\frac{-4}{3}(\frac{9}{4})=-3[/tex]
Therefore, k = -3 is the answer
