Question

find the vaule of k such that f(x)=x^2e^kx nhas a crtical ponit at x=2

Answer 1

\[f(x)=x^2e^{kx}\\
f'(x)=2xe^{kx}+kx^2e^{kx}\]
Yo're told that x = 2 is a critical value, so when you set f '(x) = 0, x is a solution:
\[2xe^{kx}+kx^2e^{kx}=0\text{ holds for $x=2$, so}\\
2(2)e^{2k}+k(2)^2e^{2k}=0\\
4e^{2k}+4ke^{2k}=0\\
e^{2k}(4 + 4k)=0\]