# [SOLVED]-1.8.254 AP Calculus find k for critical point

#### karush

##### Well-known member
View attachment 9539

$\displaystyle g'=2xe^{kx}+e^{kx}kx^2$
we are given $x=\dfrac{2}{3}$ then
$\displaystyle g'=\dfrac{4}{3}e^\left(\dfrac{2k}{3}\right)+e^\left({\dfrac{2k}{3}}\right)\dfrac{4k}{9}$

ok something is ??? aren't dx supposed to set this to 0 to find the critical point

did a desmos look like k=-3 but ???

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#### skeeter

##### Well-known member
MHB Math Helper
critical points are defined points on the function where the derivative equals zero, or where the derivative is undefined ...

$g=x^2e^{kx}$

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

the derivative, $g'$, is defined for all values of $x$ in $g$'s domain, hence any critical value will be where $g'=0$

the first factor, $xe^{kx}$ equals zero only at $x=0$, no matter what $k$ is.

the second factor ...

$k\left(\frac{2}{3} \right) + 2 = 0 \implies k = -3$

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#### karush

##### Well-known member
that was indeed very helpful Mahalo

best place to be .... MHB