1.8.1 AP Calculus find k for critical point

[karush]

$\displaystyle g'=2xe^{kx}+e^{kx}kx^2$
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 ?
[DESMOS]advanced: {"version":7,"graph":{"showGrid":false,"squareAxes":false,"viewport":{"xmin":-0.6595622289691282,"ymin":-0.7302941176470592,"xmax":2.1639671827955773,"ymax":1.1520588235294116}},"expressions":{"list":[{"type":"expression","id":"graph1","color":"#2d70b3","latex":"y_1=x^2e^{kx}"},{"type":"expression","id":"2","color":"#388c46","latex":"k=-3","hidden":true},{"id":"6","type":"table","columns":[{"values":["\\frac{2}{3}"],"hidden":true,"id":"4","color":"#000000","latex":"x"},{"values":[""],"id":"5","color":"#c74440","latex":"y_1"},{"values":[""],"id":"7","color":"#2d70b3","latex":"k"}]}]}}[/DESMOS]

 

[skeeter]

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 ...

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

 

[karush]

that was indeed very helpful Mahalo

best place to be ... MHB