Answer :
Answer:
The answer is "[tex]cirtical\ points \ (x,g(x))\equiv (e,0),(e+3,\frac{27}{e^3})[/tex]"
Step-by-step explanation:
Given:
[tex]g(x) = (x-e)^3e^{-(x-e)}[/tex]
Find critical points:
[tex]g(x) = (x-e)^3e^{(e-x)}[/tex]
differentiate the value with respect of x:
[tex]\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r} \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3][/tex]
critical points [tex]g'(x)=0[/tex]
[tex]\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3[/tex]
So,
The critical points of [tex](x,g(x))\equiv (e,0),(e+3,\frac{27}{e^3})[/tex]