Respuesta :
Answer:
The equation of the tangent line for the function [tex]f\left(x\right)=x^{2} e^{- x}[/tex] at [tex]\left(2,\:\frac{4}{e^2}\right)[/tex] is [tex]y=\frac{4}{e^{2}}[/tex]
Step-by-step explanation:
To find the tangent line to [tex]f\left(x\right)=x^{2} e^{- x}[/tex] at [tex]x_0=2[/tex]
Firstly, find the slope of the tangent line, which is the derivative of the function, evaluated at the point: [tex]m=f^{\prime}\left(2\right)[/tex]
[tex]\mathrm{Apply\:the\:Product\:Rule}:\quad \left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'\\\\f=x^2,\:g=e^{-x}[/tex]
[tex]\frac{d}{dx}\left(x^2e^{-x}\right)=\frac{d}{dx}\left(x^2\right)e^{-x}+\frac{d}{dx}\left(e^{-x}\right)x^2\\\\=2e^{-x}x-e^{-x}x^2[/tex]
Next, evaluate the derivative at the given point to find the slope.
[tex]\mathrm{Plug\:}x=2\mathrm{\:into\:the\:equation\:}2e^{-x}x-e^{-x}x^2\\\\2e^{-2}\cdot \:2-e^{-2}\cdot \:2^2=0[/tex]
[tex]m=f^{\prime}\left(2\right)=0[/tex]
Finally, the equation of the tangent line is [tex]y-y_0=m(x-x_0)[/tex]
Plugging the found values, we get that
[tex]y-\left(\frac{4}{e^{2}}\right)=0\left(x-\left(2\right)\right)\\\\y=\frac{4}{e^{2}}[/tex]
The equation of the tangent line for the function [tex]f\left(x\right)=x^{2} e^{- x}[/tex] at [tex]\left(2,\:\frac{4}{e^2}\right)[/tex] is [tex]y=\frac{4}{e^{2}}[/tex]
