Find the x-coordinate of the stationary points of the curve and determine the nature of these stationary points.

Question

The equation of a curve is $y=x^2e^{-x}$.

1. Find the x-coordinate of the stationary points of the curve and determine the nature of these stationary points.
2. Show that the equation of the normal to the curve at the point where $x=1$ is $e^2x+ey = 1+e^2$.

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This is the full question I am having difficulty solving, I simply don't know where to begin. I moved back to my home country because of covid and now I am doing self-studying I don't know how to solve this any help wold be great and much appreciated.

Answer 1

1. The stationary point of a curve is simply the point where the derivative vanishes. You can see the behavior of these points from this link. https://mathworld.wolfram.com/StationaryPoint.html

The first question asks for the $x$-coordinates of the stationary points. You can start by taking the derivative of the curve. Then, you should equate it to zero and solve the equation to find the roots.

$$\frac{dy}{dx}=2xe^{-x}-x^2e^{-x}=e^{-x}x(2-x)=0\space \Rightarrow \space x=0 \space\text{or}\space x=2$$

At $x=0$ and $x=2$, the derivative is zero which means they are the stationary points.

2. We should verify the equation for the normal line of the curve at $x=1$. At this point $y$-coordinates of the normal line and the original curve is the same. Since tangent and normal lines to a curve at some particular point are perpendecular to each other, multiplication of their slopes yields $-1$. At $x=1$, the slope of the tangent line is $\frac{1}{e}$. This means the slope of the normal line is $-e$. Below you may see the normal line equation. $$ax+b=y$$ $$x_0=1,\space \space a=-e, \space \space y_0=\frac{1}{e}$$ If we substitute these values into the line equation, we find $b$ to be $e+\frac{1}{e}$. $$-ex+e+\frac{1}{e}=y \space \rightarrow \space e^2+1=e^2x+ey$$

Answer 2

$y=x^2e^{-x} \implies y'(x)=2xe^{-x}-x^2e^{-x} \implies y'(1)=e^{-1}.$ So the slope of normal at $x=1$ is $m=-1/y'(1)=-e$. So the equation of line having slope $-e$ and passing through the point $(1,e^{-1})$ is $$y-e^{-1}=-e(x-1) \implies ey+e^2 x=1+e^2$$