Find $a>o$ so that the tangent line to the graph of $f(x)=x^2e^{-x}$ on $x=a$ goes through the origin.

Question

I believe these are the first steps: $$ f´(x)=2xe^{-x}-x^2 e^x$$ $$f´(a)= 2ae^{-a}-a^2 e^a $$ Since it goes through the origin, the tangent line on $a$ is given by: $$y-0 = f´(a)(x-0)...$$ but I don´t know what goes next in order to find the value of $a$... Thanks to everyone!

Answer 1

You made a mistake: $$ \begin{split} f(x) &= x^2 e^{-x}\\ f'(x) &= 2xe^{-x}-x^2e^{-x}\\ f'(a) &= \frac{2a-a^2}{e^a}. \end{split} $$ As you mention, you need to have the tangent line at $a$ go through the origin, so it must have equation $$ y = f'(a)x, $$ In other words, $L(a) = f'(a)a$. But since it is a tangent line, it must pass through $(a, f(a))$, i.e. $L(a) = f(a)$, so you end up with $$ f(a) = L(a) = f'(a)a. $$ Hence, you must solve $$ \frac{a^2}{e^a} = \frac{2a-a^2}{e^a}. $$

Answer 2

Hint:

As the tangent at $x=a$ also passes through the point $(a, f(a))$, you simply have to find the positive root(s) of the equation $$f(a)=af'(a).$$ Explicitly this equation becoes here $$a^2\mathrm e^{-a}=a(2a-a^2)\mathrm e^{-a}\iff a^2=a^2(2-a).$$