Question

How do you find the value `a > 0` such that the tangent line to `f(x) = x^2 * e^-x` passes through the origin `(0,0)`?

Answer 1

I assume that we want an `a > 0` such that the tangent to the graph of `f` at the point where `x=a` passes through `(0,0)`. See below.

Explanation:

`f(x) = x^2 * e^-x`

`f'(x) = 2x e^-x - x^2e^-x`

When `x = a`, the `y` coordinate of the graph is `f(a)=a^2/e^a`

and the slope of the tangent line is `m =f'(a) = (2a)/e^a-a^2/e^a`.

The tangent line has equation

`y = f(a)+f'(a)(x-a)`

`y=a^2/e^a + ( (2a)/e^a-a^2/e^a)(x-a)`

The line contains the point `(0,0)` when

`0 = a^2/e^a + ( (2a)/e^a-a^2/e^a)(-a)`.

Which is equivalent to

`(a^3-a^2)/e^a = 0`.

For all `a`, we know that `e^a != 0`, so the equation we are trying to solve is equivalent to

`a^3-a^2=0`

Clearly the solutions are `a=0` and `a=1`. We've been asked for the positive solution, so

a=1

Although we've answered the question, it might be interesting to see the graph of the function and this tangent line, so here it is:

graph{(x^2/e^x - y) (y-1/e x)= 0 [-0.15, 1.5356, -0.2965, 0.547]}

(The graph is how I double-checked my answer.)