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)#?

1 Answer
May 27, 2016

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.)