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.

#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

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.)

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Answer 2

To find the value of a > 0 such that the tangent line to f(x) = x^2 * e^-x passes through the origin (0,0), we need to determine the slope of the tangent line at the point (0,0) and then solve for a.

The slope of the tangent line at a given point can be found by taking the derivative of the function and evaluating it at that point. In this case, we need to find the derivative of f(x) = x^2 * e^-x.

Using the product rule and the chain rule, we can find the derivative as follows:

f'(x) = (2x * e^-x) + (x^2 * (-e^-x)) = 2x * e^-x - x^2 * e^-x = (2x - x^2) * e^-x

To find the slope of the tangent line at (0,0), we substitute x = 0 into the derivative:

f'(0) = (2(0) - (0)^2) * e^-(0) = 0

Since the slope of the tangent line at (0,0) is 0, we can set up the equation for the line passing through the origin:

y - y1 = m(x - x1)

Substituting (0,0) as the point and 0 as the slope:

y - 0 = 0(x - 0) y = 0

Therefore, the value of a > 0 that satisfies the condition is a = 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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