SOLUTION: For each x>0, let m(x) be the slope of the line which goes through the point (0,0) and the point (x,y) on the curve y=(x^2)(e^(-0.25x)). What is the largest possible value for the

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Question 1052798: For each x>0, let m(x) be the slope of the line which goes through the point (0,0) and the point (x,y) on the curve y=(x^2)(e^(-0.25x)). What is the largest possible value for the m(x)?
note: it is e raised to (-0.25x)
Work: I believe I've taken the correct first derivative and second derivative of the curve using chain rule and product rule, But I don't know what the question is asking for or how to answer it. y=(x^2)(e^(-0.25x)) -> 1st derivative y'=2xe^((-1/4)x) + x^2((-1/4)(e^(-1/4)x)) -> 2nd derivative:
y''= 2(e^((-1/4)x) + 2x(e^((-1/4)x)(-1/4) + 2x((-1/4)e^((-1/4)x) + x^2((-1/16)(e^(-1/4)(x)))
If possible please check my derivatives, thanks in advance.
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
For each x>0, let m(x) be the slope of the line which goes through the point (0,0) and the point (x,y) on the curve y=(x^2)(e^(-0.25x)). What is the largest possible value for the m(x)?
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y=
y' =
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y" =
y" =
y" =
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Set y" = 0

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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=128 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 13.6568542494924, 2.34314575050762. Here's your graph:

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