Point of maxima for y=xe^-2x?

Question

FInd the point where the maxmia exists for the equation y=xe^-2x? How do you know it is a maximum at this point?

Is there a way to do this problem other than just looking at the graph itself?

Answer 1

GIVEN

y=xe^-2x

Take the derivative of the above given function,

(dy/dx) = x(e^-2x)(-2) + e^-2x

and to solve for "x" that will yield the maximum value for the function, set (dy/dx) = 0, i.e.,

(dy/dx) = 0 = x(e^-2x)(-2) + e^-2x

Simplifying the above,

-2x + 1 = 0

and solving for "x",

x = -1/2.

To determine if the value for "x" determined above (x = -1/2) will yield the maximum for the function, take the second derivative of the original function (d^2y/dx^2), i.e.,

d^2y/dx^2 = -2x(e^-2x)(-2) + (e^-2x)(-2)

d^2y/dx^2 = -4x(e^-2x) - 2(e^-2x)

d^2y/dx^2 = -2(e^-2x)[2x +  1]

Since d^2y/dx^2 < 0 (negative), then the calculated value of "x" will indeed maximize the given function.

Hope this helps.