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x=0, y=0 (so it goes through the origin)
x -> infinity, y -> 0 (since e^-x will decrease at a faster rate than x increases)
x -> -(infinity), y -> -(infinity), since e^-x is now +ve (and very large), and x is negative.
dy/dx = e^(-x) - x.e^(-x) = 0 (for turning point)
e^(-x) [1-x] = 0
x=1 for a turning point.
x=1, y = 1/e
d^2y/dx^2 = -e^(-x)[1-x] - e^(-x)
Substituting x=0, you get d^2y/dx^2 = -e^(-1) < 0 => Maximum at (1, 1/e)
I think this is what it looks like. Should be a smooth curve going through the origin, peaking at (1,1/e) and levelling off to y=0.
x -> infinity, y -> 0 (since e^-x will decrease at a faster rate than x increases)
x -> -(infinity), y -> -(infinity), since e^-x is now +ve (and very large), and x is negative.
dy/dx = e^(-x) - x.e^(-x) = 0 (for turning point)
e^(-x) [1-x] = 0
x=1 for a turning point.
x=1, y = 1/e
d^2y/dx^2 = -e^(-x)[1-x] - e^(-x)
Substituting x=0, you get d^2y/dx^2 = -e^(-1) < 0 => Maximum at (1, 1/e)
I think this is what it looks like. Should be a smooth curve going through the origin, peaking at (1,1/e) and levelling off to y=0.
x=0, y=0
x -> infinity , y -> 0
x -> -(infinity) , y -> infinity
dy/dx = 2x.e^(-x) - (x^2).e^(-x) = 0
xe^(-x)[(2-x)] = 0
x=0, or x=2 for stationary points.
x=2, y = 4e^(-2)
d^2/dx^2 = -e^(-x)[2x - x^2] + e^(-x)[2-2x]
x=0, d^2y/dx^2 = 2 >0 => Minimum at (0,0)
x=2, d^2y/dx^2 = -2e^(-2) <0 => Maximum at (2, 4e^-2)
The graph should look like this:
x -> infinity , y -> 0
x -> -(infinity) , y -> infinity
dy/dx = 2x.e^(-x) - (x^2).e^(-x) = 0
xe^(-x)[(2-x)] = 0
x=0, or x=2 for stationary points.
x=2, y = 4e^(-2)
d^2/dx^2 = -e^(-x)[2x - x^2] + e^(-x)[2-2x]
x=0, d^2y/dx^2 = 2 >0 => Minimum at (0,0)
x=2, d^2y/dx^2 = -2e^(-2) <0 => Maximum at (2, 4e^-2)
The graph should look like this:
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