limits part 2

[Studyinghard]

F(x) = x^2 e^-ax    a > 0
Use limits to describe the long-term behaviour of these functions

and how would you find the max and minimum if you dont have the value of a :S

[brightsky]

As x -> infty, x^2/e^(ax) -> 0, so f(x) -> 0
x--> -infty, x^2/e^(ax) -> infty, so f(x) -> infty

[Studyinghard]

how would you find local max and min if the value of a is unknown. i keep getting different values for diff a values :s

[brightsky]

There's no global maxima for the function unless you bound it in some way. As for the global minima:
f'(x) = x^2 * -a e^(-ax) + 2x * e^(-ax) = 0
e^(-ax) x(2-ax) = 0
e^(-ax) = 0 or x = 0 or 2-ax = 0
x = 0 is the one solution, yielding f(x) = 0. So that's the minimum.
x = 2/a is the other, yielding f(x) = (2/a)^2 e^(-a(2/a)), 4/a^2 * e^(-2) = 4/a^2e^2 (note this is only a local max)

[Studyinghard]

so there are no local max and minimums?

EDIT: global max and min.

[brightsky]

so there are no local max and minimums?

EDIT: global max and min.

if you sketch the graph, you'll see that there is a global minimum, but there is no global maximum since as x-> -infty, y grows to infinity.

[Studyinghard]

how are you meant to sketch the graph if you dont know the "a" value :s
EFF THIS QUESTIONS SIGH

[xZero]

if your graphing it just to see it visually, sub a with any number > 0

[Studyinghard]

it has to be full graphed :|

[Studyinghard]

how would you go about finding the point of inflection. I double diffed it and got 2/4x-x^2