ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Studyinghard on April 18, 2011, 06:24:43 pm
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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
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As x -> infty, x^2/e^(ax) -> 0, so f(x) -> 0
x--> -infty, x^2/e^(ax) -> infty, so f(x) -> infty
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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
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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)
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so there are no local max and minimums?
EDIT: global max and min.
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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.
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how are you meant to sketch the graph if you dont know the "a" value :s
EFF THIS QUESTIONS SIGH
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if your graphing it just to see it visually, sub a with any number > 0
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it has to be full graphed :|
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how would you go about finding the point of inflection. I double diffed it and got 2/4x-x^2