Expert: Paul Klarreich - 7/1/2007

Question
find absolute max and min of y=xe^-2x^2 on interval [o,5]

Answer
Questioner:   ross
 
Question:  find absolute max and min of y=xe^-2x^2 on interval [o,5]
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Hi, Ross,

Your max and min values will occur at the critical points.  These include:

1. Points where f'(x) = 0.  [These are called Stationary Points.]
2. Points where f'(x) is undefined.  [These are called Singular Points.]
3. Endpoints of the interval of definition. [These are called, um... Endpoints.]

You find your extrema among those, usually by brute force testing.

For   y = x e^-2x^2  on  [0,5]

Differentiate using the product rule:

y' = e^-2x^2 + (x)(-4x e^-2x^2)

y' = e^-2x^2 - 4x^2 e^-2x^2

y' = e^-2x^2(1 - 4x^2)

Set that equal to zero and solve.  Since e^-2x^2 is never zero, we can ignore that factor.

1 - 4x^2 = 0
4x^2 = 1
x = 1/2   -- Well, plus-or-minus, but your interval is [0,5]

So your candidates are:

x = 0,  y = 0   << MIN
x = 1/2,  y = (1/2) e^-2(1/4) = (1/2) e^-1/2 = 0.3032  << MAX
x = 5, y = 5 e^-2(25) = 5 e^-50 is very small, but positive.