Question

State the absolute maximum value of g(t)=e^(-t^(2))

Answer 1

Is there any constraints on t or it can be any real number?

Answer 2

i'm guessing t can be any real number because the question doesn't have any constraints.

Answer 3

So take the derivative and set it equal to zero and solve since the maximum will be where the slope is zero.

Answer 4

Then we think like this:
If t is a +ve number and the value of t keeps on increasing -> we see that the g(t) keeps on reducing and tends to 0

Answer 5

@Kainui : it is absolute maximum and not LOCAL maximum

Answer 6

if t is a -ve number and t keeps decreasing the value of g(t) keeps on increasing and tends to infinity

Answer 7

I got the derivative as -2te^-2^2 but now I can not figure what t is when I set it equal to 0.

Answer 8

What I describe will find the absolute max. You should probably reread the question because in no way does this tend to infinity.

Answer 9

Perhaps I may be wrong, in that case I need to be corrected - but lets say what will be the value of g(t) when t = -50

Answer 10

@Kainnui: I am sry, I got you

Answer 11

\[\frac{ d }{ dt }e^{-t^2}=-2te^{-t^2}=0\]

So how do we solve for t? Well we know that anything to an exponent will never be 0, right? So the e^(-t^2) part will always be nonzero, so let's just divide both sides of our equation by it to get:

-2t=0

Now just divide by -2 to get

t=0.

Answer 12

So even if t is negative and keeps on decreasing the value of g(t) keeps reducing and so the maximum value must come at t=0

Answer 13

Okay, that makes a lot more sense! Thanks!

Answer 14

Yup, it's a nice even function. =)