Question

What are the extrema of `f(x) = e^(-x^2)` on `[-.5, a]`, where `a > 1`?

Answer 1

f(x) > 0. Maximum f(x) isf(0) = 1. The x-axis is asymptotic to f(x), in both directions.

Explanation:

f(x) > 0.

Using function of function rule,

`y'=-2xe^(-x^2) =0`, at x = 0.

`y''=-2e^(-x^2)-2x(-2x)e^(-x^2)=-2`, at x = 0.

At x = 0, y' = 0 and y'' < 0.

So, f(0) = 1 is the maximum for f(x), As required, . `1 in [-.5, a], a > 1`.

x = 0 is asymptotic to f(x), in both the directions.

As, `xto+-oo, f(x)to0`

Interestingly, the graph of `y = f(x)=e^(-x^2)` is the scaled `(1 unit =1/sqrt (2 pi))` normal probability curve, for the normal probability distribution, with mean = 0 and standard deviation `= 1/sqrt 2`