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

1 Answer
Apr 26, 2016

Answer:

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#