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

Apr 26, 2016

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,

$y ' = - 2 x {e}^{- {x}^{2}} = 0$, at x = 0.

$y ' ' = - 2 {e}^{- {x}^{2}} - 2 x \left(- 2 x\right) {e}^{- {x}^{2}} = - 2$, at x = 0.

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

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

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

As, $x \to \pm \infty , f \left(x\right) \to 0$

Interestingly, the graph of $y = f \left(x\right) = {e}^{- {x}^{2}}$ is the scaled $\left(1 u n i t = \frac{1}{\sqrt{2 \pi}}\right)$ normal probability curve, for the normal probability distribution, with mean = 0 and standard deviation $= \frac{1}{\sqrt{2}}$