# `f(x) = e^(-x^2)` (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of...

`f(x) = e^(-x^2)` (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of concavity and the inflection points. (e) Use the information from parts (a)-(d) to sketch the graph of `f`.

### Textbook Question

Chapter 4, 4.3 - Problem 49 - Calculus: Early Transcendentals (7th Edition, James Stewart).
See all solutions for this textbook.

Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on

and of course `+-1/sqrt(2)` are inflection points

Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on

(a) there is no vertical asymptotes, f is defined and differentiable everywhere.

When `x->+-oo,` `-x^2->-oo` and `f(x)->+0,` so there is one horizontal asymptote y=0.

(b) `f'(x) = -2x*e^(-x^2).` It is >0 for x<0 and <0 for x>0,

so f(x) is increases on `(-oo, 0)` and decreases on `(0, +oo).`

(c) therefore there is one local maximum x=0. f(0)=1.

(d) `f''(x) = e^(-x^2)*(-2 + 4x^2).`

This is <0 for x on `(-1/sqrt(2), 1/sqrt(2))` , f is concave downward there. And f''>0 on` (-oo, -1/sqrt(2))` and on `(1/sqrt(2), +oo),` f is concave upward there.

(e)

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)