# Homework Help: Describing the exponential function y=2e^-0.5x^2

1. Jun 15, 2011

### jay17jay

1. The problem statement, all variables and given/known data
Use the function y=2e^-0.5x^2 to answer the following questions
a) state the domain
b) Determine the intercepts, if any
c) Discuss the symmetry of the graph
d) Find any asymptotes
e)determine the intervals of increase and decrease
f)what is the maxima and/or minima
g) where is the curve concave upwards and downwards?
h) locate the points of inflection

2. Relevant equations

3. The attempt at a solution

im having the most difficult with understanding this function since i have never worked with anything like this but this is what i have so far, i would appreciate help me with the steps i have done wrong or dont know how to complete

a) state the domain
-22<x<22

b) Determine the intercepts, if any
y-intercept: y=2 the point is (0.2)
x-intercept: there are no x-intercepts

c) Discuss the symmetry of the graph
the graph is symmetric with respect to the y-axis, as the original equation is unchanged when x is replaced by -x

d) Find any asymptotes
vertical asymptotes: as x approaches -22, e^x decreases to 0; and as x approaches 22 e^x decreases to 0. Therefore the y-axis is a vertical asymptote.
vertical asymptotes: there are no vertical asymptotes as the graph never crosses the x axis

e)determine the intervals of increase and decrease
y=2e^-0.5x^2
dy/dx=2e^-0.5x^2 * d/dx (-0.5x^2)
dy/dx= -3xe^-0.5x^2
since the sign of the first derivative is the different then the sign of x
dy/dx<0 when x>0, and dy/dx<0 when x>0

( im pretty sure that ones wrong)

f)what is the maxima and/or minima
Max: (0,2)
Min: 0 (not sure about that)

g) where is the curve concave upwards and downwards?
dy/dx=-3xe^-0.5x^2
second derivative:
-3x d/dx(e^-0.5x^2) + e^-0.5x^2 d/dx (-3x)
-3x * e^-0.5x^2 * 2x + e^-0.5x^2 * -3
= -3e^-0.5x^2 (3x^2 +1)
the second derivative is negative meaning the graph is concave down

h) locate the points of inflection
there are no inflection points

i know this is so long but i would really appreciate help on this problem i would like to learn how to do this right.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 15, 2011

### Staff: Mentor

Do you mean y = 2e^(-.5x^2)?
How did you get this? The domain is all real numbers.
We usually represent points in the plane as a pair of numbers separated by a comma. The y-intercept is (0, 2).
You have two sections for vertical asymptotes and none for horizontal asymptotes. There are several things wrong here:
1) The function is not e^x, which is positive for all real x. e^(-22) is larger than zero. e^22 is a very large number.
2) Whether the graph crosses the x-axis has nothing to do with vertical or horizontal asymptotes.
Your derivative is incorrect. It should be dy/dx = -2xe^(-.5x^2)

3. Jun 15, 2011

### jay17jay

thankyou for your help yes it is 2e^(-o.5x^2)
and iv looked over your comments and tried this again this is what i came up with still have difficulty on some parts

a) state the domain
- the domain is all real numbers

b) Determine the intercepts, if any
- y-inter (0,2)
- x-inter are non

c) Discuss the symmetry of the graph
- the graph is symmetric with respect to the y-axis, as the original equation is unchanged when x is replaced by -x

d) Find any asymptotes
- im really confused how to do this now??

e)determine the intervals of increase and decrease
- y=2e^-0.5x^2
dy/dx=2e^-0.5x^2 * d/dx (-0.5x^2)
dy/dx= -2xe^-0.5x^2
since the sign of the first derivative is the different (neg) then the sign of the function

-the graph rises when xE (-22,0) and falls when xE(22,0). the turning point occurs when x=0, y=2.

f)what is the maxima and/or minima
Max: (0,2)

g) where is the curve concave upwards and downwards?
????
im having trouble with the second derivative

i have:
= -2x d/dx (e^-0.5x^2) + e^-0.5x^2 d/dx(2x)
= -2x * (-0.5x^2)e + e^(-0.5x^2) * -2
(x^3)e + -2e^(-0.5x^2)

but im not sure if thats right or where to go from there

i do know that the second derivative should be negative for all x, becuase the graph is concave downwards

h) locate the points of inflection
inflection point at (0,2) ???

4. Jun 15, 2011

### Ray Vickson

Why do you restrict x to between -22 and 22? What prevents me from substituting x = 175 into the formula for f(x), Why can't I put x = -500? Maple can do it:
f:=exp(-x^2/2):
evalf(subs(x=75,f));
-1221
0.3521840206 10
(this is 0.3521840206e(-1221))
evalf(subs(x=-500,f));
-54286
0.1547968410 10
(this is 0.1547968410e(-54286))

Of course, in something like a spreadsheet the program will print out "0" (or else give an underflow warning), but it is not really zero; it is just smaller then the smallest number that program can handle.

Theoretically, f(x) = exp(-x^2/2) is defined for ALL real x; there is not upper or lower limit.

RGV

5. Jun 15, 2011

### Staff: Mentor

For vertical asymptotes, are there any values of x for which y is undefined?
For horizontal asymptotes, do the y values approach a specific value as x approaches infinity or neg. infinity?
As already mentioned, you need to write this using parentheses.
y = 2e^(-.5x^2)
You didn't complete your thought here.

dy/dx consists of three factors: -2, x, and e^(-.5x^2).
-2 is (obviously) always negative.
x can be negative, zero, or positive.
e^(-.5x^2) is ALWAYS positive.

On what interval(s) is dy/dx negative?
On what interval(s) is dy/dx positive?
Where is dy/dx equal to zero?
I think that you are using a graphing calculator, and that it is leading you astray. The numbers 22 and -22 play NO ROLE WHATSOEVER in this function.
First off, never start your work with =. You are finding the second derivative, so the first line could start of with y'' = ...

In your first line, it should be
= -2x d/dx (e^-0.5x^2) + e^-0.5x^2 d/dx(-2x)

In your second line, things go downhill. The derivative of e^(-.5x^2) is -xe^(-.5x^2).
Neither statement is true. The 2nd derivative is NOT always negative, and the graph is NOT concave down everywhere.