Question

5.(15points) Letf(x)=e^−x^2.
(a) What horizontal or vertical asymptotes does f(x) have? (b) Find the intervals of increase or decrease.
(c) Find the local maximum and minimum values. (d) Where is f(x) concave downward?
(e) Where are the inflection points?

i have that th ederivatice is 2x e^(-x^2)

Answer 1

it does not have vertical asymptotes since e^(x^2) is never zero

Answer 2

u can find the horizontal asimptotes finding the limit when x goes to INF
and u will see that the horizontal asymtota is y=0

Answer 3

yeah i got that. I'm having a hard time finding the critical numbers other than 0

Answer 4

it just have 0 like its critical number

Answer 5

so for every negative x the derivative is <o

Answer 6

the derivative is always negative
so it is decreasing always except in x=0

Answer 7

decreas = -infinity, 0
increase 0, infinity

Answer 8

u forgot the (-2x) in the derivative

Answer 9

ohright
(-infinity, 0) and (0, infinity)

Answer 10

wait there is a probkem i made a mistake

Answer 11

the derivative is positive in a interval

Answer 12

so was my first answer right?

Answer 13

i know teh steps to solve this but what is throwing me off is the
e^-x^2 because i don't know how this affects the fuction.

Answer 14

\[\huge 0=-2xe^{-x^2}\]

\[\large 0=-2x \qquad\rightarrow\qquad x=0\]\[\large 0=e^{-x^2} \qquad\rightarrow\qquad e^{-x^2}\ne0\]

Yah you've got the right idea melo, x=0 is our only critical point.
The exponential can never be zero.
It just makes the function ...curvy.. and.. stuff :p

Answer 15

ok what about my increasing and decreasing intervals

Answer 16

So our critical point is at x=0.
So the only place the function `can` change from increasing to decreasing or the other way around, is at x=0.

So let's plug in a test point to see what's happening on the left side of x=0,
and then on the right side as well.

f'(-1)=?

Answer 17

-2 e^-1

Answer 18

Woops you missed an x there! D:

\[\huge f'(-1)=-2(-1)e^{-1}\]

Answer 19

We don't actually care about the `value` this is producing.
We only care about it's `sign`.

That being said, we can actually ignore the exponential part, since it will always be positive.
So just plug in for the other part, and figure out the sign.

\[\large f'(-1)=-2(-1)\]

Answer 20

oh i missed the first negative sign actually -_- ok yes

Answer 21

oh i see :o

Answer 22

so the exponential sign is always positive.

Answer 23

x<o is increasing and x>o is decreasing

Answer 24

Yah that sounds right!
Understand how to find inflection points?

Answer 25

which mean at o it is a local max

Answer 26

why are you using o? lol .. you mean 0? :p

Answer 27

yes thats what i mean sorry
x=0 is local max
the second derivative is -2e^(-x^2)+xe^(-x^2)-2x right

Answer 28

\[\large f''(x)=-2e^{-x^2}-2xe^{-x^2}(-2x)\]

Answer 29

You apply the product rule.
It looks like you did the first term correctly, simply differentiating the x giving you 1 in it's place.

I'm not sure what you did beyond that though :o
The next term should just give you another -2x, similar to the way you got your first derivative.

Answer 30

\[\large f'(x)=-2xe^{-x^2}\]

\[\large f''(x)=\color{royalblue}{\left(-2x\right)'}e^{-x^2}+(-2x)\color{royalblue}{\left(e^{-x^2}\right)'}\]

\[\large f''(x)=\color{orangered}{\left(-2\right)}e^{-x^2}+(-2x)\color{orangered}{\left[e^{-x^2}(-2x)\right]}\]

Understand how that works? :o

Answer 31

yes sometimes i confuse myself when there is something in front of the x.I see my mistake.
so to find the concavity with the second derivative i just need to see where it would be negative and positive.
the inflection points im not sure.

Answer 32

\[\large f(x)=0 \qquad\rightarrow\qquad\;\; \text{roots}\]\[\large f'(x)=0 \qquad\rightarrow\qquad \text{critical points}\]\[\large f''(x)=0 \qquad\rightarrow\qquad \text{inflection points}\]

Answer 33

Critical points tell us where the function can change from increasing to decreasing.
Inflection points can tell us where a function changes from CCU to CCD.

Answer 34

Set the second derivative equal to zero.
Find those inflection points! :)
\[\large 0=-2e^{-x^2}+4x^2e^{-x^2}\]

Answer 35

i dont know how to with the e's there

Answer 36

See how the e's are the same in both terms?
Factor it out, then we'll do similar step we did with the first derivative ~ Setting each factor equal to zero.

\[\large 0=(-2+4x^2)e^{-x^2}\]

Answer 37

We already determined earlier that \(\large e^{-x^2}\ne0\) right?

So we only have to worry about the other factor,\[\large 0=-2+4x^2\]

Answer 38

x= \[\sqrt{2}/2\]

Answer 39

Careful, on the left side we had \(\large \sqrt{x^2}\) which will produce both a negative and positive solution.

Answer 40

Or right side, depending on where you moved the stuff heh.

Answer 41

so both the negative nd the positive are the inflection poins

Answer 42

Yes, it appears we have 2 inflection points :)

Answer 43

so lets say since at 0 the second derivative is negative then it is CCD

Answer 44

in teh interval between the inflection points?

Answer 45

yes, very good.

Answer 46

thank you so much for the help

Answer 47

np \c:/