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# (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 $f$.$f(x) = e^{-x^2}$

## (a) $\lim _{x \rightarrow \pm \infty} e^{-x^{2}}=\lim _{x \rightarrow \pm \infty} \frac{1}{e^{x^{2}}}=0,$ so $y=0$ is a HA. There is no VA.(b) $f(x)=e^{-x^{2}} \Rightarrow f^{\prime}(x)=e^{-x^{2}}(-2 x), \quad f^{\prime}(x)=0 \Leftrightarrow x=0, \quad f^{\prime}(x)>0 \Leftrightarrow x<0$ and $f^{\prime}(x)<0 \Leftrightarrow$$x>0 . \quad So f is increasing on (-\infty, 0) and f is decreasing on (0, \infty)(c) f changes from increasing to decreasing at x=0, so f(0)=1 is a local maximum value. There is no local minimum value.(d) f^{\prime \prime}(x)=e^{-x^{2}}(-2)+(-2 x) e^{-x^{2}}(-2 x)=-2 e^{-x^{2}}\left(1-2 x^{2}\right)$$f^{\prime \prime}(x)=0 \Leftrightarrow x^{2}=\frac{1}{2} \Leftrightarrow x=\pm 1 / \sqrt{2} . \quad f^{\prime \prime}(x)>0 \Leftrightarrow$$x<-1 / \sqrt{2} or x>1 / \sqrt{2} and f^{\prime \prime}(x)<0 \Leftrightarrow-1 / \sqrt{2}<x<1 / \sqrt{2}$$f$ is $\mathrm{CU}$ on $(-\infty,-1 / \sqrt{2})$ and $(1 / \sqrt{2}, \infty),$ and $f$ is $\mathrm{CD}$ on $(-1 / \sqrt{2}, 1 / \sqrt{2})$There are inflection points at $\left(\pm 1 / \sqrt{2}, e^{-1 / 2}\right)$

Derivatives

Differentiation

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

All right. So we are first being asked to find the vertical and horizontal ask until so, um, dysfunction can be rewritten as one over e to the X Square. And now, if you haven't already noticed, um, this bottom function, I'm dysfunction denominator cannot equal zero because that's what defines vertical. I think that's one of the ways. And since exponential function never equal zero, there is no vertical Jacinto. So no vertical ascent toe. And so now we gotta find the horizontal axis until we find that I find the limit. So the limit is X goes to infinity, and this will be one over e to the X squared. And as you can see as, uh, this number guests, uh, the denominator gets exponentially higher. You get one of the infinity, which is a one of a really big numbers, just zero. And this is a case also for negative numbers because negative numbers squared. It's the same thing as positive number squared. So this was also a plus or minus infinity. So we have a horizontal attitude at y equals zero. Yeah, and then now, to find the intervals in which function increases or decreases, we apply the first derivative cath that I've been taking the first derivative. In this case, you will have to apply the chain rule, so I'll come out to be negative. Two X equals negative. Two x 10 e to the negative x squared. Uh, sorry about that. Just bring this down. I was just negative two x or eat the minus X squared and this can be rewritten at negative two X over E to the X squared. Then you set this equal to zero. The denominator cannot equal zero, so that's not one of the critical number. Negative two X equals zero when X is equal to zero. So we have a signed chart evaluation here, so it's going to be X, and you put it at zero, and then you bring it down and they were looking at the sign of a crime. But people are numbers less than zero. You get positive numbers and you fucking greater than zero. You get negative numbers, so we noticed increasing and decreasing. So we have a local max occurring at X equals zero. Uh, and we know that it is increasing from negative infinity to zero, and it is decreasing from zero to infinity. Now, to find where the funk says, uh, find the functions can cavity would take the second derivative tests that they have been trying in the second derivative. So again, this is a little bit more chain rule and, uh, combination of some protocol, and this will come out to be negative one plus two x squared statistical zero. Um, this part of the function cannot obviously equals zero. So we have to set this part of the pumpernickel zero. So this will be negative one plus two x squared because you have to find a critical number. You add one divide by Tuesday. You got X square called 1/2. This will be X equals. The square root of one is one. So we one over root two plus or minus because we took a square root. Now we can do a sign chart evaluation. This will be negative. 1/2. It would be positive one over. Richer, not a line, but not a line. Looking at the sign of a double prime your fucking values less than negative one or two. You get positive numbers between these two negative and positive. Still can't give up. Down, up. So we have a con cave up interval occurring from negative infinity. Two negative 1/2. And from one over to to infinity. We have concrete down occurring between negative one over to positive 1/2. And we have inflection points occurring when sign change occurs. So that's occurred that both plus and minus route to I mean plus or minus, uh, 1/2. So inflection point point occurred that plus or minus 1/2. Now we have enough information to draw a graph. So it looks something like this. Uh, you know that there is a local max at zero, and we have a concave up shape from here from the negative. So we're kind of coming up like I use so this is like, like up you, and then I'll turn into a concave down shape. So now we start decreasing after zero, you come down and and then I'll go off to zero. And this this max occurring a gray here. Um, the peak is supposed we had zero Mhm. Sorry, if that wasn't clear enough, but this is the graph of F WebEx and it's symmetrical. That's it.

#### Topics

Derivatives

Differentiation

Volume

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp