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Problem 72 Hard Difficulty

The family of bell-shaped curves
$$ y = \frac{1}{\sigma \sqrt{2\pi}} e^{-(x - \mu)^2 /(2\sigma^2)} $$
occurs in probability and statistics, where it is called the normal density function. The constant $ \mu $ is called the mean and the positive constant $ \sigma $ is called the standard deviation. For simplicity, let's scale the function so as to remove the factor $ 1/(\sigma \sqrt{2\pi}) $ and let's analyze the special case where $ \mu = 0 $. So we study the function
$$ f(x) = e^{-x^2 /(2\sigma^2)} $$
(a) Find the asymptote, maximum value, and inflection points of $ f $.
(b) What role does $ \sigma $ play in the shape of the curve?
(c) Illustrate by graphing four members of this family on the same screen.

Answer

a)HA: $y=0, \quad$ local $\&$ abs. $\max : f(0)=1$ inflection points: $f( \pm \sigma)=\frac{1}{\sqrt{e}}$
b) since we have IP at $x=\pm \sigma,$ the inflection points move away from the $y-$ axis as $\sigma$ increases.
c) SEE GRAPH

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Video Transcript

we're told the family are bell shaped curve or described by dysfunction. And this is called a normal density function and theirs to constantly have them. You, which is described demean. And we have, Ah, another continent called Sigma. And that tells us two standard deviations. And we're going to be looking at the case where knew is you could zero giving us this more simple function and we're being asked a couple of things are being told to find the Athen toe maxim by inflection point. So the story starts. So to find the ascent of this function we're looking for when x goes tio positive or negative infinity. And so if you look at this function is e to the minus X wherever treason, we're square. So this could be re written as why you call one over, um e to the X squared over two to sigma squared over to think much word. So if you look at this as X goes to really, really big numbers say e to the one hundred over two Sigma squared sigma still positive. So it's going to eat in a one hundred or eat it, you get one over a really big numbers to get one over infinity. Well, that's just really, like gets closer to zero. So and since exponential functions since it's being squared, it doesn't matter if he's going to positive or a negative infinity. Ill struggle approaches zero. So it will have a horizontal ascent or that Y equals zero. And then we're told, then we're being asked to find the maximum value. So this means by taking the first derivative. So if we take the first driven and we're gonna have a problem, Max, and come down to B minus e ah, negative X squared over two Sigma squared, sometimes to act. So there's a time to act over two Sigma squared and said This equals zero and you can see that this from every calls, you know. So if you plug in X equals your you get zero so you get X equals zero and you plug in fo zero. You get one. So we have maximum local Mac one. And in this case it is the absolute max, too. And then we're told, find the inflection point. It involves taking a second to evidence and this is a bit more work. So this is going to come out to be minus a to the minus. X squared over two Sigma squared. That's a signal Times one over Sigma squared Truth X over Sigma squared. Um, times e to the minus X squared that's squared over two. Take my squared Our time's too x time to act over two sigma squared And then you can pull out a common factor of the, um minus X squared. So you e to the minus x squared over two sigma squared And you can also pull out a factor of I think, Mr Ford. And then you'LL get X square minus sigma squared because zero and you can't This cannot go zero. So we set experiments Tecnicos Sigma squared equals zero. So this gives us X square equal Sigma squared Take the square root you get X eagles off plus or minus sigma So that's the inflection point in collection point. And then we're being asked to evaluate what role the signal play in the shape of our curve. So he looked at this function again, so we had remember the function with y equals one over Run E to the X squared over two Sigma squared and is being asked to look at how signal Please rule. So as signal home and the value of sigma gets closer to zero. But this number gets really literally closest zero you get, you get X squared Reminder. I'm just looking writing up the exponents of X squared over to signal square. So as the denominator gets really, really small, you get a really, really big numbers. Sigma approaches, you know, you get a big number. So you get a really, really, really, really, really big number. And the fact that as he goes really close to infinity so you get one over e Teo the infinity again. They got one of the gates infinity and remember, one over a really big number approaches zero. So sigma is close to zero. Um, what's happening, toe? Why is that? Why is approaching zero? So signal my girls to deal. Why approaches why is also getting closer and closer? It is, you know. And so what happens is is this makes this are the function itself gets a lot closer to the Y axis, so it becomes more and more narrow and more more closer and compacted together. And the improvised versa. So a Sigma Gesche. So stigma approaches. Um, another signal gets bigger, so it's just a signal. Goes to infinity was just a significant infinity. Um, X squared over two things squared. Well, what happens to this? So Sigma Squared goes to really, really big number. So you get get X squared over Sigma squared and this approaches. It's approaches zero and then you get and then zero. And so what's happening? Teo it It's the opposite. It's going to happen. So a signal goes closer and closer. Why Asset Purchase one And this will become a bit more clear when I showed the graph, so actually have taken the liberty of drawing from the function before. So, as you can see, the green curve right here is AA minus X square over four. So if I increase it, that's their increasing to sixteen. This purple graph shows that is much more widespread and much more further out, which is what we predicted. And as they get closer and closer to a smaller number of the denominator, you get a more more sophisticated, more compacted graph, and that's exactly what we predicted.