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Sketch the graph and show alllocal extrema and inflection points_ y = e" 3e * 4...

Question

Sketch the graph and show alllocal extrema and inflection points_ y = e" 3e * 4

Sketch the graph and show alllocal extrema and inflection points_ y = e" 3e * 4



Answers

Find the points of inflection and discuss the concavity of the graph of the function. $$y=e^{-3 / x}$$

We wish to know where this function as concave up. Concave down and where the inflection points are. So let's get to the second derivative. The derivative uh 40 of negative X is negative one. The derivative of negative three X. To the fourth is negative 12 X. Cute. The second derivative then would be negative 36 X square for the inflection point. This is never not differentiable for the second derivative. So the only possible inflection points would be where that second derivative is zero. And we have negative 36 X squared 20 It's quite obvious that that's going to be true If X is zero. So let's check the sign of the second derivative Around the x value of zero. We filmed a number less than zero like -1. We have negative 36 times negative one squared. That's negative 36 times one. That's a negative answer. We fill in a number like one. We'd have negative 36 times one. Also a negative so it never switched. Uh can cavity. It's always concave down. So we'll say this is con caved down from negative infinity to infinity.

We are looking for can cavity regions as well as points of inflection. So let's get to the second derivative. The derivative of the function would have to use the product rule. That would be first times the derivative of the second and under the derivative of the second. I would bring the three out front. Subtract one from the old power and multiply by the derivative the inside since that's chain rule. But the derivative of the inside is just one. I add to that the second term times the derivative of the first, which is one for our second derivative. The first set, we're going to use the product rule on that be first times the derivative of the second. I'll bring the two out front. Subtract one from the old power. So that's the first multiply by the derivative. The inside which is one plus the second term times the derivative of the first, which is three. And then for the x minus four cubed. I will bring the power out front. Subtract one from the old power and multiply by the derivative of the inside but the derivative of the inside is just one. Now, the last two terms I could actually combine it says three times x minus four quantity squared so added together. That's going to be six times x minus four quantity squared. So I just put a six in front. Okay, let's simplify. This is going to be six X times x minus four plus six times x minus four squared. There is a common six times x minus four. So let's pull that out front. And that would leave us with X plus x minus four. And we could get that to simplify too. Then six times x minus four times two X minus four. This is going to be zero. Either if x minus 40 or if two x minus four is zero. X minus forest zero of x is four and two X minus 40 effects is to So the number line check will focus around points around X equals two and X equals four. If we fill in these values if I fill in one and it's probably easiest to fill it in right here. If I feel in one I'd have six times negative three times negative two. That's a positive answer by filling three I'd have six times negative one times two that is negative. And if I fell in five I'd have six times one times six which is a positive. So for the concave regions we've got concave up from negative infinity to to as well as for to infinity. And it's concave down in the region from two 24 Both X equals two and X equals four. Our inflection points. Yeah. To find the function values for those, we would fill those values in. If I fill in two, I would have two times negative two cube. That's two times negative eight or negative 16. I'm filling this into the function and if I fell in four, I get four times zero, so that's zero.

Okay, suppose given the function F of X, I want to find the different points of inflection. Right. When do I have points of inflection. Well, this is related to when my second derivative is either equal to zero or when my second derivative is undefined. Okay, so it wouldn't be in defined here because I have just a polynomial function. It's not a rational polynomial with the denominator with the accident. So this is just with my second group of secret zero. Okay, what else do I know with my second derivative? Well, on con cave up when my second derivative is positive. My graph will open down or be concave down when my second derivative it is naked. So I can tell a lot from the second derivative here. First thing I want to do is I'm just going to distribute this X. The third to both functions to make simpler to differentiate. So third time's Xx The fourth minus four X. First step find my first derivative. Okay, Power will bring down the four. Strike one from the exponent four times three is 12 truck one from the exponents From the first derivative. I can find my second derivative taking the derivative of my first four times three is 12 truckloads. The exponents 24 X. Okay, so now I have my second derivative. First. I'm going to set it equal to zero to find my possible points of inflection. Right where my con cavity could be changing from concave out to contact down. I can factor out a 12 x vaccines to equal 00 product property 12 X equals zero or divided by 12 to both sides, X equals zero. My next minus two equals zero. This happens when X equals two. Okay, so I have possible points of inflection at zero and two. So I'm going to create different intervals between these points and on these air rules, I'm going to pick chest points. Does not matter which test points I pick as long as there may be as easy as possible. And we're making a plug these checkpoints in up my original my first derivative, my second derivative, well no con cavity and puts an inflection are all determined with my second derivative. So I'm gonna plug it into my second derivative function or 12 X times X minus two. Okay, So if I were to plug a negative one, I'd get a negative times a negative which would be a positive. Okay, so that means I'm con cave up From Negative Infinity to zero. If I plugged in one, I get a positive and then 1 -2 is a negative positive times a negative is a negative. Okay, so that means that my graph is concave down. Okay, so my graph is changing from concave up the concave down. So that means X equals zero is a point of inflection because I'm changing con cavity. And then when I plug in three I get positive times a positive, which is a positive. Concave up. So since I go from concave down to concave up X equals two is also a point of inflection because I'm changing con cavity. If I wasn't changing con cavity, X equals two would not be appointed perfection.

We need to find the extreme, um, the points of inflection where the function is increasing or decreasing and where it is con cave up in Khan Cave down So we're given the function is three x to the fourth plus four x cute. So first, let's find the extreme, um, when you find that by setting the first derivative equal to zero, so the first derivative is 12 x cubed plus 12 x squared. This gives us that X equals zero and negative one. So we have a point. Negative one negative one. It's a coordinate and we have the coordinates. Zero common zero. So these air both points on the graph where the inflection point is where the second derivative is equal to zero. The second derivative is 36 x squared plus 24 x So we find that this gives us X equals zero and X equals negative to over three. So 00 is also an inflection point, and another inflection point is negative to over three. And the function is equal to negative 16/27. At this point, where the first derivative is greater than zero tells us where the function is increasing so we can see from X equals negative infinity to negative one. The function is decreasing from X equals negative one to infinity, the function is increasing. Where the second derivative is greater than zero Tells us where the function is con cave up so we can see that the function is con cave up from X equals negative infinity too negative to over three It is con cave down X equals negative to over 3 to 0 and it is again Khan cave up from zero to infinity Using the information we just found. This is what the sketch with a graph will look like. We can see that we have a minimum At negative 11 we have another extreme, um at 00 our inflections. Our inflection is at 00 and at negative to over three negative 16/27. The function is decreasing from negative infinity to negative one. It's increasing from negative one to infinity and it is con cave up in Khan cave down on the appropriate intervals


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