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Numerade Educator



Problem 48 Hard Difficulty

Use the guidelines of this section to sketch the curve.

$ y = e^x/x^2 $




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Video Transcript

So for this graph, you know that the domain is, um everything else is good, but we have a restriction that X equals zero. So ex cannot be zero for the domain because we can't zero in the denominator. There are no intercepts, and there is no symmetry for Assam tools we have, Let's says, for a vertical ass in tow. So setting the denominator equal to zero X squared equals zero, so we can see that we have an ask himto X equals zero. So it's a vertical asking, too. And let's test for a horizontal as into it. Now for horizontal, We can see, um, by testing either the limit as X approaches infinity or negative infinity. But for this one, um, limit as X approaches negative infinity because that's where we're going to see the ass in two, um, e to the X or function over X squared. We're gonna have to use low Patel's rule. So taking the first derivative of the top, he did the X over. Sorry, My bad credit limit limit as X approaches negative infinity of the derivative of the top you to the ex over x X squared the derivatives to X So now do the rule again The derivative of the top. Any to the ex Because, right, lim, over the derivative of the bottom, which is just too so here. If we substitute negative infinity into X, we see that the limit zero. So we have a horizontal ass in tow at y equals zero. Next thing we have to do is check for increasing and decreasing intervals. So we take the first derivative. Why prime it was e to the x x times X squared minus two e to the x Times X over X to the fourth. That's just quotient rule. And if we set our first derivative to zero, we confined critical points at X equals zero and X equals two. So now if we make a little line for our first derivative test, this is f prime and we are critical points on it. We can test around or critical points by substituting values into the first derivative. So substituting values smaller than zero, we're going to see that it's positive outputs between zero and two. It's gonna give us negative, and anything greater than two will give us positive. So here it's increasing and then decreasing around zero. So here we have. Ah, Max. And here it's decreasing and an increasing. So we have decreasing to increasing. This is a minimum value. And now what we normally would do is test for con Cavity using the second derivative. But, um, if we try to do that, we'll see that it's inconclusive, meaning that there's no solutions when we set our second derivative to zero. So we can we can just use the first derivative to Graff or function. We're just doing a rough sketch. So we said we have ah, vertical. Ask him to X equals zero and a horizontal ass in tow at Y equals zero. So our function can approach but not touch these lines. So now if you graph the function in red, we have, um our function is gonna increase until it reaches zero. So anything smaller than zero, it's gonna be increasing. So the only way this can happen is like this, and it's gonna increase, but it cannot touch until it reaches your But I cannot touch this line is gonna increase and be close to it. But it's never going to touch. And now here, um, after zero, it's gonna be decreasing. So this is a decreasing interval is gonna decrease until it hits, too. So here, let's mark that two over here. And let's just see what the Y value is. Firmin So f of two is gonna be approximately, um, 1.8. So let's say this is about 1.8 1.8. So that's our minimum value right here. So around her minimum value, it's gonna be after the zero, but not touching the ass until obviously, it's gonna decrease until it hits our minimum value. And then beyond that, it's gonna increase. And this is the graph for the function.