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Numerade Educator

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Problem 45 Easy Difficulty

Use the guidelines of this section to sketch the curve.

$ y = \dfrac{1}{x} + \ln x $

Answer

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Video Transcript

So for the domain, we know that it's gonna be, um X is greater than zero because it has to be first of all, ex cannot be zero because of first part. And the second part tells us that it has to be positive. So x is greater than zero intercepts. There are none symmetry, there is none. And for Assam totes, if we look for a horizontal, ask himto, for example, we have to set limits. Um, as X approaches infinity or limit as X approaches. Negative infinity. We see that there's no horizontal. However, there is a vertical ass until at X equals zero. Because we set our denominator for one of her ex. We said the denominator to zero. So x zero is our, ah, vertical ass in two. And now we look for increasing decreasing intervals. So you find the first derivative y prime, and that is one of her ex minus one over X squared. And if we set our first derivative to zero, we confined critical points and we see that X is one is a critical point. So if we do a first derivative test, so this is for F prime and we substitute, um, and research for inside of our derivative. We're gonna plug values that are smaller than one. What we have to remember to stay in our domain. So if it's smaller than one, it's going to be negative. If it's greater than one, it's gonna be positive. So that means that it's decreasing until it gets to one and increasing after that. So if it's going from negative to positive, so decreasing to increasing its gonna be men a local minimum value. And if you want to know the why value of our critical point we go F one. So we're putting it back into her original equation to find one. So 11 is our minimum local minimum value for our graph. And now we can find, um, we can look at kong cavity for this graph by finding the second derivative. So why double prime equals negative, and this negative is on the outside of the entire fraction. So negative X minus two over X cubed. So if we set that equal to zero, we see that X is equal to two and now our second derivative test, so double prime, and this is too and we have to make sure to test values in our domain. So, for example, if it says 0.5 for over here so smaller than two. But in still inside of her domain, we're going to get positive. So Khan came up, and beyond that, it's gonna be Khan came down so negative. Okay? And now we can graph with the information that we got. So first thing I'm gonna put on the graph is the, um, vertical ascent. Tow X equals zero, and we can see that, um, we have a minimum value at 11 So let's say this is one. This is one. So that's our local minimum value. And, um, we see that we haven't Oh, yeah. This is an inflection point because it's switching signs from Khan gave up to conquer it down. So we can say that this is an inflection point at two. All right, so, um, say this is about to So this is a rough sketch of our graph. It's gonna be decreasing until it hits our minimum value so it cannot touch the ass in tow. So decreasing until hits the minimum value, then it's gonna increase from then on. So it's so we have con cave up. That's correct. And then it's gonna be calling came down after it gets to our inflection permit. So it's Kong cave up, but still increasing until it's our inflection point that it's gonna be con cave down but still increasing, and that's a rough sketch.