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(a) Graph the function.(b) Explain the shape of the graph by computing the limit as $ x \to 0^+ $ or $ x \to \infty $.(c) Estimate the maximum and minimum values and then use calculus to find the exact values.(d) Use a graph of $ f" $ to estimate the $ x $-coordinates of the inflection points.

$ f(x) = x^{1/x} $

see solution

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 6

Graphing with Calculus and Calculators

Derivatives

Differentiation

Volume

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

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four of the function F of X is equal to X to the one of her ex. We want you graph the function, explain the shape of the draft by community element as X approaches zero from the right or X approaches Infinity. We're going to estimate the maximum mannitol values using or grab and then use calculus to find out. Exactly. And lastly, we want to use the graph of our second road to ex debate the X coordinate of our inflection point. So over here on the left, I went ahead and graft the function of two different viewing windows. So the 1st 1 helps us find what our maximum and minimum should be. And so it looks like our maximum value occurs around 2.718 and the output for that is about 1.4447 Uh, so at least when we try to calculate this exactly this looks really close. And actually color 2.718 is approximately the value for E. So if we do this correct, then we would think when we actually do part c finding this exactly are value for X should be about all right s so that gives us our local packs. So are you, Max. And over here at 00 that gives us our minimum. Now, the reason why I had these two different doing with those is because in part B, we want to explain the behavior for each of those limits. So on the right of there, we want to show that it's actually going to approach zero. And as X approaches infinity. If we look at this second graf here it looks like it's getting really close to Why is a good one? So we would expect that. Why is he with the one is a horizontal acid. So let's actually see if we get those same results. So if we were to just plug this in directly, we would get well X approaches your open right, just zero and one over zero is infinity. And that's an undetermined. So we don't know what that means. Released. I surely don't know what that means. So we need to rewrite it into a form that we do know. So unicycle to do is to introduce the natural log and e so e to the So if I take the natural longer. That is going to be natural. Log of X over X now. And if we were to go ahead and apply the limit, remember, we can move it inside using the chain rule for limits, so this could be the limit. As X approaches, you're from the right of natural longer pipes. Over. Remember, we could only do that since we know E to the X is a continuous spunky. All right, so we have that now natural log of X. If I, approaching zero from the right, goes to negative infinity. And if I approached X from the right, that's going to go to zero. Uh, and since these values are going to be positive, if I'm dividing into negative infinity and I keep getting closer and closer zero that's going to go to negative infinity still is just gonna go a lot faster. So this should be e to the negative infinity, which is zero. So the behavior as expert Israel from the right checks out. So let's check for this other behavior soaping you infinity. First, it's just gonna be infinity to zero. And once again, that's an interment form. I don't know what that's supposed to mean? Let's go ahead and rewrite this using the same trick that we did core. And I'm pretty much just going to skip too. This step right here. So we should be able to rewrite this as e to the limit, as I approach is a journey of natural log of X over X. Now, if we were to just apply this directly, we're going tow E to the infinity over infinity. And so again, I don't know what that's supposed to mean, but this implies that we can apply Loki college rule. So by local calls, this will be equal to so you limit as X approaches infinity. Oh, so we think the derivative of the numerous Dominic. So the derivative of natural log is going to be one of her ex. The derivative of X is just going to be what? And so we can rewrite that as the to the limit as X approaches a fairly off one of X. And we know one of her ex approaches zero has exclusive infinity e to zero or just one so are in behaviors for both. End up checking now toe finish off part. See, So we estimated. What? Our maximum entropy. Now we want to find the mixer. Remember, To do that, we need to take the derivative and find a critical values. And before we actually do that, I'm going to re bite this to make a little bit easier for us to take the drip. And I'm going to write this as e to the natural log of X over X, just like what we had before. Now if we wanted to take the first derivative so first derivative, we're going to need to use changeable. So General says, take the driven are outside puncture, which is e Well, that doesn't change anything. So it's just gonna be e to the natural log of X rex. But then we need to take the derivative with respect to X natural log of X over X and to take the derivative this function, we're going to need to use change for question. I'm sorry. And now I think I'm just going to do is rewrite this again just as X to the one over X just so I don't need to keep on writing that e. And then let's go ahead and use caution will take that to rebut it there. So remember, quotable says we're going to take what we have in the denominator so low d high, the derivative of natural log of X is going to be one over X and then minus high d low. So this should the natural log of x times the derivative of heaven did on there, which is just one all over what we have in the denominator squared. So now we can go ahead and just you little bit algebra to rewrite this as X to the one of her ex and one minus natural longer vex over X squared. Now what we want to do is set this here equal to zero. And so we already know that this part here only equal zero if X equals zero. But since here in the denominator, we can't have an equal to zero. We would only set one minus natural log of X equals zero and then we just need to solve Rex let go a little bit. So we're going to add natural longer vexed over get one is a natural log x exponentially eight on the upside and we're going to get e to the first power. It's ex, which is just, um e Now we go ahead and plug this into epidemics so f of e is going to be e to the one over. So this here is going to be our actual value for our Max, all right. And in part D. We need to know what the second derivative is going to be. So let's just go ahead and find the second derivative since we're here. So if I want to take the derivative of this, what we're going to do is so deep, I actually let me go ahead and just right. It's out again. So do you. By the x o x to the one over X times one minus natural under vexed over expert. Now, to take this, we're going to use the product on the question, so I'll go ahead and do it in this order. So one by this natural vex over X squared times, the derivative with respect to X of X to the one of rex, plus them in the opposite or so x to the one over x times the derivative with respect to X of one minus natural log of X vulgar X squared. So we already found this derivative in our last parts. Let's go ahead and write that down. So that should just be X to the one of her ex times. One Maya's natural longer vex all over export and taking the derivative of one buys natural defects over X squared. We're going to need to is questionable. Remembers as low so x squared de high. So the derivative of having the numerator here derivative 10 derivative of negative natural log of X is going to be negative and then minus than the opposite or so well, one minus natural log of X times the derivative of X squared, which is two X and then all over what we have in it A dominator squared so before it was X squared is gonna be ecstasy before and once you do a little bit of algebra. This here likely simplify too, what I have here for our second derivative. All right, now, after we did this, I want a head in graft, the function and since we just want to estimate where these points of inflection are, we should end up with, so this he's a different color, actually So we have this point here and this point here. And so the value for each of those. So the 1st 1 is about 0.580 point 582 and this other one I went ahead in, zoomed in a little bit closer just to make sure that we actually do have a change in Con Cavity. Because remember, we wanted to go from positive, negative, negative to positive and that value. There we end up with about 4.368 So 4.368 And remember, we just want to check that there is a change from concave up to conquer down or down to up. And so to the right, our first value it is can't give up because after department next district larger than zero and to the left is going to be conquered down. So that is appointed inflection. And our other one is Kong K down to left. Calm cable to the right. So we have our other point of inflection. Uh,

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