Video: AP Calculus AB Exam 1 β€’ Section I β€’ Part A β€’ Question 15

Let 𝑓 be the function defined by 𝑓(π‘₯) = ln π‘₯/π‘₯. What is the absolute minimum value of 𝑓?

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Video Transcript

Let 𝑓 be the function defined by 𝑓 of π‘₯ is equal to the natural log of π‘₯ over π‘₯. What is the absolute minimum value of 𝑓?

Here, we haven’t been given a closed interval in which to find the absolute minimum. Instead, we’re interested in looking at absolute extrema over the entire domain of our function. So we begin by recalling the fact that the domain of the function, the natural log of π‘₯, is π‘₯ is greater than zero. If we restrict our domain to π‘₯-values which are greater than zero, then with the function, the natural log of π‘₯ divided by π‘₯, we will always be dividing by values which are greater than zero. And this means the domain of the natural log of π‘₯ over π‘₯ is also π‘₯ is greater than zero. It’s also important that we do recall that not all functions have an absolute minimum or maximum over their entire domain.

So next, we’re going to find the critical points of our function. And we do this by evaluating its derivative and setting that equal to zero. Since our function is itself the quotient of two functions, we’re going to use the quotient rule. This says that, for two differentiable functions 𝑒 and 𝑣, the derivative of their quotient is 𝑣 times d𝑒 by dπ‘₯ minus 𝑒 times d𝑣 by dπ‘₯ over 𝑣 squared. The numerator of our fraction is the natural log of π‘₯. So we let 𝑒 be equal to the natural log of π‘₯ and 𝑣 be equal to π‘₯. This then means that d𝑒 by dπ‘₯ is equal to one over π‘₯. And d𝑣 by dπ‘₯ is equal to one. And then, the derivative 𝑓 prime of π‘₯ is therefore π‘₯ times one over π‘₯ minus the natural log of π‘₯ times one all over π‘₯ squared. This simplifies to one minus the natural log of π‘₯ all over π‘₯ squared.

To find our critical point, we set this equal to zero. And for this statement to be true, the numerator of our fraction one minus the natural log of π‘₯ must be equal to zero. We solve for π‘₯ by adding the natural log of π‘₯ to both sides and then raising both sides of our equation as a power of 𝑒. And when we do, we see that π‘₯ is equal to 𝑒 or just simply 𝑒. So we have one critical point. And that lies at the point when π‘₯ is equal to 𝑒. But how do we establish whether this is a local maximum or a local minimum? We’re going to perform the first derivative test. We choose a point that lies either side of this critical point. Let’s clear some space.

Since 𝑒 is between two and three, we’re going to choose the point π‘₯ equals one and π‘₯ equals three. We’re going to evaluate the nature of the first derivative at these points. 𝑓 prime of one is one minus the natural log of one over one squared. Well, the natural log of one is zero. So 𝑓 prime of one is one minus zero over one squared, which is simply one. This is greater than zero. Since our derivative is greater than zero at the point when π‘₯ equals one, then the function 𝑓 is increasing at this point. We’ll repeat this process for 𝑓 prime of three. That’s one minus the natural log of three over three squared.

Since the natural log of 𝑒 is one, we know that the natural log of three must be greater than one. This means one minus the natural log of three must be less than zero. And that in turn means one minus the natural log of three divided by three squared must also be less than zero. This means the function is decreasing at the point π‘₯ equals three. And in fact, since we said the graph 𝑓 of π‘₯ has one critical point only at π‘₯ equals 𝑒, we can extend this idea. We can say that the function must be increasing over the open interval zero to 𝑒. And it must be decreasing over the open interval 𝑒 to positive ∞.

And next, we’re going to imagine ourselves walking along the graph of 𝑓 of π‘₯ is equal to the natural log of π‘₯ over π‘₯. We’ll start at the left end of our domain, which we know to be zero. And we’ll go all the way to the right end which is positive ∞. According to our evaluation of the derivative, we will go up and up until we reach π‘₯ equals 𝑒. Then, we will be forever moving in a downwards direction. This means the critical point π‘₯ equals 𝑒 must be an absolute maximum. It gives the highest possible value of 𝑓 of π‘₯ along our entire domain. And we can imagine this in the opposite direction. And we’d still ensure that π‘₯ equals 𝑒 is an absolute maximum. This also means that 𝑓 has no absolute minimum. And we will look at a quick sketch to understand why.

The graph of 𝑓 of π‘₯ equals the natural log of π‘₯ over π‘₯ is as shown. We can see that as π‘₯ tends to zero, the function 𝑓 of π‘₯ tends to negative ∞. It has no absolute minimum.

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