Video: MATH-DIFF-INT-2018-S1-Q10

If 𝑓(π‘₯) = π‘₯(π‘Ž βˆ’ ln(π‘₯)) such that π‘Ž is a constant and the curve of the function has a critical point at π‘₯ = 𝑒, find π‘Ž.

03:47

Video Transcript

If 𝑓 of π‘₯ is equal to π‘₯ lots of π‘Ž minus ln of π‘₯ such that π‘Ž is a constant and the curve of the function has a critical point at π‘₯ equals 𝑒, find π‘Ž.

First, let’s remember what a critical point is. A critical point is a point, such as π‘₯ is equal to 𝑐, such that the differential of 𝑓 at 𝑐, which we can also call 𝑓 dash of 𝑐, is equal to zero. So let’s start answering this question by differentiating 𝑓 of π‘₯ with respect to π‘₯. And this is also called 𝑓 dash of π‘₯. And it’s equal to d by dπ‘₯ of π‘₯ times by π‘Ž minus ln of π‘₯. In order to differentiate this, we’ll need to use the product rule. The product rule tells us that if we have some function 𝑦, which is equal to a product such as 𝑒 multiplied by 𝑣, then when we differentiate 𝑦, we’ll get d𝑦 by dπ‘₯ is equal to 𝑒 d𝑣 by dπ‘₯ plus 𝑣 d𝑒 by dπ‘₯.

Now, in our case, we can let π‘₯ be 𝑒 and let π‘Ž minus ln of π‘₯ be 𝑣, then we obtain that this differential is equal to π‘₯ times by d by dπ‘₯ of π‘Ž minus ln of π‘₯ plus π‘Ž minus ln of π‘₯ times d by dπ‘₯ of π‘₯. And, immediately, we can evaluate d by dπ‘₯ of π‘₯, since this is just equal to one. In order to differentiate π‘Ž minus ln of π‘₯ with respect to π‘₯, we’ll need to use a rule. And this rule tells us that d by dπ‘₯ of ln of π‘₯ is equal to one over π‘₯. Now, in order to answer this question, you do not need to know how to derive this differential. But for those of you who are interested, I shall write down the derivation of this differential. And you can pause the video and read through how it’s done.

So this is how you can prove that d by dπ‘₯ of ln of π‘₯ is equal to one over π‘₯. However, it’s not necessary to know this in order to answer the question. All you need to remember is that the differential of the natural logarithm of π‘₯ is equal to one over π‘₯. So now we can use this in order to resolve the differential of π‘Ž minus ln of π‘₯. Now, we’re told in the question that π‘Ž is a constant. And any constant will differentiate to zero. So the π‘Ž will go to zero, and then we have minus ln of π‘₯. And since ln of π‘₯ differentiates to one over π‘₯, negative ln of π‘₯ will differentiate to negative one over π‘₯. And then, we need to add on the π‘Ž minus ln of π‘₯ times by one.

Now, we can simplify this to give us that 𝑓 dash of π‘₯ is equal to negative one plus π‘Ž minus ln of π‘₯. Now that we found 𝑓 dash of π‘₯, we need to use the fact that this function has a critical point at π‘₯ equals 𝑒. If we remember back to the definition, if a critical point of a function is at π‘₯ equals 𝑐, then this means that 𝑓 dash of 𝑐 is equal to zero. Since our critical point is at 𝑒, we need to substitute π‘₯ equals 𝑒 into this equation and set it equal to zero.

This gives us that 𝑓 dash of 𝑒, which is equal to negative one plus π‘Ž minus ln of 𝑒, is equal to zero. Now, since ln is the natural logarithm or log to base 𝑒, ln of 𝑒 is simply one. And, therefore, our equation becomes π‘Ž minus two is equal to zero. This can be arranged for π‘Ž to give us that π‘Ž is equal to two. And this is our solution.

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