Video: AP Calculus AB Exam 1 β€’ Section I β€’ Part A β€’ Question 2

Find the values of π‘š and 𝑛 that make 𝑓(π‘₯) = ln(π‘₯ βˆ’ 2) if π‘₯ < 3 and 𝑓(π‘₯) = 2π‘š + 𝑛π‘₯ + 1 if π‘₯ β‰₯ 3 differentiable at π‘₯ = 3.

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Video Transcript

Find the values of π‘š and 𝑛 that make 𝑓 of π‘₯ which is equal to ln of π‘₯ minus two if π‘₯ is less than three and two π‘š plus 𝑛π‘₯ plus one if π‘₯ is greater than or equal to three differentiable at π‘₯ equals three.

For this question, weβ€²ve been given a piecewise defined function 𝑓 of π‘₯. And weβ€²ve been asked to find the values of two constants which would make our piecewise defined function differentiable at the boundary between its two subfunctions. To begin this question, we recall the following general rule, which is if a function is differentiable at some point, which we’ll call π‘₯ nought, it is also continuous at π‘₯ nought. Taking our value of three as π‘₯ nought, our steps to solve this problem can be as follows.

We’ll first check that our function 𝑓 is continuous at the point where π‘₯ equals three. If this condition is not satisfied, then an implication of our general rule is that 𝑓 will also not be differentiable at π‘₯ equals three. If this condition is satisfied, we’ll then check that the limit as π‘₯ approaches three of the derivative of our function 𝑓 dash of π‘₯ exists. If the second condition is not satisfied, we again conclude that 𝑓 is not differentiable at π‘₯ equals three. However, if this condition is satisfied, we conclude that 𝑓 is differential at π‘₯ equals three. This is because in order for a function to be differentiable at a point, we must be able to define the derivative at this point.

Letβ€²s begin on our first step, checking the continuity of 𝑓 at π‘₯ equals three. Here, we recall the condition for continuity which states that the limit as π‘₯ approaches π‘₯ nought of our function 𝑓 of π‘₯ must be equal to the function evaluated at π‘₯ nought. Our value of π‘₯ nought is three, so letβ€²s begin by evaluating 𝑓 of three. Here, we see that when π‘₯ is greater than or equal to three, 𝑓 of π‘₯ is defined by our second subfunction: two π‘š plus 𝑛π‘₯ plus one. 𝑓 of three is therefore two π‘š plus 𝑛 times three plus one. Here, we can rewrite this as three 𝑛 to tidy things up.

To check our limit, we’ll need to recall the following facts. If the left- and the right-sided limits of our function both exist and are equal to some value 𝐿, then the normal limit will also exist and be equal to the same value 𝐿. Since we have a piecewise defined function of which the two subfunctions join up at π‘₯ equals three, the left- and the right-sided limits of 𝑓 of π‘₯ will be defined by our two different subfunctions. As π‘₯ approaches three from the negative direction, π‘₯ is less than three. And hence, 𝑓 of π‘₯ is defined by ln of π‘₯ minus two.

Here, we can take a direct substitution approach of π‘₯ equals three. And we find that our limit is equal to ln of three minus two, which is equal to ln of one. And we know that any logarithm of one is equal to zero. Weβ€²ve now found the value for our left-sided limit. So letβ€²s work on our right-sided limit. When π‘₯ approaches three from the positive direction, π‘₯ is greater than three. And hence, 𝑓 of π‘₯ is defined by our second subfunction, which is two π‘š plus 𝑛π‘₯ plus one. To solve this limit, we again take a direct substitution approach. And here, we may recognize that weβ€²ve already substituted π‘₯ equals three into this subfunction when evaluating 𝑓 of three. We hence know that our answer is the same which is two π‘š plus three 𝑛 plus one.

Here, we see that both our left and our right limits exist. But in order to satisfy our continuity condition, we need both of these things to be equal to 𝑓 of three. We can immediately see that 𝑓 of three and our right-sided limits are equal. However, we also need these to be equal to zero. And hence, we can form the following equation that two π‘š plus three 𝑛 plus one is equal to zero. Now, this equation gives us a relationship between π‘š and 𝑛. But it doesnβ€²t yet solve our question since there are infinitely many pairs of values which would satisfy the equation.

To move forward, we’ll set this equation aside and move on to our next condition for differentiability. We now need to make sure that the limit as π‘₯ approaches π‘₯ nought of the derivative of our function exists. Again, since we have a piecewise defined function, this will involve making sure that our left- and our right-sided limits are equal to each other. First, letβ€²s define 𝑓 dash of π‘₯ by differentiating our two subfunctions.

First, we must take the derivative of ln of π‘₯ minus two. Here, we may recall that a standard result is that the derivative of ln of π‘₯ is one over π‘₯. Since we have a composition of functions with ln of π‘₯ minus two, here we must use the chain rule. However, since the derivative of π‘₯ minus two is just one, our answer is one over π‘₯ minus two. Next, we must differentiate our second subfunction, which is two π‘š plus 𝑛π‘₯ plus one. Looking at this expression, we see that we have one π‘₯ term and two constant terms. The constant terms when differentiated are equal to zero. But using the power rule to differentiate our 𝑛π‘₯ term, this becomes 𝑛.

Now that we’ve differentiated our two subfunctions, we can put both of these back into our definition of 𝑓 dash of π‘₯. We then have that 𝑓 dash of π‘₯ is equal to one over π‘₯ minus two if π‘₯ is less than three and 𝑛 if π‘₯ is greater than three. Itβ€²s worth noting that we do not ordinarily make claims as to the value of our derivative when π‘₯ is equal to the value π‘₯ nought until the very end of the question. This is because weβ€²re not yet sure whether our function is differentiable at the point in question. In this instance, since we are making 𝑓 of π‘₯ differentiable at π‘₯ equals three, weβ€²ll allow ourselves to put this in.

To check that the normal limit exist, we’ll need to check the left and right limits as π‘₯ approaches three. Much like the continuity condition, we’ll be using different subfunctions for the left- and the right-sided limits of our derivative 𝑓 dash of π‘₯. Much like when checking the continuity condition, the left- and the right-sided limits of 𝑓 dash of π‘₯, weβ€²ll use different subfunctions because we have a piecewise defined function.

As π‘₯ approaches from the negative direction, π‘₯ is less than three. And hence, 𝑓 dash of π‘₯ is defined by one over π‘₯ minus two. Directly substituting π‘₯ equals three into this, we get that this limit is equal to one over three minus two which is of course just one over one which is one. Moving on to the right-sided limit, here π‘₯ approaches three from the positive direction. Hence, π‘₯ is greater than three and 𝑓 dash of π‘₯ is defined by our second subfunction 𝑛. Of course, the limit as π‘₯ approaches anything of 𝑛 is just 𝑛. And hence, we have the value of our right-sided limit. In order to satisfy this condition, the normal limit as π‘₯ approaches three of 𝑓 dash of π‘₯ must exist. And hence, both the left- and the right-sided limits must both exist and be equal to each other.

Since weβ€²ve just calculated both of these values, we can substitute them in to form an equation. And this equation tells us that 𝑛 is equal to one. We may now recognize that with this piece of information alongside our earlier equation, we’re now able to answer our question. Given that we now know 𝑛 is one, we substitute this into our first equation. We then find that two π‘š plus four is equal to zero, two π‘š is negative four, and hence π‘š is negative two.

Upon finding these values, we have now solved our question. And we have found that when π‘š is equal to negative two and when 𝑛 is equal to one, the function 𝑓 of π‘₯ is differentiable at the point where π‘₯ is equal to three.

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