Video: APCALC04AB-P2B-Q03-592128586059

The given figure shows the graph of 𝑓′, which is the first derivative of the function 𝑓. In the region βˆ’6 ≀ π‘₯ ≀ 6 this graph consists of a semicircle and four line segments. In the region 6 ≀ π‘₯ ≀ 10 the graph is part of the parabola with equation 𝑦 = (π‘₯ βˆ’ 7)Β² βˆ’ 3. i) If 𝑓(0) = 10, what is the value of 𝑓(βˆ’3)? Please give your answer in terms of πœ‹. ii) Evaluate ∫_(2) ^(9)𝑓′(π‘₯) dπ‘₯. iii) For βˆ’6 ≀ π‘₯ ≀ 10, on what open intervals, if any, is the graph of 𝑓 both increasing and concave downward? Give a reason for your answer. iv) Find the π‘₯-coordinate of each point of inflection, if any, for the graph of the function 𝑓 for βˆ’6 ≀ π‘₯ ≀ 10. Give a reason for your answer.

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Video Transcript

The given figure shows the graph of 𝑓 prime, which is the first derivative of the function 𝑓. In the region π‘₯ is greater than or equal to negative six and less than or equal to six, this graph consists of a semicircle and four line segments. In the region π‘₯ is greater than or equal to six and less than or equal to 10, the graph is part of the parabola with equation 𝑦 equals π‘₯ minus seven squared minus three. Part i), if 𝑓 of zero equals 10, what is the value of 𝑓 of negative three? Please give your answer in terms of πœ‹.

And there are three further parts to this question which we’ll consider in a moment. We need to be extra careful here. We’ve been given the graph of 𝑓 prime. That’s the graph of the derivative of 𝑓, not the function 𝑓 itself. It is simply not enough then to look at our graph and read off the value of the function when π‘₯ is equal to negative three.

Instead, we’ll use a part of the fundamental theorem of calculus. This says that the definite integral between π‘Ž and 𝑏 of 𝑓 of π‘₯ with respect to π‘₯ or sometimes, alternatively, 𝑓 of 𝑑 with respect to 𝑑, is equal to capital 𝐹 of 𝑏 minus capital 𝐹 of π‘Ž, where capital 𝐹 is the antiderivative of our original function 𝑓.

Now, by definition, it follows that 𝑓 must be the antiderivative of 𝑓 prime. So, we can reword this somewhat and say that the definite integral between π‘Ž and 𝑏 of 𝑓 prime of π‘₯ with respect to π‘₯ is equal to 𝑓 of 𝑏 minus 𝑓 of π‘Ž. We’re going to let π‘Ž be equal to negative three and 𝑏 be equal to zero. And so, we now see that the definite integral between negative three and zero of 𝑓 prime of π‘₯ with respect to π‘₯ is equal to 𝑓 of zero minus 𝑓 of negative three. We’re trying to find the value of 𝑓 of negative three. So, let’s rearrange this equation to make that the subject.

When we do, we see that 𝑓 of negative three is equal to 𝑓 of zero minus that definite integral. We were told, though, that 𝑓 of zero is 10. So, what about our definite integral? Well, we recall that the definite integral tells us the net area between the curve and the π‘₯-axis. In this case, we’re interested in the area bounded by the lines π‘₯ equals negative three and π‘₯ equals zero. So, we’ll find the area between π‘₯ equals negative two and zero and then subtract the area of the triangle since that lies below the π‘₯-axis.

This first bit is a quarter of a circle, so its area is πœ‹π‘Ÿ squared divided by four or πœ‹ times two squared divided by four, which is equal to πœ‹ square units. The triangle has an area of a half times base times height. That’s a half times one times 0.5, which is 0.25 or a quarter square units. The net area, and therefore our definite integral, is given by πœ‹ minus a quarter.

So, we substitute this into our earlier formula for 𝑓 of negative three. And we find that 𝑓 of negative three is equal to 10 minus πœ‹ minus a quarter. Distributing our parentheses, that’s 10 minus πœ‹ plus a quarter, which simplifies to 41 over four minus πœ‹. So, we found the value of 𝑓 of negative three. Let’s clear some space and look at part ii) of this question.

Evaluate the definite integral between two and nine of 𝑓 prime of π‘₯ with respect to π‘₯.

The function 𝑓 prime, we can see, is piecewise over the closed interval π‘₯ equals two to nine. Therefore, we’re going to split it up into the following integrals. It’s the definite integral between two and four of 𝑓 prime of π‘₯ with respect to π‘₯ plus that same definite integral, but this time between the limits of four and six, plus the definite integral between six and nine of 𝑓 prime of π‘₯ with respect to π‘₯.

Now, it should be quite clear to us that this first integral is equal to zero. And that’s because there’s actually no area between the graph of 𝑓 prime of π‘₯ and the π‘₯-axis. This second integral is the negative area of this triangle. And it’s negative, of course, since it lies below the π‘₯-axis. So essentially, we’re going to subtract a half times two times two.

Finally, for our third definite integral, we’re actually going to evaluate the definite integral between six and nine of the function. So, that’s π‘₯ minus seven all squared minus three with respect to π‘₯. So, how do we evaluate this definite integral? Well, we could distribute the parentheses, simplify, and then integrate each term. Alternatively, we can use the reverse of the chain rule, also called integration by substitution. Here, we let 𝑒 be equal to π‘₯ minus seven. This means d𝑒 by dπ‘₯ equals one. Or alternatively, d𝑒 equals dπ‘₯.

Now remember, d𝑒 by dπ‘₯ isn’t a fraction. But when working with integration by substitution, we do treat it a little like one. This means we can replace π‘₯ minus seven with 𝑒 and dπ‘₯ with d𝑒. We are, however, going to need to change our limits. We substitute the lower limit into our original substitution. 𝑒 is equal to π‘₯ minus seven. So, when π‘₯ is equal to six, 𝑒 is equal to six minus seven, which is negative one. And when π‘₯ is equal to nine, 𝑒 is equal to nine minus seven, which is two.

We’re now going to evaluate the definite integral between negative one and two of 𝑒 squared minus three. It’s 𝑒 cubed over three minus three 𝑒. We then substitute our limits in and this becomes two cubed over two minus three times two minus negative one cubed over three minus three times negative one, which is equal to negative six. Our definite integral then simplifies to negative two plus negative six which is, of course, negative eight. The definite integral between two and nine of 𝑓 prime of π‘₯ with respect to π‘₯ is negative eight. Let’s clear some space and consider the third part of this question.

Part iii) says, for the open interval negative six to 10, on what open intervals, if any, is the graph of 𝑓 both increasing and concave downward? Give a reason for your answer.

We begin by recalling what it means for the graph of a function to be increasing. Its first derivative must be greater than zero. 𝑓 prime of π‘₯ must be greater than zero. Let’s check our graph for these locations. Well, we have two of these. One is here; it’s this semicircle bit. And the other is up here. So, what about the concave downward part?

Well, for a graph to be concave downward, the second derivative of the function 𝑓 double prime of π‘₯ must be less than zero. In other words, the first derivative, 𝑓 prime, must be decreasing. Well, there’s only one place on our graph where 𝑓 prime is positive but also decreasing. And that’s here. So, the graph of 𝑓 is both increasing and concave downward on the open interval zero to two. And this is because 𝑓 prime of π‘₯ is greater than zero and decreasing. We’ll now consider the fourth and final part of our question.

Part iv), find the π‘₯-coordinate of each point of inflection, if any, for the graph of the function 𝑓 for π‘₯ is greater than negative six and less than 10. Give a reason for your answer.

Remember, for a point to be an inflection point, its concavity changes. It either changes from concave up to down or vice versa. This means its second derivative must change sign. Alternatively, we can think about that as its first derivative going from increasing to decreasing or vice versa. What will that look like on our graph of 𝑓 prime of π‘₯?

Well, we’ll have a turning point. And there are two turning points on our graph of 𝑓 prime of π‘₯. They are here at π‘₯ equals zero and here at π‘₯ equals seven. So, the π‘₯-coordinates of our turning points are at π‘₯ equals zero and π‘₯ equals seven. And the reason behind this is 𝑓 prime of π‘₯ changes from being increasing to decreasing and vice versa.

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