Video: APCALC02AB-P2B-Q05-574131023031v2

The given figure shows the graph of 𝑓′, the derivative of a twice differentiable function 𝑓, on the interval between [βˆ’6, 7]. The graph has horizontal tangent at π‘₯ = βˆ’3, π‘₯ = 2, and π‘₯ = 5.5, and the areas of the regions bounded by the π‘₯-axis and the graph on the intervals [βˆ’5, 2] and [2, 7] at 27 and 9, respectively. i. Find all the π‘₯-coordinates at which 𝑓 has a relative minimum. Give a reason for your answer. ii.On what open interval contained in βˆ’6 < π‘₯ < 7 is the graph of 𝑓 both concave up and increasing? Give a reason for your answer. iii. Find the π‘₯-coordinate of all points of inflection for the graph of 𝑓. Give a reason for your answer. iv. Given that 𝑓(2) = βˆ’2, write an expression for 𝑓(π‘₯) that involves an integral. Find 𝑓(βˆ’5) and 𝑓(7).

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Video Transcript

The given figure shows the graph of 𝑓 prime, the derivative of a twice differentiable function 𝑓, on the interval between negative six and seven. The graph has horizontal tangent at π‘₯ is equal to negative three, π‘₯ is equal to two and π‘₯ is equal to 5.5 and the areas of the regions bounded by the π‘₯-axis and the graph on the intervals between negative five and two and two and seven at 27 and nine, respectively. Part one) Find all the π‘₯-coordinates at which 𝑓 has a relative minimum. Give a reason for your answer. Part two) On what open interval contained in the interval where π‘₯ is greater than negative six and less than seven is the graph of 𝑓 both concave up and increasing? Give a reason for your answer. Part three) Find the π‘₯-coordinate of all points of inflection for the graph of 𝑓. Give a reason for your answer. Part four) Given that 𝑓 of two equals negative two, write an expression for 𝑓 of π‘₯ that involves an integral. Find 𝑓 of negative five and 𝑓 of seven.

Now since there’s not much space left on the screen to answer the questions, I’ll remove parts two to four in order to make space to answer part one. In part one, we’re required to find all the π‘₯-coordinates for which 𝑓 has a relative minimum. Let’s first note that relative minimums will occur at critical points of the graph. So that is when 𝑓 prime of π‘₯ is equal to zero. However, not all critical points are relative minimums. In order to find whether a critical point is a relative minimum, we must consider the second derivative of 𝑓 with respect to π‘₯.

Now at a relative minimum, 𝑓 will go from decreasing to a slope of zero at the critical point and then to increasing. However, this is what happened on 𝑓 of π‘₯ not 𝑓 prime of π‘₯. So let’s translate what this means for 𝑓 prime of π‘₯. Well, when 𝑓 of π‘₯ is decreasing, it has a negative slope and therefore 𝑓 prime of π‘₯ is also negative. And so, we can say 𝑓 prime of π‘₯ is less than zero. Then, as we know, at the critical point 𝑓 prime of π‘₯ is equal to zero. Then to the right of the critical point, 𝑓 of π‘₯ is increasing. And so, it has positive slope. Therefore, 𝑓 prime of π‘₯ must be greater than zero.

So we’re looking for points on 𝑓 prime of π‘₯ where 𝑓 prime goes from being negative to being equal to zero to then being positive, there are three points on our graph when 𝑓 prime is equal to zero. And these are at negative five, two, and seven. However, only at one of these points, does 𝑓 of π‘₯ go from being negative to being equal to zero to being positive. And so that gives us the solution to the first part. 𝑓 has a relative minimum at π‘₯ is equal to negative five.

For part two, we’re required to find the open intervals between negative six and seven, for which 𝑓 is both concave up and increasing. Now if 𝑓 is increasing, this means that it has a positive slope. And so, this means the 𝑓 prime of π‘₯ or the derivative of 𝑓 with respect to π‘₯ must be positive too. So we can say that 𝑓 prime of π‘₯ must be greater than zero. Now we’re also looking for when 𝑓 is concave up. When 𝑓 is concave up the second derivative of 𝑓 with respect to π‘₯ or 𝑓 double prime of π‘₯ must be positive. So we can say that 𝑓 double prime of π‘₯ is greater than zero.

And now we can link this back to 𝑓 prime of π‘₯ since 𝑓 double prime of π‘₯ is the derivative of 𝑓 Prime of π‘₯ with respect to π‘₯. So 𝑓 double prime of π‘₯ tells us about the slope of 𝑓 prime of π‘₯. And so, we can say that 𝑓 prime of π‘₯ must be increasing. So now we found the conditions we require for 𝑓 prime of π‘₯ in order for 𝑓 to be both concave up and increasing. And that is for 𝑓 prime to be greater than zero and 𝑓 prime to be increasing. We can see for our graph that 𝑓 prime of π‘₯ is greater than zero between the values of negative five and seven except for at the value where π‘₯ is equal to two since 𝑓 prime of π‘₯ touches the π‘₯-axis here and it’s therefore equal to zero.

Okay, so now we just need to find for which π‘₯-values on this range between negative five and seven excluding two 𝑓 prime of π‘₯ is increasing. We can see that it’s going increasing between the values of negative five and the first turning point of the graph. Now we know that at a turning point, the slope of the curve will be equal to zero. And when the slope is equal to zero, the graph will have a horizontal tangent. We have been given the π‘₯-values of the horizontal tangent for 𝑓 prime of π‘₯. And these are at π‘₯ is equal to negative three, π‘₯ is equal to two, and π‘₯ is equal to 5.5.

Therefore, this turning point which we’re interested in must occur at π‘₯ equal to negative three. There is in fact another interval for which 𝑓 prime of π‘₯ is increasing. And that is between the values of π‘₯ is equal to negative two and our final turning point. So here’s you know π‘₯-values of the horizontal tangent, we know that this point must be π‘₯ is equal to 5.5. So now we have found the endpoints of the integrals over which 𝑓 is both concave up and increasing. We need to decide whether or not to include or exclude these endpoints.

At π‘₯ is equal to negative five and π‘₯ is equal to two, 𝑓 prime of π‘₯ is equal to zero. And we require 𝑓 prime of π‘₯ to be strictly greater than zero. Therefore, both of these points can be excluded from their integrals. Now at π‘₯ is equal to negative three and π‘₯ is equal to 5.5, 𝑓 prime of π‘₯ has a turning point. And therefore, the derivative of 𝑓 prime of π‘₯ or 𝑓 double prime of π‘₯ will be equal to zero. And we require for 𝑓 double prime of π‘₯ to be strictly greater than zero. So we can exclude negative three and negative five from their respective intervals. So we can say that 𝑓 is both concave up and increasing when π‘₯ is strictly greater than negative five and less than negative three and π‘₯ is strictly greater than two and strictly less than 5.5.

For part three, we’re required to find the π‘₯-coordinates of all points of inflection of 𝑓. Now a point of inflection will occur when the second derivative of 𝑓 with respect to π‘₯ is equal to zero. However, this does not guarantee that there is an inflection point at π‘₯. In order for π‘₯ to be an inflection point of 𝑓, 𝑓 double prime of π‘₯ must change from negative to positive or from positive to negative at π‘₯. And this is a condition on top of 𝑓 double prime of π‘₯ being equal to zero. Now let’s relate these two conditions to 𝑓 prime of π‘₯.

If 𝑓 double prime of π‘₯ is equal to zero, this means that 𝑓 prime of π‘₯ must have a critical point. And if 𝑓 double prime of π‘₯ is changing from negative to positive or positive to negative, then this means 𝑓 prime must change from decreasing to increasing or increasing to decreasing. Now we can combine these two conditions on 𝑓 prime of π‘₯. When we combine them, we can say that at any π‘₯-coordinate, where 𝑓 has an inflection point, 𝑓 prime of π‘₯ must have a relative maximum or relative minimum. So let’s find all the π‘₯-values for which 𝑓 prime of π‘₯ has a relative minimum or a relative maximum.

We can see that our graph has three critical points. And these critical points will occur when the graph has a horizontal tangent. And we know that the horizontal tangent occurs at π‘₯ is equal to negative three, π‘₯ is equal to two, and π‘₯ is equal to 5.5. We can see from our graph that at π‘₯ equals negative three and π‘₯ is equal to 5.5, we have relative maximums. And at π‘₯ is equal to two, we have a relative minimum. So we find that the π‘₯-coordinates of the points of inflection of 𝑓 are at π‘₯ is equal to negative three, π‘₯ is equal to two, and π‘₯ is equal to 5.5.

For part four, we’ve been given that 𝑓 of two is equal to negative two. We’ve been asked to find an expression for 𝑓 of π‘₯ involving an integral. We’ve also been asked to find the values of 𝑓 of negative five and 𝑓 of seven. In order to find an expression for 𝑓 of π‘₯, let’s first recall the fundamental theorem of calculus. This tells us that the integral from π‘Ž to 𝑏 of 𝑓 prime of 𝑑 with respect to 𝑑 is equal to 𝑓 of 𝑏 minus 𝑓 of π‘Ž.

Now since we know the value of 𝑓 of two, if we substitute in either π‘Ž or 𝑏 is equal to two, then we’ll know either 𝑓 of 𝑏 or 𝑓 of π‘Ž. And we’re also asked to find 𝑓 of π‘₯. So we can let the other bound β€” either π‘Ž or 𝑏 β€” be equal to π‘₯. We can choose them either way around. However, I’m gonna choose π‘Ž is equal to two and 𝑏 is equal to π‘₯, giving us that the integral from two to π‘₯ of 𝑓 prime of 𝑑 with respect to 𝑑 is equal to 𝑓 of π‘₯ minus 𝑓 of two. Now we know the value of 𝑓 of two. So can substitute in 𝑓 of two is equal to negative two. And now, we can rearrange this equation in order to make 𝑓 of π‘₯ the subject. We obtain that 𝑓 of π‘₯ is equal to the integral from two to π‘₯ of 𝑓 prime of 𝑑 with respect to 𝑑 minus two. So we have answered the first section of part four.

Now, let’s find 𝑓 of negative five. We can substitute π‘₯ is equal to negative five into our formula, giving us that 𝑓 of negative five is equal to the integral from two to negative five of 𝑓 prime of 𝑑 with respect to 𝑑 minus two. Now we notice that our bounds are in fact the wrong way around. Since two is greater than negative five, however currently two is our lower bound and negative five is our upper bound. However, we can use a rule which allows us to switch the bounds.

This rule tells us that the integral from π‘Ž to 𝑏 of 𝑓 of 𝑑 with respect to 𝑑 is equal to the negative of the integral from 𝑏 to π‘Ž of 𝑓 of 𝑑 with respect to 𝑑. Essentially, when we switch our bounds, we need to multiply our integral by negative one. This tells us that 𝑓 of negative five is equal to the negative of the integral from negative five to two of 𝑓 prime of 𝑑 with respect to 𝑑 minus two. Next, we can use the fact that this integral will tell us the area between 𝑓 prime of 𝑑 and the π‘₯-axis between the π‘₯-values of negative five and two.

Now, our line is of the function 𝑓 prime of π‘₯. And although we are integrating 𝑓 prime of 𝑑 with respect to 𝑑, the fact that our graph is of 𝑓 prime of π‘₯ does not matter. We can simply switch the variable of π‘₯ for 𝑑 and it will be the same. Therefore, we can say that the integral from negative five to two of 𝑓 prime of 𝑑 with respect to 𝑑 is equal to the shaded area under the curve between negative five and two.

Now if we look back in the question, we have been given the area of the region bounded by the π‘₯-axis and the graph on the interval between negative five and two. And it is equal to 27. So we know that this area is equal to 27. And we can substitute this in for our integral, giving us that 𝑓 of negative five is equal to negative 27 minus two. And therefore, we have found that 𝑓 of negative five is equal to negative 29.

Now all that remain to do is to find the value of 𝑓 of seven. We can again substitute π‘₯ is equal to seven into our formula for 𝑓 of π‘₯, giving us that 𝑓 of seven is equal to the integral from two to seven of 𝑓 prime of 𝑑 with respect to 𝑑 minus two. Now the integral from two to seven of 𝑓 prime of 𝑑 with respect to 𝑑 is equal to the area between the curve of 𝑓 prime of π‘₯ and the π‘₯-axis between the π‘₯-coordinates of two and seven. So that’s this shaded area here. We’ve been told that the area of the region bounded by the π‘₯-axis and the graph on the interval between two and seven is equal to nine. Therefore, we can set our integral equal to nine. Simplifying this, we find the last part of our solution which is that 𝑓 of seven is equal to seven. This completes part four of the question. Let’s quickly recap our solutions.

In part one, we found that 𝑓 has a relative minimum at π‘₯ is equal to negative five. In part two, we found that 𝑓 is both concave up and increasing on the intervals where π‘₯ is greater than negative five and less than negative three and where π‘₯ is greater than two and less than 5.5. In part three, we found that 𝑓 has inflection points at π‘₯ is equal to negative three, π‘₯ is equal to two, and π‘₯ is equal to 5.5. And in part four, we found that 𝑓 of π‘₯ is equal to the integral from two to π‘₯ of 𝑓 prime of 𝑑 with respect to 𝑑 minus two. 𝑓 of negative five is equal to negative 29. And 𝑓 of seven is equal to seven.

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