### Video Transcript

A function π is continuous and
twice differentiable for all values of π₯. The given figure shows the graph of
π prime, the derivative of the function π on the closed interval between negative
three and two. The graph has horizontal tangents
at π₯ is equal to negative 1.5 and π₯ is equal to one. The areas of the regions A, B, and
C are three, one, and 12, respectively. And π of negative one is equal to
negative three. Part one, find all π₯-values on the
closed interval between negative three and two such that the function π has a local
maximum. Justify your answer. Part two, find all π₯-values on the
closed interval between negative three and two such that the function π has a point
of inflection. Justify your answer. Part three, evaluate the limit as
π₯ tends to negative two of four π of π₯ over π₯ minus two. Explain your reasoning. Part four, evaluate the limit as π₯
tends to two of 15 plus π of π₯ over π₯ minus two. Explain your reasoning.

In order to give us some more space
to answer the question, Iβm just going to clear up parts two to four. And weβll focus on part one for
now. For part one, weβre required to
find all π₯-values on the closed interval between negative three and two such that
π has a local maximum. Now if π has a local maximum, that
means there must be a critical point. And a critical point will occur
when the differential π is equal to zero. Or we could say that π prime of π₯
is equal to zero.

Now, not all critical points are
local maximums. In order to find out what our
critical point is, we can look at the second derivative of π with respect to π₯ or
π double prime of π₯. Now if π has a local maximum, this
tells us that, at π₯, our function π will be concave down, as we can see in this
little sketch. And if π is concave down, that
means that the second derivative of π with respect to π₯ must be less than
zero. Now, π double prime of π₯ is the
derivative of π prime of π₯ with respect to π₯. So π double prime of π₯ tells us
about the slope of π prime of π₯. Therefore, if π double prime of π₯
is negative, π prime of π₯ has a negative slope and is therefore decreasing.

Combining these two pieces of
information, we know that π will have a local maximum when π prime is equal to
zero and decreasing. From our graph, we can see that π
prime is equal to zero when π₯ is equal to negative two, π₯ is equal to negative
one, and π₯ is equal to two. And the only one of these three
points where it is decreasing is at π₯ is equal to negative one. Therefore, our solution to the
first part is that the only π₯-value for which π has a local maximum on the
interval between negative three and two is when π₯ is equal to negative one.

Part two, find all π₯-values on the
closed interval between negative three and two such that the function π has a point
of inflection. Justify your answer.

Now, point of inflection occurs
when the concavity of π changes from concave up to concave down or concave down to
concave up. And we know that the concavity of
π is represented by the second derivative of π with respect to π₯ or π double
prime of π₯. We know that when π double prime
of π₯ is less than zero, π is concave down. And when π is greater than zero,
π is concave up. So what weβre looking for is a sign
change of π double prime of π₯. Now our graph is of π prime of π₯,
not π double prime of π₯. So letβs try to work out what this
means for π prime of π₯.

Well, π double prime of π₯ tells
us about the slope of π prime of π₯. So when thereβs a sign change to π
double prime of π₯, this means thereβll be a sign change in the slope of π prime of
π₯. So we can say that π prime of π₯
changes from increasing to decreasing or decreasing to increasing. When a function changes from
increasing to decreasing, we have a local maximum. And when it changes from decreasing
to increasing, we have a local minimum. So weβre looking for local maximum
and local minimum on π prime of π₯. From our graph, we can see that
there are two.

Now in order to find the values of
these points, we need to refer back to the question. Since weβve been told that the
graph has horizontal tangent at π₯ is equal to negative 1.5 and π₯ is equal to
one. And these horizontal tangents will
occur at the critical points of our graph and the critical points of the local
minimum and the local maximum on the graph. Therefore, we can deduce that the
point of inflection of π will be at π₯ is equal to negative 1.5. And π₯ is equal to one.

Part three, evaluate the limit as
π₯ tends to negative two of four π of π₯ over π₯ minus two. Explain your reasoning.

Letβs start by trying to find this
limit using direct substitution. We obtain four timesed by π of
negative two over negative two minus two, which can be simplified to give negative
π of negative two. Now, we do not know the value of π
of negative two. We can use the fundamental theorem
of calculus in order to help us find this value. The fundamental theorem of calculus
tells us that the integral from π to π of π prime of π₯ with respect to π₯ is
equal to π of π minus π of π. And the reason why weβre going to
use the fundamental theorem of calculus here is because weβve been given the graph
of π prime of π₯. And weβre trying to find π of
negative two. And if we choose our bounds right,
weβll be able to find π of negative two using this method. We have also been given the value
of π of negative one in the question. So the two bounds we will use here
will be negative one and negative two.

Substituting in π is equal to
negative two and π is equal to negative one, we obtain that the integral from
negative two to negative one of π prime of π₯ with respect to π₯ is equal to π of
negative one minus π of negative two. And we can substitute in this value
of π of negative one since itβs equal to negative three, to give us that the
integral is equal to negative three minus π of negative two. Now, letβs think about what this
integral is represented by on our graph. Itβs the area between the curve and
the π₯-axis between the π₯-values of negative one and negative two. So thatβs this region here, which
is also labelled as region B.

Now, if you look back at the
question, weβve in fact been given the area of region B. And itβs equal to one. Therefore, we know that the value
of this integral must be one. Therefore, we can say one is equal
to negative three minus π of negative two. Rearranging this, we obtain that π
of negative two is equal to negative four. Now, we simply substitute this
value of π of negative two back in, in order to find the value of the limit. In doing so, we evaluate the limit
to be equal to four.

Part four, evaluate the limit as π₯
tends to two of 15 plus π of π₯ over π₯ minus two. Explain your reasoning.

Now, letβs try to solve this limit,
again using direct substitution. We obtain that itβs equal to 15
plus π of two over two minus two. Now, the denominator of this
fraction is equal to zero. Therefore, we know that this
fraction will be undefined. However, if the numerator is also
equal to zero, then weβll be able to use LβHΓ΄pitalβs rule. So letβs check what π of two is
equal to. Similarly to the last part, weβll
use the fundamental theorem of calculus. However, this time, weβre trying to
find the value of π of two, not π of negative two. So we know that one of our bounds
will be two. And since weβve been given π of
negative one in the question, we can again use negative one as the other bound.

Letting π be equal to negative one
and π be equal to two, we find that the integral from negative one to two of π
prime of π₯ with respect to π₯ is equal to π of two minus π of negative one. We know that π of negative one is
negative three. So we can substitute this in
here. The right-hand side of our equation
becomes π of two plus three. Now, the integral from negative one
to two of π prime of π₯ with respect to π₯ is represented by the area between the
curve of π prime of π₯ and the π₯-axis between π₯ is equal to negative one and π₯
is equal to two. So thatβs this area here,
represented by the region C. Weβve been given in the question
that the area of region C is 12. Now we must be careful here since
this region is below the π₯-axis. Therefore, the value of its
integral will be negative. And so we can say that the integral
from negative one to two of π prime of π₯ with respect to π₯ must be equal to
negative 12.

We obtain that negative 12 is equal
to π of two plus three, which can be rearranged to give π of two is equal to
negative 15. Now, we can substitute this value
of π of two back in to our evaluation of the limit. We obtain that our limit is equal
to 15 minus 15 over two minus two, which is equal to zero over zero. And as mentioned earlier, this
enables us to use LβHΓ΄pitalβs rule on this limit.

LβHΓ΄pitalβs rule tells us that the
limit as π₯ approaches π of π of π₯ over β of π₯ is equal to the limit as π₯
approaches π of π prime of π₯ over β prime of π₯. In our case, we have that π of π₯
is equal to 15 plus π of π₯. And β of π₯ is equal to π₯ minus
two. Differentiating π of π₯ with
respect to π₯, we find that π prime of π₯ is equal to π prime of π₯. And differentiating β of π₯ with
respect to π₯, we get that β prime of π₯ is equal to one.

Now, we can substitute π prime and
β prime into LβHΓ΄pitalβs rule. We obtain that the limit as π₯
approaches two of 15 plus π of π₯ over π₯ minus two is equal to the limit as π₯
approaches two of π prime of π₯ over one. And here, we can use direct
substitution, giving us that our limit is equal to π prime of two. In order to find the value of π
prime of two, we can simply read it off our graph. We see that when π₯ is equal to
two, π prime of π₯ is equal to zero. Therefore, the value of this limit
and the answer to part four must be zero.

Weβve now completed all four parts
of the question. Letβs quickly go over our
solutions. In part one, we found that π has a
local maximum at π₯ is equal to negative one. In part two, we found that π has
an inflection point at π₯ is equal to negative 1.5 and π₯ is equal to one. In part three, we found that the
limit as π₯ tend to negative two of four π of π₯ over π₯ minus two is equal to
four. And in part four, we found that the
limit as π₯ tends to two of 15 plus π of π₯ over π₯ minus two is equal to zero.