### Video Transcript

Consider the function π of π₯ is
equal to four π₯ squared over π₯ squared plus three. Find π prime of π₯. Find and classify all critical
points of π and all intervals of increase and decrease of π. Which of the following could be the
graph of π? Options (A), (B), (C), and (D).

In this question, we are given a
function π of π₯. And we can see that this function
is the quotient of two polynomials. In other words, it is a rational
function. And we need to answer four
questions about this function. The first three parts lead to
giving us information about the graph of this function. To start, letβs clear some space
since we do not need the graphs until the fourth part, and we can look at each graph
separately when we arrive at that question.

We want to find an expression for
π prime of π₯, that is, the first derivative of π of π₯ with respect to π₯. We can differentiate π of π₯ in
many different ways. However, since π of π₯ is the
quotient of polynomials, we will do this by using the quotient rule. We recall that this tells us if we
have two differentiable functions π of π₯ and β of π₯, then the derivative of π of
π₯ over β of π₯ with respect to π₯ is π prime of π₯ β of π₯ minus β prime of π₯ π
of π₯ all over β of π₯ squared, provided β of π₯ is nonzero. We set π of π₯ to be the
polynomial in the numerator of π of π₯, that is, four π₯ squared. And we set β of π₯ to be the
polynomial in the denominator of π of π₯, that is, π₯ squared plus three.

Since both of these functions are
polynomials, we can differentiate them term by term by applying the power rule for
differentiation. We multiply by the exponent of π₯
and then reduce this exponent by one. We obtain that π prime of π₯ is
eight π₯ and β prime of π₯ is two π₯. We can now find an expression for
π prime of π₯ by substituting these expressions into the quotient rule. Doing this gives us that π prime
of π₯ is equal to eight π₯ times π₯ squared plus three minus two π₯ multiplied by
four π₯ squared all over π₯ squared plus three all squared.

There are many ways we could
simplify this expression. For instance, we could take the
shared factor of eight π₯ out of the numerator. However, we are going to distribute
the parentheses in the numerator. This gives us eight π₯ cubed plus
24π₯ minus eight π₯ cubed all over π₯ squared plus three all squared. We can then note that eight π₯
cubed minus eight π₯ cubed is equal to zero. We cannot simplify this expression
any further. Hence, we have shown that π prime
of π₯ is equal to 24π₯ over π₯ squared plus three all squared.

Letβs now clear some space and move
on to the second part of the question. We want to find and classify all of
the critical points of π. We can start by recalling that the
critical points of a function occur when its derivative is zero or when its
derivative does not exist. We can check for values of π₯ where
the derivative does not exist by considering our expression for π prime of π₯. We know that this expression is
valid for all values of π₯ in the domain of π such that the denominator is
nonzero. We can note that π₯ squared plus
three is always greater than or equal to three, so it is never zero. Thus, the denominator of π prime
of π₯ is always nonzero. Therefore, the derivative of π
exists for all values of π₯ in the domain of π. So, we only need to look for values
of π₯ where the derivative is zero.

Before we do this, it is worth
noting that the domain of π of π₯ is all real values since its denominator is
nonzero and it is a rational function. To find the critical points of π,
we can solve π prime of π₯ equals zero. We get the equation 24π₯ over π₯
squared plus three all squared equals zero. For a quotient to be equal to zero,
we need the numerator to be equal to zero and the denominator to be nonzero. We know that the denominator is
nonzero for all real values of π₯. So we only need to solve 24π₯
equals zero, which occurs only when π₯ equals zero. Since we want the critical points
of π, we need to find its π¦-coordinate by substituting π₯ equals zero into π of
π₯ and evaluating. We obtain that π of zero equals
four times zero squared over zero squared plus three, which we can calculate is
equal to zero.

Therefore, we have shown that the
only critical point of π is the point zero, zero. There are many ways that we can
classify this critical point. However, since we have an
expression for π prime of π₯, we will use the first derivative test. We can start by recalling that the
denominator of π prime of π₯ is always positive. In fact, it is greater than or
equal to three. This means that the sign of π
prime of π₯ is determined entirely by the sign of the numerator. We can then note that if π₯ is
positive, 24π₯ is positive. So π prime of π₯ is the quotient
of two positive numbers and hence must be positive. Similarly, if π₯ is negative, then
the numerator of π prime of π₯ is negative, so π prime of π₯ is negative.

We can now apply the first
derivative test to classify the critical point at the origin. First, we show that when π₯ is less
than zero, π prime of π₯ is negative. This means that the graph must
decrease to the origin from the left. We know that the slope is zero at
the origin, and we also showed that the derivative is positive when π₯ is greater
than zero. So, the graph must increase to the
right of zero. This means that the origin is a
local minimum of the graph of this function.

We can now move on to the third
part of this question. However, we have already considered
this when we used the first derivative test. We recall that a differentiable
function is increasing when its derivative is positive and decreasing when its
derivative is negative. We showed that π prime of π₯ is
defined for all real values of π₯ and that it is positive for π₯ greater than zero
and negative for π₯ less than zero. We can write this in interval
notation as π is increasing on the open interval from zero to β and π is
decreasing on the open interval from negative β to zero.

We can now move onto the final part
of the question. We need to determine which of the
four given graphs are possible graphs of π¦ equals π of π₯. We will look at each option
separately and use our answers to the previous parts to aid us. Letβs start with option (A). First, we can note that the origin
is not a local minimum of the graph, since the graph is negative when π₯ is less
than zero and positive when π₯ is greater than zero. This is enough to say that this is
not a possible graph of π. However, we can also note that the
graph appears to have two critical points. And the graph decreases for a
section to the right of zero and increases for a section to the left of zero. None of this must be true for the
graph of π¦ equals π of π₯.

Letβs follow the same process for
option (B). Since it is difficult to see this
graph with the given scaling, we can zoom in on the critical point at the origin to
better see what is occurring. In this case, we can see that the
origin is a local maximum of the function. And as such, the graph increases to
zero from the left and decreases from zero to the right. Hence, this is not the correct
graph of π¦ equals π of π₯.

We can follow the same process for
the graph in option (C). We note that the origin in this
graph is a local maximum, not a local minimum. So this cannot be the correct
option.

This only leaves the graph in
option (D). For due diligence, letβs check that
it matches the information we have found. First, we note that the origin
appears to be the only critical point and it is a local minimum. Second, we can also note that the
graph slopes downward to the left of zero and the graph slopes upward to the right
of zero. So, the graph is increasing and
decreasing on the correct intervals. Hence, the answer is that only the
graph in option (D) can be a graph of the given function π of π₯.