Video Transcript
In this video, we’ll learn how
finding the derivative can make graphing a function much more simple. Up to this stage, it’s likely we’ll
have focused on particular aspects of curve sketching, such as limits, continuity,
and concavity. It’s now time to combine all of
this information to help us sketch graphs that reveal the full extent of the
features of the various functions.
The given checklist is intended to
be used like a set of guidelines when curve sketching. We won’t need to use every single
point every single time. But it’s a good idea to begin by
considering each of these features in turn.
The first is domain and range. It can be useful to consider the
domain of the function. In other words, the set of values
of 𝑥, for which the function is defined. And from this, we might consider
the range, the complete set of resulting function values, though this usually
develops through the other steps.
The second is intercepts. We look for both 𝑦- and
𝑥-intercepts. Remember, the 𝑦-intercept is
obtained by letting 𝑥 equal to zero and solving for 𝑦, whereas the 𝑥-intercepts
are obtained by setting 𝑦 equal to zero and solving for 𝑥. We look for symmetry. Are the functions even? In other words, does 𝑓 of negative
𝑥 equal 𝑓 of 𝑥? Are they odd? Does 𝑓 of negative 𝑥 equal
negative 𝑓 of 𝑥? Or neither. And then we might consider
asymptotes.
Remember, if the limit as 𝑥 tends
to either positive or negative infinity of the function is equal to some value 𝐿,
then the line 𝑦 equals 𝐿 is a horizontal asymptote of the curve. If it turns out that the limit as
𝑥 tends to infinity of the function is either positive or negative infinity, then
we don’t have an asymptote. But this fact is still useful for
sketching the curve.
We also recall that the line 𝑥
equals 𝑎 is a vertical asymptote if at least one of the following is true. The limit as 𝑥 tends to 𝑎 from
the right of the function is either positive or negative infinity. Or the limit as 𝑥 approaches 𝑎
from the left of the function is either positive or negative infinity. The fifth point is intervals of
increase or decrease.
We recall that a function is
increasing when its derivative is greater than zero and decreasing when its
derivative is less than zero. This tells us the shape of the
graph over various intervals. We also look for local extrema. Remember, critical values occur
when the derivative is equal to zero or does not exist. Then the first derivative test can
tell us the nature of these critical points.
We can consider concavity and
points of inflection. And this time, we evaluate the
second derivative. If it’s greater than zero over some
interval, then the curve is concave upward. And if it’s less than zero over
some interval, the curve is concave downward. When the concavity changes, we know
we have a point of inflection. Finally, we might consider the end
behavior of the function. And this we usually come about as a
result of the other work we’ve done.
This is, of course, a hugely
extensive list. And there’ll be times where many of
the features can be predicted by using a graphing calculator. But one of these will not always be
accessible. We’ll now consider a number of
examples which require consideration of these features.
Consider the function 𝑓 of 𝑥
equals 𝑥 minus one squared times 𝑥 plus two. Find 𝑓 prime of 𝑥. Find and classify the critical
points of 𝑓. Find the intervals of increase and
decrease for 𝑓. Find the limit as 𝑥 approaches
infinity of 𝑓 of 𝑥.
And then there’s one more part of
this question, which asks us to identify the correct graph for our function. So we’ll look at our options when
we’ve completed the first four parts. So how are we going to evaluate 𝑓
prime of 𝑥?
That’s the first derivative of our
function with respect to 𝑥. Well, taking a look at our
function, there are a number of ways we can do this. We could distribute our parentheses
and differentiate term by term. Alternatively, notice that we have
a product of two functions, one of which is a composite function. So we could use the product rule
alongside the general power rule.
The product rule says that the
derivative of the product of two differentiable functions 𝑢 and 𝑣 is 𝑢 times d𝑣
by d𝑥 plus 𝑣 times d𝑢 by d𝑥. So let’s let 𝑢 be equal to 𝑥
minus one all squared and 𝑣 be equal to 𝑥 plus two. We’ll need to use the general power
rule to find the derivative of 𝑥 minus one all squared. This is a special case of the chain
rule. And it says we can multiply the
entire function by the exponent, reduce the exponent by one, and then multiply all
of this by the derivative of the inner function.
Well, the derivative of 𝑥 minus
one is just one. So d𝑢 by d𝑥 is two times 𝑥 minus
one to the power of two minus one times one. And that simplifies to two times 𝑥
minus one. d𝑣 by d𝑥 is much more straightforward. It’s simply one. So we substitute what we have into
the product rule. It’s 𝑢 times d𝑣 by d𝑥 — that’s
𝑥 minus one all squared times one — plus 𝑣 times d𝑢 by d𝑥. Distributing our parentheses, and
the first derivative of our function simplifies really nicely to three 𝑥 squared
minus three.
In the second part of this
question, we need to find and classify the critical points of our function. Remember, these occur when our
first derivative is either equal to zero or does not exist. Well, we actually know that
polynomial functions are all differentiable. So we’re just going to consider the
case for which our first derivative is equal to zero. That is, three 𝑥 squared minus
three is equal to zero.
Let’s solve for 𝑥. We divide through by one. And then we use the difference of
two squares to factor 𝑥 squared minus one. It’s 𝑥 plus one times 𝑥 minus
one. Now for the product of these two
terms to be equal to zero, either 𝑥 plus one has to be equal to zero or 𝑥 minus
one has to be equal to zero. And solving each equation for 𝑥,
we obtain 𝑥 must be either negative one or 𝑥 equals one. So we have critical points where 𝑥
is either equal to positive or negative one.
Our next job is to classify
these. We can perform the first derivative
test and evaluate the first derivative just before and just after each critical
point. Let’s add a table. We know that the first derivative
is equal to zero when 𝑥 is equal to negative one or positive one. The first derivative evaluated at
𝑥 equals negative two is three times negative two squared minus three, which is
nine. The first derivative evaluated when
𝑥 is equal to zero is three times zero squared minus three, which is negative
three. And the first derivative evaluated
at 𝑥 equals two is also nine. The function is increasing before
𝑥 equals negative one and decreasing after. So the critical point 𝑥 equals
negative one is a local maximum. The opposite is true when 𝑥 equals
one. So that must be our local
minimum.
In the third part of this question,
we’re looking to find the intervals of increase and decrease for our function. Let’s clear some space. Now this might seem awfully similar
to what we just did. But at that point, we were simply
considering the nature of the function at specific points. We now want to know over what
intervals the function is increasing or decreasing. So we’re going to work out when the
first derivative is less than zero, decreasing, or greater than zero,
increasing.
We know the graph of 𝑦 equals
three 𝑥 squared minus three looks a little something like this. We can see that the graph of the
first derivative is greater than zero here and here. In other words, when 𝑥 is less
than negative one or greater than one. The first derivative is less than
zero here. That’s when 𝑥 is greater than
negative one or less than one. So the intervals of increase are
the open intervals from negative infinity to negative one and one to infinity. And the interval of decrease is the
open interval from negative one to one.
And finally, we need to find the
limit as 𝑥 approaches infinity of the function. We can do this by direct
substitution. We see that as 𝑥 approaches
infinity, the function itself also grows larger and larger. So the function also approaches
infinity. Let’s now combine everything we’ve
done to identify the graph of our function.
Which of the following is the graph
of 𝑓? If we were to distribute the
parentheses of our function, we’d see we have a cubic graph with a positive leading
coefficient. In other words, the coefficient of
𝑥 cubed is positive. That tells us the shape of the
graph will be a little something like this. We know it has critical points 𝑥
equals negative one and one and a maximum and minimum in those locations,
respectively.
We can work out the value of our
function at these points by substituting negative one and one in. And when we do, we see that 𝑓 of
negative one is four and 𝑓 of one is zero. So we know the graph has a maximum
at negative one, four and a minimum at one, zero. We also saw that it had intervals
of increase over the open interval negative infinity to negative one and one to
infinity and decrease over the open interval negative one to one. The only graph that satisfies all
of these conditions is A. In fact, we could have very quickly
spotted they could not have been D or E as these are graphs of quadratic
functions. C is a cubic graph which has a
negative coefficient of 𝑥 cubed. So we really only had two
options.
Notice that we didn’t worry about
evaluating the limits of our functions to look for asymptotes. As we know, a cubic graph doesn’t
have any.
In our next example, we will
consider a graph which has at least one asymptote.
Sketch the graph of 𝑓 of 𝑥 equals
three 𝑥 squared over 𝑥 squared minus four.
Let’s begin by considering the
domain and range of our function. We know that a function which is a
quotient won’t exist at points where the denominator of the quotient is zero. So we set 𝑥 squared minus four
equals zero and solve for 𝑥 to find the domain of our function. We can add four to both sides of
this equation. Then we take the square root of
both sides of our equation, remembering to take both the positive and negative
square root of four. And we obtain that when 𝑥 squared
minus four is equal to zero, 𝑥 is equal to positive or negative two. So the domain of our function is
all real values, except 𝑥 is equal to positive or negative two. Now the range itself is a set of
possible outputs on the function. And we can infer this through the
graphing process.
Next, we’ll work out whether there
are any intercepts. By setting 𝑥 equal to zero and
solving for 𝑦, we’ll find the 𝑦-intercept. When we do, we obtain 𝑦 to be
equal to three times zero squared over zero squared minus four, which is zero. So there’s a 𝑦-intercept at 𝑦
equals zero. Next, we set 𝑦 or 𝑓 of 𝑥 equal
to zero and solve for 𝑥. That is, zero equals three 𝑥
squared over 𝑥 squared minus four.
Now for this to be true, we know
that the numerator of this fraction must itself be equal to zero. And for three 𝑥 squared to be
equal to zero, 𝑥 must be equal to zero. So we actually only have one
intercept altogether. And that’s at the origin: zero,
zero. Next, we can check for
symmetry. An even function is one for which
𝑓 of negative 𝑥 equals 𝑓 of 𝑥, whereas an odd function is one for which 𝑓 of
negative 𝑥 equals negative 𝑓 of 𝑥. Well, 𝑓 of negative 𝑥 is three
times negative 𝑥 squared over negative 𝑥 squared minus four, which is equal to 𝑓
of 𝑥. So our function is even. And that means it’s going to be
symmetrical about the 𝑦-axis.
Next, we’ll look for
asymptotes. We’ll look for horizontal
asymptotes by considering what happens to our function as 𝑥 approaches either
positive or negative infinity. We can’t use direct substitution as
when we substitute in 𝑥 equals either positive or negative infinity, we end up with
infinity over infinity, which is undefined. So instead, we divide both the
numerator and the denominator of our expression by 𝑥 squared. That gives us three over one minus
four over 𝑥 squared.
And now we can use direct
substitution. As 𝑥 approaches either positive or
negative infinity, four over 𝑥 squared approaches zero. So the limit of our function is
three over one, which is just three. And we see that the line 𝑦 equals
three must be a horizontal asymptote.
Remember, we said that when 𝑥 is
equal to positive or negative two, the denominator is zero. So we find the following
limits. The limit as 𝑥 approaches two from
the right of our function is infinity. And the limit as 𝑥 approaches two
from the left of the function is negative infinity. And the limit as 𝑥 approaches
negative two from the right of the function is negative infinity. And as it approaches negative two
from the left, it’s positive infinity. So we obtain 𝑥 equals positive or
negative two to be vertical asymptotes.
Let’s clear some space and look for
intervals of increase and decrease. We’ll begin by using the quotient
rule to find the first derivative of our function. By letting 𝑢 be equal to three 𝑥
squared and 𝑣 be equal to 𝑥 squared minus four, we obtain expressions for d𝑢 by
d𝑥 and d𝑣 by d𝑥. And so the first derivative of our
function is 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 over 𝑣 squared. And distributing our parentheses,
we find the first derivative to be negative 24𝑥 over 𝑥 squared minus four all
squared.
Now since 𝑥 squared minus four all
squared must be greater than zero for all 𝑥, this means the first derivative of our
function must be less than zero when 𝑥 is greater than zero. So for values of 𝑥 greater than
zero, we get negative 24 multiplied by a positive, which gives us a negative
value. And so for 𝑥 greater than zero,
the first derivative is less than zero. The opposite is true. When 𝑥 is less than zero, the
first derivative is greater than zero. Since we know the function doesn’t
exist when 𝑥 is equal to negative two, we can say that its intervals of increase
are negative infinity to negative two and negative two to zero. And it has open intervals of
decrease over zero to two and two to infinity.
Now, actually, if we look
carefully, we also see that the first derivative is equal to zero when 𝑥 is equal
to zero. So there’s a critical point at the
point zero, zero. Coincidently, that’s also the point
where our curves cross the axes.
We now have everything we need to
complete our sketch. We know that there’s a horizontal
asymptote at 𝑦 equals three and vertical asymptotes at 𝑥 equals negative two and
𝑥 equals two. There’s a critical point and an
intercept at zero, zero. And that’s important because we see
that the curve won’t be able to cross the 𝑥-axis here or here. That helps us decide where these
two parts of the curve go. The curve is increasing over the
open interval negative infinity to negative two and decreasing over the open
interval two to infinity. And of course, we have these
asymptotes. The function is also even and
symmetrical about the 𝑦-axis. So this confirms we’re probably on
the right track. We then use the remaining pieces of
information to sketch the final part of our curve. And we’ve successfully sketched the
graph of 𝑓 of 𝑥 equals three 𝑥 squared over 𝑥 squared minus four.
In this video, we’ve seen that
using derivatives can help us to graph complicated functions. We have a rather extensive set of
guidelines, including finding the domain or range, the intercepts, the symmetry of
the curve. We look for asymptotes, intervals
of increase or decrease, any local extrema, concavity, or inflection points. And we also consider the end
behavior of our function. And of course, this list is
extremely extensive. But we won’t need to use every
single point every single time.
