Curve Sketching

Our goal is to understand how the graph of a function relates to the graph of its derivative, and vice versa. In particular, we want to be able to pick out key characteristics from each graph and use that information to sketch a graph of either the derivative or anti-derivative.

Armed with the power rule for polynomials which states that , let us begin by looking at a particular function and its derivative.

**Example 1**: Graph *f*(*x*)
= *x*^{3} *x*^{2}
+ *x*
+ 1 and sketch its derivative.

**Solution**: When
sketching graphs of functions and derivatives, it is often useful to stack them
on top of each other, with the same *x*-axis
being used. That way, we can see how attributes of one graph appear in the
other.

Notice that if *f*(*x*) = *x*^{3}
4*x*^{2}
+ 2*x* + 1, then we have that* f*′(*x*)
= 3*x*^{2} 8*x* +
2.

Graph of

The graph of

*f*(*x*)*f*(*x*) appears to the left
with a graph of *f*′(*x*)
below it. (We shall construct *f*′(*x*)
shortly, but for the time being, we can use it to help illustrate some
relationships.)

Notice that *f*(*x*) has a local max.
around *x* = 1/3 and a local min. around *x*
= 7/3.

Graph of

*f*¢(*x*)

If we look at the graph of *f*′(*x*),
we notice that it has *x*-intercepts at
precisely the same points.

This should not surprise us,
since, by definition, the slope of *f*(*x*) is zero at its local max/min and the
graph of *f*′(*x*)
is a graph of the slopes of *f*(*x*).

** Thus, the points where f(x) has
zero slope corresponds to x-intercepts on f′(x)**.

** Figure 1: **Graphs of *f*(*x*)
and *f*′(*x*)

Knowing the *x*-intercepts
is one piece of the graph. The question remains: what does the graph do between
the *x*-intercepts?

Recall, one interpretation of the derivative is the slope of
the tangent line at a point. Using that property, we can determine the
intervals where *f*′(*x*)
is positive and where *f*′(*x*)
is negative.

Looking back at our graph of *f*(*x*), we see that the
function *f*(*x*) is increasing (i.e. *f*(*x*) has positive slope, i.e. *f*′(*x*)
is positive) on the intervals (-∞, 1/3) and
(7/3, ∞).
Similarly, we see that *f*(*x*) is decreasing (i.e. *f*(*x*)
has negative slope, i.e. *f*′(*x*)
is negative) on the interval (1/3, 7/3).

So, we know that *f*′(*x*)
lies above the *x*-axis to the left of
the first *x*-intercept and to the
right of the second *x*-intercept; in
between the *x*-intercepts, *f*′(*x*)
lies below the *x*-axis.

So, let us make a crude sketch of what *f*′(*x*) should look like, given the
information we have found thus far. (Compare this to the actual graph above.)

**
Figure 2: **A rough sketch of *f*′(*x*)

When we sketch our graph, we need to follow the general feel
of the graph that is given. What do we mean by that? Well, if *f*(*x*)
appears to have nice, smooth curves, then we should draw something similar for
the graph of *f*′(*x*).
That is, *f*′(*x*)
should not be rough or jagged, as is shown above, since the graph of *f*(*x*)
does not look like this.

What’s more, we can get a better idea of what the graph
should look like. Notice that *f*(*x*) appears to be the graph of a cubic
(third degree polynomial). The coefficient out front of the *x*^{3} term is positive, since *f*(*x*)
∞ as *x*
∞.

Now, using the fact that ,
we can see that the derivative of our function should look like *bx*^{2}, where *b* is positive. (Taking the derivative of
a polynomial reduces its degree by one and it does not affect the sign out front.)

So, let us reconsider our graph above. Notice that it looks *similar* to a parabola, though it is far
too rough. Smoothing out the graph, we have the following graph:

**
Figure 3: **A better sketch of *f*′(*x*)

Comparing this graph to that of the actual graph of *f*′(*x*),
we see that the two are almost identical, except for a few small differences.
Those differences include a higher local minimum (around *y* = -2, instead of *y* =
-3) and slightly different *y*-intercept
(*y* = 1.5, instead of *y* = 2).

These leads into the next property of curve sketching. ** We
can be certain about x-values, but not certain about y-values without knowing
the actual function**. When we sketch a curve, we are illustrating the
basic shape, not the exact graph. With that in mind, both the actual graph and
the sketched graph would be considered accurate.

Let us try our hand at another example of sketching the
derivative given the function, but this time instead of being given the
function *f*(*x*) explicitly, we only have a graph of the function to work with.

**Example 2**: Given
the following graph of *f*(*x*), sketch the graph of *f*′(*x*).

**Figure 4**: Graph of *f*(*x*)

**Solution**: We
begin by stacking the graph of *f*(*x*) and *f*′(*x*) on top of each other. Also, we shall
draw in the *x*-intercepts of *f*′(*x*).
Those occur wherever *f*(*x*) has local maximums or local minimums.
In this case, we see that happens at *x*
= -1, *x* = 2, and *x* = 4.

**Figure 5**: Graph of *f*(*x*) and the *x*-intercepts of *f*′(*x*)

Let us get a feel for what the graph of *f*′(*x*) looks like. To the left of *x* = -1, we see that *f*(*x*) has negative slope.
Hence, for *x* < -1, we have that *f*′(*x*)
will be negative. For -1 < *x* <
2, we have that the slope of *f*(*x*) is positive, i.e. *f*′(*x*)
is positive. For 2 < *x* < 4, we
have that the slope of *f*(*x*) is negative, i.e. *f*′(*x*)
is negative. And for *x* > 4, we see
that *f*(*x*) has positive slope, i.e. *f*′(*x*)
is positive.

Now, armed with that, we can draw a rough sketch of what *f*′(*x*)
should look like. We can place this information on the graph of *f*′(*x*).

**Figure 6**: Rough sketch of *f*′(*x*)

At this point, you may be wondering why our graph looks the
way it does. If we were just trying to make a smooth graph, then we probably
would have drawn something similar, but the local maximum would be to the
right, probably closer to halfway between -1 and 2. (The general shape,
however, should not come as a surprise. We used the information about *f*′(*x*)
and the fact that the graph is smooth to get the general shape.)

To understand why we put the local max/min on *f*′(*x*)
where we did, we need to look at the second derivative of *f*(*x*).

Why do we need to talk about the second derivative of *f*(*x*)?
Well, above, we have the sketch of *f*′(*x*).
On it, we drew places where the slope of *that*
graph was equal to zero. That is, we drew in places where the derivative of *f*′(*x*)
is equal to zero. But what is the derivative of *f*′(*x*)? That is precisely the second
derivative of *f*(*x*).

So, that means that we should place our local max/min on *f*′(*x*)
wherever *f*″(*x*)
equals zero. The question remains: where does *f*″(*x*) equal zero?

Recall, we say that if *f*″(*x*)
= 0, then *f*(*x*) has an inflection point, i.e. *f*(*x*) changes concavity.
At this point, let us return out attention to the graph of *f*(*x*).

**Figure 7**: The graph of *f*(*x*),
again

Notice that up until around *x* = 1, the graph of *f*(*x*) is concave up. (That is, the graph of
*f*(*x*)
looks like an *upwards* pointing
parabola.) Between *x* = 1/2 and *x* = 3, *f*(*x*) is concave down.
(The graph of *f*(*x*) looks like a *downwards*
pointing parabola.) And beyond *x* = 3,
*f*(*x*)
is concave up again.

The points *x* = 1/2
and *x* = 3 correspond to places where *f*(*x*)
changes concavity. By definition, those are inflection points.

So, that means that the graph of *f*′(*x*) should have a local max/min at *x* = 1/2 and *x* = 3. Whether it is a max or a min depends on the general shape of
the graph.

In the above example, we knew that *f*′(*x*) was positive between -1 and 2, which
meant that *x* = 1/2 would be a local
maximum, since the graph rose up from *x*-axis
(at *x* = -1) to some high-point and
then back down to it (at *x* = 2).

Similarly, we have *x*
= 3 should be a local minimum since the graph of *f*′(*x*) goes below that *x*-axis at *x* = 2 and
reaches it again at *x* = 4. Thus,
there has to be some low-point in between.

At this point, take a moment to look back through the
previous example. In order to sketch the graph of *f*′(*x*), we had to use information from *f*(*x*),
including local max/min, slopes, and inflection points. We talked about the
functions *f*(*x*), *f*′(*x*),
and *f*″(*x*).
It is a lot to keep straight.

Now, we turn our attention to working backwards. That is,
given the graph of *f*′(*x*),
we can sketch the graph of *f*(*x*).

**Example 3**: Given
the graph of *f*′(*x*),
sketch a possible graph of *f*(*x*).

**Figure 8**: The graph of *f*′(*x*)

**Solution**: Recall,
wherever *f*′(*x*)
has an *x*-intercept, we know that the
graph of *f*(*x*) has zero-slope. That means that *f*(*x*) has zero-slope at *x* = 1. Let’s determine whether that
point corresponds to a local maximum or a local minimum.

Looking to the left of *x*
= 1, we see that *f*′(*x*)
is negative, which means that *f*(*x*) is decreasing. To the right of *x* = 1, we see that *f*′(*x*) is positive, which means that *f*(*x*)
is increasing.

Using the relationship that the derivative of a polynomial
is one degree less than the function, we see that if *f*′(*x*) has degree *n*, then *f*(*x*) should have degree *n* + 1.

In this case, *f*′(*x*)
has degree 1, which means that *f*(*x*) should have degree 2, which is a
parabola. What do we know about parabolas? They have either a local maximum or
a local minimum.

Here, we see that *f*(*x*) decreases to *x* = 1 and then increases. If we draw that shape, we have something
like: .
From that general shape, we can see that we have an upwards pointing parabola.

Alternatively, we could recognize that the slope of the line is positive, thus the coefficient out front of the parabola should be positive as well, i.e. it is an upwards pointing parabola.

Regardless, armed with this knowledge, we can now proceed to
sketch the graph of *f*(*x*). It is an upwards pointing parabola
that has a local minimum at *x* = 1.
Can we exactly what the *y*-values are
going to be of *f*(*x*)? Nope. So, we just make a sketching using what we have. One
possible graph is given by:

**Figure 9**: Sketch of the graph of *f*(*x*)

Above, we said that where the graph of *f*′(*x*) had an *x*-intercept (i.e. *f*′(*x*)
= 0), *f*(*x*) had a local maximum or local minimum. That is not *always* the case, though.

Notice that we said that knew the *x*-intercept corresponded to a local minimum since *f*′(*x*)
was negative to the left (hence *f*(*x*) was decreasing) and positive to the
right (hence *f*(*x*) was increasing). The following example shows that just because
we have an *x*-intercept does not mean
that the function has as local maximum or a local minimum.

**Example 4**:
Consider the graph of *f*′(*x*)
= 3*x*^{2} (below). Sketch a
possible graph of *f*(*x*).

**Figure 10**: The graph of *f*′(*x*)
= 3*x*^{2}

**Solution**:
Thinking about derivatives in a slightly different way, we have that the
function could equal *f*(*x*) = *x*^{3}.
Why is this? Well, we know that the derivative of *x*^{3} is 3*x*^{2},
by the power rule we stated above. (Recall, .)

Let us consider the graph of *f*(*x*) = *x*^{3}.

**Figure 11**: The graph of *f*(*x*)
= *x*^{3}

We have that derivative, *f*′(*x*),
is zero at *x* = 0, but why doesn’t the
function, *f*(*x*), have a local maximum or local minimum at *x* = 0? The answer lies in what happens around the *x*-intercept on *f*′(*x*).

Notice that to the left of *x* = 0, *f*′(*x*)
> 0, which means that *f*(*x*) is increasing. Similarly, we have
that to right of *x* = 0, *f*′(*x*)
> 0, so *f*(*x*) is again increasing.

If we look at the graph of *f*(*x*), we see that indeed,
at *x* = 0, there is a horizontal
tangent line, but it does not correspond to local maximum or local minimum.

Thus, it is important to pay attention to what the
derivative is doing to the left and right of any *x*-intercepts.

Let’s put all of our concepts together and find the graph of
both *f*(*x*) and *f*″(*x*)
given the graph of *f*′(*x*).

**Example 5**: Given
the following graph of *f*′(*x*),
sketch possible graphs for *f*(*x*) and *f*″(*x*).

**Figure 12**: The graph of *f*′(*x*)

**Solution**: We
shall break this problem into two parts. We start by treating this as a
derivative to some function that we want to find. Then, we shall treat this as a
new function, call it *g*(*x*), and we want to sketch the graph of
its derivative.

To sketch the graph of *f*(*x*), we begin by looking for the places
where *f*′(*x*)
= 0. Notice that the function has *x*-intercepts
at *x* = -1, *x* = 1, *x* = 2 and *x* = 4. Those points correspond to either
local maximums or local minimums (or neither) on the graph of *f*(*x*).

For *x* < -1, we
see that *f*′(*x*)
< 0, so *f*(*x*) is decreasing. For -1 < *x*
< 1, *f*′(*x*)
> 0, so *f*(*x*) is increasing. For 1 < *x*
< 2, *f*′(*x*)
< 0, so *f*(*x*) is decreasing again. For 2 < *x* < 4, *f*′(*x*)
> 0, so *f*(*x*) is increasing. And for *x*
> 4, *f*′(*x*)
< 0, so *f*(*x*) is decreasing.

Putting that together in a very crude sketch, we see the function looks like: .

Thus, we have that *x*
= -1 and *x* = 2 are local minimums while
*x* = 1 and *x* = 4 are local maximums. (Notice that *f*′(*x*) is a negative fourth degree
polynomial, so we expect *f*(*x*) to be a negative fifth degree
polynomial. Thus, it should have four humps, which we see it does.)

So, at this point, we can sketch a graph of *f*(*x*).
We have the following:

**Figure 13**: Possible sketch of *f*(*x*)

Now, to find the graph of *f*″(*x*), we shall think of *f*′(*x*)
as a function *g*(*x*) and we are asked to sketch a graph of *g*′(*x*). Proceeding in that manner, we see
that *g*(*x*) (= *f*′(*x*))
has local maximums/minimums around *x*
= -0.4, *x* = 1.5 and *x* = 3.2.

For *x* < -0.4,
the slope of *g*(*x*) is positive, hence *g*′(*x*)
> 0. For -0.4 < *x* < 1.5, the
slope of *g*(*x*) is negative, hence *g*′(*x*)
< 0. For 1.5 < *x*< 3.2, the
slope of *g*(*x*) is positive, hence *g*′(*x*)
> 0. And for *x* > 3.2, we have
that the slope of *g*(*x*) is negative, hence *g*′(*x*)
< 0.

Since *g*(*x*) was a negative fourth degree
polynomial, we expect *g*′(*x*)
to be a negative third degree polynomial (a cubic). Using the information we
have just gathered, we can now sketch a graph of *g*′(*x*).

Again, to help us see the relationship between the graph and its derivative, let us stack the two graphs on top of each other.

**Figure 14**: Graph of *g*(*x*)
and *g*′(*x*)

(Graph of *f*′(*x*)
and *f*″(*x*))

Again, we see that the *x*-intercepts
line up, as does the general shape of the graph (it is a negative cubic). So,
this is a possible graph for *f*″(*x*).
Again, we cannot say anything about the *y*-values
of the graph without knowing more about the function.

In all of the above graphs, I knew what the formulas were, and hence I was able to make the graphs agree perfectly. But, as I said earlier, all we are looking for when we sketch a curve is the general shape, not the specific function.