Sketching Graphs

Now that we've seen what kind of information is represented in the first and second derivatives, we can put everything together to obtain a comprehensive understanding of functions. In this section, we'll demonstrate this by sketching the graphs of a few functions.

Example 1

Let's consider the function $ f(x) = x^5 - 9x^3 $ . This is a simple polynomial and we should be able to say quite a lot about its graph.
Let's begin with some simple observations. First, the graph will have some symmetry: since only odd powers occur in the polynomial, we see that $ f(-x) = - f(x) $ . We call this type of function odd : it implies that the graph is symmetric when rotated by 180 degrees about the origin. In essence, this will reduce the amount of work we really have to do by half since if we know what the graph looks like for positive , we also understand its behaviour for negative as well.
Secondly, we can find some interesting points on the graph---namely, where it crosses the $x$ axis. These are points for which
$f(x) = x^5 -9x^3 = x^3(x^2 - 9) = x^3(x-3)(x+3) = 0$
Notice that the values $x = 0, \pm 3$ are zeroes of $ f $ and hence the graph crosses the axis at these points.
We can find out where the function is increasing and decreasing by studying the function's first and second derivatives. If we compute, we find that
$\begin{eqnarray*} f^\prime(x) & = & 5x^4 - 27x^2 \\ f^{\prime\prime}(x) & = & 20x^3 - 54x. \end{eqnarray*}$
To find where the function increases and decreases, we can factor the first derivative as $f^\prime(x) = x^2(5x^2 - 27)$ . The function has critical points for those values of where $f^\prime(x) = 0$ . These occur at $x = 0, \pm \sqrt{\frac{27}{5}}$ . Now if we consider how the derivative behaves in between these critical points, we see that

$f^\prime(x)$ + 0 - 0 - 0 +

$-\sqrt{\frac{27}{5}}$
0
$\sqrt{\frac{27}{5}}$

This shows that the function is increasing when $x < -\sqrt{\frac{27}{5}}$ and $x > \sqrt{\frac{27}{5}}$ and decreasing when $-\sqrt{\frac{27}{5}} < x < \sqrt{\frac{27}{5}}$ .
Now the second derivative $f^{\prime\prime}(x) = x(20x^2 - 54) = 0$ when either $x = 0, \pm \sqrt{\frac{27}{10}}$ . In between the zeroes of the second derivative, we have

$f^{\prime\prime}(x)$ - 0 + 0 - 0 +

$-\sqrt{\frac{27}{10}}$
0
$\sqrt{\frac{27}{10}}$

This says that the graph is concave up when either $-\sqrt{\frac{27}{10}} < x < 0$ and $\sqrt{\frac{27}{10}} < x$ and concave down in the other areas. Notice that the points $x = 0, \pm\sqrt{\frac{27}{10}}$ are all inflection points since the second derivative changes sign at each of them.
We can now put all of this together to sketch the graph. We will denote the special points (i.e. points where either the first or second derivative is zero) by putting a ball on the graph at those points.
Notice that the graph looks like $ -9x^3 $ near the origin and $ x^5 $ far away from the origin. This agrees with our discussion of powers of x.

Example 2

For our next example, we'll consider the function
$f(x) = \frac{1}{x^2 + 1}.$
Once again, this graph will have symmetry since $ f(-x) = f(x)
$ . This type of function is called even and it implies that the graph will look the same if reflected in the $ y
$ axis.
Notice that, for this function, there are no points which cross the axis since the numerator is never equal to 0.
Now let's compute the first and second derivatives: we see that
$\begin{eqnarray*} f^\prime(x) & = & \frac{-2x}{(x^2 + 1)^2} \\ f^{\prime\prime} & = & \frac{2(3x^2-1)}{(x^2 + 1)^3} \end{eqnarray*}$
Now, the first derivative provides us with a critical point when $ x = 0 $ . When $ x < 0 $ , $f^\prime(x) > 0$ which means that $ f $ is increasing. When $ x >
0 $ , then $f^\prime(x) < 0$ which means that $ f
$ is decreasing.
For the second derivative, we see that $f^{\prime\prime}(x) = 0$ when $x=\pm \frac{1}{\sqrt{3}}$ . The graph is concave up for $x < -\frac{1}{\sqrt{3}}$ and $x > \frac{1}{\sqrt{3}}$ and concave down between $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$ .
To finish off the graph, notice the following. As becomes very large, the denominator looks approximately like $ x^2
$ and hence the function looks like $\frac{1}{x^2}$ . This means that, for large values of , the function becomes very small. In terms of the graph, this means that it becomes very close to the axis.
We can now sketch the graph using this information.