A cubic curve has one of the six characteristic shapes shown in Fig. 9-2. There’s always one, but only one, inflection point. The overall trend of the curve is either upward (negative-to-positive) or downward (positive-to-negative), although the curve might reverse its direction in the vicinity of the inflection point.
Example
Let’s examine the cubic function
y= 5x 3+ 3x 2+ 5x+ 7
We can plug in a few x-values, calculate the y-values, and tabulate the resulting numbers. Table 9-1 is a list of coordinate values that can give us a good idea of what the curve looks like when we plot the corresponding points in the xy-plane. Figure 9-5 shows the graph. On the x axis, each division represents 1/2 unit. On the y axis, each division represents 10 units. This axis distor- tion allows us to graph the function within a region of reasonable dimensions.
Now let’s differentiate our cubic function to see how the slope varies with the value of x. The first derivative is
dy /dx= 15x 2+ 6x+ 5
The slope follows a quadratic function. This fact makes the cubic curve inherently more com- plicated than any quadratic curve, where the slope always follows a linear function.
The second derivative of this cubic function tells us how fast, and in what direction, the slope changes as we increase the value of x. When we differentiate the first derivative, we get
d 2y /dx 2= 30x+ 6
Table 9-1. Selected values for graphing the function
yⴝ 5x3ⴙ 3x2ⴙ 5xⴙ 7. x 5x3ⴙ 3x2ⴙ 5xⴙ 7 −3 −116 −2 −31 −1 0 0 7 1 20 2 69
We can find the x-value of the inflection point, where the concavity reverses, by figuring out where the second derivative is 0. We create a linear equation by setting d 2y /dx 2= 0, like
this:
30x+ 6 = 0
The solution to this equation is x= −1/5. When we plug −1/5 into the original function for x and then calculate y, we get
y = 5x 3+ 3x 2+ 5x+ 7 = 5 · (−1/5)3+ 3 · (−1/5)2+ 5 · (−1/5)+ 7 = 152/25
The coordinates of the inflection point, written as an ordered pair, are therefore (x,y ) = (−1/5,152/25)
This curve is concave downward to the left of the inflection point, and concave upward to the right of the inflection point. This is a geometric indication of the facts that
d 2y /dx 2 < 0 when x < −1/5
Graph of a Cubic Function 145
x y x= –1/5 Inflection point d2y/dx2= 0 Concave downward d2y/dx2< 0 Concave upward d2y/dx2 > 0 1 2 3 –1 –2 –3 30 60 –30 –60 –90 –120 Figure 9-5 Graph of y= 5x 3+ 3x 2+ 5x+ 7. On thex axis, each division represents 1/2 unit. On the y axis, each division represents 10 units.
and
d 2y /dx 2> 0 when x> −1/5
Are you confused?
It’s reasonable to wonder whether there are any local minima or local maxima in this curve. There is no
absolute minimum and no absolute maximum, because cubic curves never have them! When we examine Fig. 9-5, we can see that the overall trend of this curve is upward (negative-to-positive). But is this true at every point along the entire curve? Or does the curve reverse its direction near the inflection point? We can’t be sure by merely looking at Fig. 9-5.
To figure this problem out by brute force, we could plot hundreds of points with the help of a com- puter and a good graphing program. We could also evaluate the first derivative at hundreds of x-values, again with the help of a computer. But there’s a more elegant way to resolve this mystery. (Mathematicians use the term “elegant” to describe a proof, derivation, or process that uses finesse, rather than force, to solve a problem.)
If the slope at the inflection point is positive, then we can be sure that the curve in Fig. 9-5 trends upward all the time (like the one in Fig. 9-2C), never reverses direction, and therefore has no local mini- mum or maximum. If the slope at the inflection point is negative, then we can be sure that the curve in Fig. 9-5 reverses direction momentarily, trending downward (positive-to-negative) in a region near the inflection point. In that case, the curve has a local maximum slightly to the left of the inflection point, and a local minimum slightly to the right of it (like the one in Fig. 9-2E). If the slope at the inflection point is 0, then we know that the curve in Fig. 9-5 levels off at that point (like the one in Fig. 9-2A), but doesn’t reverse direction and therefore has no local minimum or maximum. Let’s see which of these three situations is the case here.
Here’s a challenge!
Look again at Fig. 9-5 and the function that this graph represents. What is the slope of the curve at the inflection point? What does this tell us about the general nature of the curve? Which of the generic profiles in Fig. 9-2 does this curve most closely resemble?
Solution
To find the slope of the curve at the inflection point, we plug x = −1/5 into the formula for the first derivative, getting
dy /dx = 15x 2+ 6x+ 5 = 15 · (−1/5)2+ 6 · (−1/5)+ 5 = 22/5
That’s a positive number. We have dy /dx > 0 at the inflection point. The overall trend of the curve is positive, and so is the slope at the inflection point. Based on all this knowledge, we can deduce that the curve does not reverse its direction at the inflection point, and doesn’t level off there, either. It therefore has no local minimum or maximum. Of the six generic profiles shown in Fig. 9-2, this curve most nearly resembles the one shown in drawing C.