Today, we will be discussing concepts related to concavity and the second
derivative.
As you know, the second derivative measures how quickly the first derivative
is changing. If the second derivative is positive, the first derivative is increasing,
and if the second derivative is negative, the first derivative is decreasing.
Take a look at the picture in the geogebra file. The function graphed in
the file has positive second derivative everywhere. Notice that the slopes of the
red and green lines are negative, but the green line is flatter so it has a more
positive slope. The blue line has the highest slope of the three lines shown- it
has a positive slope because the function is increasing there.
So the point is that if the second derivative is positive and the first derivative
is negative, the graph gets flatter as you go to the right. If the second derivative
and first derivative are positive, the graph gets steeper as you go to the right.
Think of the bottom half of a circle, or an upwards-opening parabola as an
example of a function with positive second derivative.
If the second derivative is positive, we say that the function is concave up.
If the second derivative is negative, we say that the function is concave down.
You can identify the concavity by looking at where the tangent line lies
relative to the graph. If the second derivative is positive, the first derivative
is increasing, so the tangent line lies below the graph. If the second derivative
is negative, the first derivative is decreasing, so the tangent line lies above the
graph.
A point where the concavity changes from positive to negative is called a
point of inflection or inflection point. This is the point where the graph is locally
at its steepest or flattest (but not a local maximum or local minimum). You
can identify points of inflection by noticing that the second derivative changes
sign at a point of inflection. In order to identify points of inflection, we look for
places where the second derivative is equal to zero, and check the concavity on
each side of the point.
Example 0.1. Determine where f (x) = x4 − 3x3 + x2 − 5x is concave up, and
where the function is concave down.
To find the concavity of this function, we need to take the second derivative.
The first derivative is 4x3 −9x2 +2x−5. The second derivative is 12x2 −18x+2.
So we just need to determine where 12x2 − 18x + 2 is positive and where it’s
negative. We can do this just be factoring or using the quadratic formula:
√
18 ± 324 − 96
x=
24
this comes out to
x=
√
18 ± 228
.
24
Let’s divide everything by 2:
√
9 ± 57
x=
12
1
√
Note that 57 is much less than 9, so 0 is to the left of both inflection points.
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2 lies between the inflection points, and 2 lies to the right of both inflection
points.
We get f ′′ (0) = 2, which is positive, so the graph is concave up for x <
√
9− 57
′′ 1
24 . f ( 2 ) = −4, which is negative, so the function is concave down between
the inflection points. At 2, the second derivative is f ′′ (2) = 14, which is positive.
So both of these points are inflection points.
Please look at the geogebra file to see what these inflection points look like.
Finally, I would like to mention the second derivative test for classifying
critical points. Suppose that f ′ (c) = 0. Then c is a critical point of f . What
can we infer about c from looking at f ′′ (c)?
Suppose first that f ′′ (c) is positive. Notice that to the left of c, we expect
that f ′ (c) is negative, and to the right of c, we expect that f ′ (c) is positive.
Therefore, f (x) has a minimum at x = c.
Similarly, if f ′′ (c) is negative, f (x) has a maximum at x = c.
This test isn’t too important, but it can be useful if the first derivative is
difficult to compute.
If f ′′ (c) = 0, the test fails: f (x) may have a maximum, minimum, or inflection point at x = c.
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