#### Calculus Of One Real Variable – By Pheng Kim Ving Chapter 5: Applications Of The Derivative Part 1 – Section 5.4: Concavity And Inflection

5.4
Concavity And Inflection

 1. Concavity

Let f be differentiable on (a, b). See Fig. 1.1. The graph of f is bending upward. We say that the graph of f is concave up.
We note that as x increases from the near right of a to the near left of b, the tangent line to the graph of f at x turns
counterclockwise, which means that its slope increases. The slope of the tangent line at x is f '(x). So, as x increases in
(a, b), f '(x) increases.

Let g be differentiable on (a, b). See Fig. 1.2. The graph of g is bending downward. We say that the graph of g is
concave down. We note that as x increases from the near right of a to the near left of b, the tangent line to the graph of
g at x turns clockwise, which means that its slope decreases. So, as x increases in (a, b), g'(x) decreases. # The graph of f is concave up on (a, b). # The graph of g is concave down on (a, b).

##### Definition 1.1
 Suppose the function f is differentiable on (a, b). We say that (the graph of ) f is concave up on (a, b) if f '(x) is increasing there. Similarly, we say that (the graph of ) f is concave down on (a, b) if f '(x) is decreasing there.

Remarks 1.1

a. Suppose f is concave up, so that f '(x) is increasing. See Fig. 1.1. Now f '(x) is also the rate of change of f. Thus if f is
decreasing or its rate of change is negative then the magnitude of the rate of change is decreasing (a negative
increasing quantity has a decreasing magnitude) and consequently f is decreasing slower and slower, and if f is
increasing or its rate of change is positive then the magnitude of the rate of change is increasing and consequently f is
increasing faster and faster.

b. Suppose f is concave down, so that f '(x) is decreasing. See Fig. 1.2 and think of g as f. Now f '(x) is also the rate of
change of f. Thus if f is increasing or its rate of change is positive then the magnitude of the rate of change is
decreasing and consequently f is increasing slower and slower, and if f is decreasing or its rate of change is negative
then the magnitude of the rate of change is increasing (a negative decreasing quantity has an increasing magnitude)
and consequently f is decreasing faster and faster.

c. Concavity is defined only for differentiable functions.

d. It can be shown that if a function is concave up on an open interval, then any tangent line to its graph on that interval
lies below the graph, and any chord line on that interval lies above the graph. For a proof of the tangent-line case, see
Problem & Solution 4. The proof of the chord-line case is similar.

e. Analogously, if a function is concave down on an open interval, then any tangent line to its graph on that interval lies
above the graph, and any chord line on that interval lies below the graph.

 2. Inflection Points

The graph of y = f(x) = x3 is shown in Fig. 2.1. We have f '(x) = 3x2 and thus f ''(x) = 6x. Clearly f ''(x) < 0 for all x < 0
and f ''(x) > 0 for all x > 0. Remember, f '' is the (first) derivative of f '. So, by the first-derivative test, as presented in The graph of h is shown in Fig. 2.3. We see that h changes concavity at x = 0. Fig. 2.1   x = 0 is an inflection point of y = f(x) = x3. Fig. 2.2   x = 0 is an inflection point of y = g(x) = x1/3. Fig. 2.3   x = 0 isn't an inflection point of y = h(x).

Now we want to call the point where the function changes concavity an inflection point, like the point x = 0 for f and g,
but unlike the point x = 0 for h because h has a sharp point there. We remark that:

i.  All three functions change concavity at x = 0. Of course we'll include the change of concavity as a requirement in the definition of an inflection point. If we include
differentiability, we'll eliminate h but we'll also eliminate g. We see that the inclusion of the existence of a tangent line is
appropriate. Therefore, we'll define an inflection point  as a point where the function changes concavity and has a
tangent line.

Definition 2.1

 A point x1 in the domain of a function f is called an inflection point of f if f changes concavity at x1 and has a tangent line there.

Remarks 2.1

Suppose x1 is an inflection point of f.

i.  f either is differentiable at x1 or has a vertical tangent line there; this is clear from the definition.

ii.  The tangent line to the graph of f at x1 crosses the graph of f at that point. If the tangent is vertical, this property is
obvious. If the tangent is non-vertical, this property means that the tangent is above the graph on one side of x1 and
below it on the other side. The proof that the tangent is above or below the graph is analogous to that in
Problem & Solution 4.

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 3. Concavity, Inflection Points, And The Second Derivative

Refer to Fig. 3. For x > 0, f ''(x) = 6x > 0 and f is concave up. For x < 0, f ''(x) < 0 and f is concave down.

Now suppose x1 is an inflection point of f. As seen in Part 2 above, x1 = 0 is an inflection point of f(x) = x3 and also of
g(x) = x1/3; we see that f ''(x) = 6x exists at x1 = 0 while g''(x) = –2/(9x5/3) doesn't exist at x1 = 0. We observe that
f ''(x1) = f ''(0) = 0.

Theorem 3.1

 i.   Let f be twice differentiable on (a, b).      a.  If f ''(x) > 0 on (a, b), then f is concave up on (a, b)      b.  If f ''(x) < 0 on (a, b), then f is concave down on (a, b). ii.  If x1 is an inflection point of f and f ''(x1) exists, then f ''(x1) = 0.

Proof
i.  If f ''(x) > 0 on (a, b), then f ' is increasing there, so f is concave up on there. If f ''(x) < 0 on (a, b), then f ' is
decreasing there, so f is concave down on there.

ii.  Suppose x1 is an inflection point of f and f ''(x1) exists. Since f changes concavity at x1, f ' changes from increasing to
decreasing or vice versa as x passes thru x1. As f ' is differentiable at x1, it's continuous there. Thus, f '(x1) is a local
maximum or a local minimum of f '. Consequently, by Section 5.1 Theorem 2.1, f ''(x1) = 0.
EOP

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 4. Determining Concavity And Finding Inflection Points

Part i of the above theorem gives us a straightforward way to determine the intervals of concavity of a function f.

Part ii of the theorem provides us this insight: inflection points of a function f can occur only at points x where:

either f ''(x) = 0
or       f ''(x) doesn't exist.

( f ''(x) either exists or doesn't exist; if it exists then it must be 0.) That's similar to the situation where a local extremum can
occur only at points x where f '(x) = 0 or f '(x) doesn't exist (critical point) or x is an endpoint; see Section 5.2 Remark 4.2.
There, we saw that not every such point yields a local extremum. Here, the situation is similar: not every point x where f ''(x)
= 0 or f ''(x) doesn't exist is an inflection point. For example, for f(x) = x3 and x1 = 0, as shown in Fig. 2.1, we have f ''(x1) =
0 and x1 is an inflection point of f; however, for g(x) = x4 and x1 = 0, as shown in Fig. 4.1, Fig. 4.1 x = 0 isn't an inflection point of y = g(x) = x4.

we have g'(x) = 4x3 and g''(x) = 12x2, so g''(x1) = 0, but x1 isn't an inflection point of g, because g doesn't change
concavity at x1 ( g is concave up on both sides of x1 and as a matter of fact on every interval).

So, to find inflection points of a function f, first we identify all points x where f ''(x) = 0 or f ''(x) doesn't exist, then we
check to see at which of those points f changes concavity and has a tangent line. The existence of the tangent line is
implied and thus doesn't have to be explicitly mentioned if f '(x) exists. The determining of the intervals of concavity and
the finding of the inflection points of a function is illustrated in the following example.

Example 4.1 Solution  Fig. 4.2 Chart showing local extrema of f. LMAX = local maximum, LMIN = local minimum.  Fig. 4.3 Chart showing concavity and inflection points of f. IP = inflection point. d. Fig. 4.4 Chart for f.

The graph of f is sketched in Fig. 4.5. Fig. 4.5 Graph of y = f(x) = x4 – 4x2.

EOS

Chart Showing Concavity And Inflection Points Chart Summarizing The Behavior Of A Function

The chart for f in Fig. 4.4 summarizes the behavior of f: intervals of increase and decrease, local extrema, intervals of
concavity, and inflection points. All the critical points and all the points x where f ''(x) = 0 are placed in the row for x in
increasing order. The meanings of the arcs in the row for f(x) in the charts in Figs. 4.3 and 4.4 are shown in the table in
Fig. 4.6. Fig. 4.6 Table showing meanings of arcs.

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 5. Distinction Between Inflection Points, Critical Points, And Local     Extrema

The following distinction should be clear. ## Problems & Solutions

1.  Let: Solution      d.   2.  Discuss the concavity of the linear function f(x) = ax + b. Does f have inflection points?

##### Solution

We have f '(x) = a, which is constant and so is neither increasing nor decreasing on any interval. Thus f is neither
concave up nor concave down on any interval, ie, it has no concavity, and consequently no inflection points. 3.  Sketch the graph of a twice differentiable function f where f ''(x) < 0 for x < 2, f(2) = – 1, f '(2) = 1, f ''(2) = 0,
f ''(x) > 0 for x > 2, and f(4) = 3.

Solution The equation of the tangent line to the graph at x = 2 is y = (–1) + (1)(x – 2), or y = x – 3. The graph is below the
tangent line to the left of x = 2, and above it to the right of x = 2.  4.  Prove, using the mean-value theorem, that for a twice differentiable function f, one cannot have f(2) = –1, f '(2) = 1,
f ''(x) > 0 for all x > 2, and f(4) = 1.

Note

See the graph in the previous Problem & Solution.

Solution  5.  Prove that if a function is concave up on an open interval, then any tangent line to its graph on that interval lies below
the graph.

Note

The graphs below will help in the proof. Solution

Let f be a concave-up function on (a, b) and x1 where a < x1 < b be arbitrary. The equation of the tangent line to the
graph of f at x1 is y = t(x) = f(x1) + f '(x1)(xx1). Let g(x) = f(x) – t(x) = f(x) – ( f(x1) + f '(x1)(xx1)) on (a, b).
Note that g(x) = f(x) – f '(x1)x + C, where C = x1 f '(x1) – f(x1) is a constant, and that g(x1) = 0. Now, suppose a < x < x1. If g(x) = g(x1), then, since g is continuous on [x, x1] and differentiable on (x, x1), Rolle's
theorem would imply that there exists c where x < c < x1 such that g'(c) = 0, which contradicts the fact that it must be
that g'(c) < 0. So we must have g(x) > g(x1). Similarly, we have g(x) > g(x1) if x1 < x < b.

Therefore, g(x) > 0 on (a, x1) U (x1, b) and g(x1) = 0, or f(x) > t(x) on (a, x1) U (x1, b) and f(x1) = t(x1), which mean
that the tangent line to the graph of f at x1 lies below the graph. Since x1 in (a, b) is arbitrary, the proof is complete.