Calculus Of One Real Variable
– By Pheng Kim Ving

5.4 
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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.
Fig. 1.1
The graph of f is concave
up on (a, b).

Fig. 1.2
The graph of g is concave
down on (a, b).

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 
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 tangentline case, see
Problem & Solution 4. The proof of the
chordline 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.
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2. Inflection Points 
The graph of y
= f(x) = x^{3} is shown in Fig. 2.1. We have f '(x) = 3x^{2} 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 firstderivative test,
as presented in
The graph of h is shown in Fig. 2.3. We see that h changes concavity at x = 0.
x = 0 is an inflection point of y = f(x) = x^{3}. 
Fig. 2.2 x = 0 is an inflection point of y = g(x) = x^{1/3}. 
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 x_{1} in the
domain of a function f is called
an inflection point of f if f changes
concavity at x_{1} and has a
tangent line 
Remarks 2.1
Suppose x_{1} is an inflection point of f.
i. f
either is differentiable at x_{1} or has a
vertical tangent line there; this is clear from the definition.
ii. The tangent line to the
graph of f at x_{1} crosses the
graph of f at that point. If the tangent is
vertical, this property is
obvious.
If the tangent is nonvertical, this property means that the tangent is above
the graph on one side of x_{1} 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 x_{1} is an
inflection point of f. As seen in
Part 2 above, x_{1} = 0
is an inflection point of f(x) = x^{3} and also of
g(x) = x^{1/3}; we see that f ''(x)
= 6x exists at x_{1}
= 0 while g''(x) = –2/(9x^{5/3}) doesn't
exist at x_{1} = 0. We
observe that
f ''(x_{1}) = f ''(0)
= 0.
Theorem 3.1
i. Let f be
twice differentiable on (a, b). 
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 x_{1} is an
inflection point of f and f ''(x_{1}) exists. Since
f changes concavity at x_{1}, f ' changes from increasing to
decreasing or vice versa as x passes thru x_{1}. As f ' is differentiable at x_{1}, it's
continuous there. Thus, f '(x_{1}) is a local
maximum or a local minimum of f '. Consequently, by Section
5.1 Theorem 2.1, f ''(x_{1}) = 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) = x^{3} and x_{1} = 0, as shown in Fig. 2.1, we
have f ''(x_{1}) =
0 and x_{1} is an inflection point of f; however, for g(x) = x^{4} and x_{1} = 0, as shown in Fig. 4.1,
Fig. 4.1 
we have g'(x) = 4x^{3} and g''(x)
= 12x^{2}, so g''(x_{1}) = 0, but x_{1} isn't an inflection point of g,
because g doesn't change
concavity at x_{1} ( g is concave
up on both sides of x_{1} 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 
Fig. 4.3 
d.
Fig. 4.4 
The graph of f
is sketched in Fig. 4.5.

Fig. 4.5 
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 
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5. Distinction Between Inflection Points, Critical Points, And
Local 
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?
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.
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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.
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4. Prove, using the meanvalue
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
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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 concaveup function on (a,
b)
and x_{1} where a < x_{1} < b be
arbitrary. The equation of the tangent line to the
graph of f
at x_{1} is y = t(x) = f(x_{1}) + f '(x_{1})(x – x_{1}). Let g(x) = f(x) – t(x) = f(x) – ( f(x_{1}) + f
'(x_{1})(x – x_{1})) on (a, b).
Note that g(x) = f(x) – f '(x_{1})x + C, where C = x_{1} f '(x_{1}) – f(x_{1}) is a
constant, and that g(x_{1}) = 0.
Now, suppose a < x < x_{1}. If g(x) = g(x_{1}), then, since g is
continuous on [x,
x_{1}] and
differentiable on (x,
x_{1}), Rolle's
theorem would imply that there exists c where x < c < x_{1} such that g'(c) = 0, which
contradicts the fact that it must be
that g'(c) < 0. So
we must have g(x) > g(x_{1}). Similarly,
we have g(x) > g(x_{1}) if x_{1} < x < b.
Therefore, g(x) > 0 on
(a, x_{1}) U (x_{1}, b) and g(x_{1}) = 0, or f(x) > t(x) on (a, x_{1}) U (x_{1}, b) and f(x_{1}) = t(x_{1}), which mean
that the tangent line to the graph of f at x_{1} lies below the graph. Since x_{1} in (a, b) is arbitrary, the proof is
complete.
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