Analysis of Functions I
Increase, Decrease, and Concavity
Definition. Let f(x) be defined on an interval, and let x1 and x2
denote points in that interval.
(a) f is increasing on the interval if f(x1) < f(x2) whenever x1< x2.
(b) f is decreasing on the interval if f(x1) > f(x2) whenever x1< x2.
(c) f is constant on the interval if f(x1) = f(x2) for all x1and x2.
decreasing
constant
increasing
increasing
Here are some other examples.
Increasing functions Decreasing functions
Notice that the shape of the functions can vary.
Theorem. Let f be a function that is continuous on a closed
interval [a, b], and differentiable on the open interval (a, b).
(a) If f ′(x) > 0 for every value of x in (a, b), then f is increasing
in [a, b].
(b) If f ′(x) < 0 for every value of x in (a, b), then f is decreasing
in [a, b].
(c) If f ′(x) = 0 for every value of x in (a, b), then f is constant
on [a, b].
Positive derivative = increasing
Negative derivative = decreasing
Zero derivative = constant
Example. Let
f ( x) = x2 − 2 x+1
Find the intervals on which f is increasing and the intervals on
which f is decreasing.
Solution. f ′(x) = 2x − 2 = 2(x −1)
It is clear that this derivative is positive when x > 1 and
negative when x < 1. Thus the function is decreasing in the
interval (−∞, 1) and is increasing in the interval (1, ∞).
The graph confirms this analysis.
Example. Let
f ( x) = x4 −8x2 +16
Find the intervals on which f is increasing and the intervals on
which f is decreasing.
Solution.
f ′(x) = 4x3 −16 x = 4( x3 − 4 x)
= 4 x( x2 − 4) = 4x(x − 2)( x + 2).
If x < −2, all three factors are negative, so the product is
negative.
If −2< x <0, then x + 2 is positive, while x and x −2 are
negative. Thus the product is positive.
If 0 < x < 2, then x −2 is negative, while x and x + 2 are
positive, so again the product is negative.
If x > 2, then all factors are positive, and so the product is
positive
This is the graph of the function, which confirms the
analysis.
Concavity
Concavity refers to the shape of a curve, rather than its
direction. A curve is concave up or down, if it will “hold
water, or “spill water”.
Concave up
increasing – decreasing
Concave Down
increasing – decreasing
Definition. If f is differentiable on an open interval I, then f is
said to be concave up on I if f′ is increasing on I, and f is said to
be concave down on I if f′ is decreasing on I.
We know that a function is increasing or decreasing according
to whether its derivative is positive or negative. This result can
be applied to the function f′ if that function has a derivative; in
other words if f has a second derivative on I. This leads to the
following useful theorem.
Theorem. Let f be twice differentiable on an open interval I.
(a) If f′′ (x) > 0 on I, then f is concave up on I.
(b) If f′′ (x) < 0 on I, then f is concave down on I.
+
+
+
+
+
+
Concave up – holds water
Second derivative > 0
−
−
−
−
−
−
Concave down – spills water
Second derivative < 0
Example. Find the open intervals on which the function
f ( x) = x3 + x2 − 2x−1
is concave up, and the open intervals on which it is concave
down.
Solution. The derivative of f is 3x2 + 2x − 2. Thus the second
derivative is 6x + 2 = 2(3x +1).
This means that f is concave down if
Similarly, f is concave up if x>−1.
3
2(3x +1) < 0 or x <−1.
3
f ( x) = x3 + x2 − 2x−1
f ( x) = x3 + x2 − 2x−1
Concave Down
in this half
Concave Up
in this half
Definition. If f is continuous on an open interval containing the
point x0, and if f changes the direction of its concavity at that
point, then we say that f has an inflection point at x0, and we call
the point (x0, f(x0)) on the graph of f an inflection point of f.
f ( x) = x3 + x2 − 2x−1
Inflection
point
Example. Find the inflection points of xex and cos( x) , and
confirm the results by looking at the graphs of these functions.
Solution.
 xe x ′′ =  e x + xe x ′ = 2ex + xex = ex 2 + x .

 


 

(
)
Since the exponential is never negative or 0, we see that xex
is concave down if x < −2 and concave up if x > −2. Thus −2 is
an inflection point.
cos( x)′′ = (−sin( x))′ =−cos( x).
Thus the cosine is concave up where it is negative and
concave down where it is positive. Thus the inflection points
come where cos (x) is 0.
The graphs of these two functions confirm these results.
Example. Use the graph of the function y = f(x), shown below,
dy and d 2 y
to determine the signs of
at the points A, B, and C.
dx
dx 2
B
f ( x)
A
C
2y
dy
d
<0
Solution. At A, >0 and
dx
dx 2
2y
dy
d
At B, <0 and
<0
dx
dx 2
2y
dy
d
At C, >0 and
>0
dx
dx2
Example. Use the graph of the derivative f′ (x), shown below,
2y
dy
d
to determine the signs of
at the points A, B, and C.
and
dx
dx 2
f ′(x)
B
A
C
2y
dy
d
>0
Solution. At A, >0 and
dx
dx 2
2y
dy
d
At B, >0 and
<0
dx
dx 2
2y
dy
d
At C, <0 and
>0
dx
dx2
Example. Use the graph of y = f(x), shown below to identify all
intervals where the function is increasing, decreasing, concave
up, concave down. Find all values of x at which f has an
inflection point.
3
1
2
4
5
6
7
3
1
2
4
5
6
7
Solution. The function f is increasing on the closed intervals
[1, 2], and [4, 7]. It is decreasing on the closed interval [2, 4].
The function f is concave up on the open intervals (3, 5) and
(6, 7). It is concave down on the open intervals (1, 3) and
(5, 6).
Example. For the function f ( x) = 5 +12x − x3 find all intervals
on which f is increasing, all intervals on which f is decreasing,
all open intervals on which f is concave up, all intervals on
which f is concave down, and all points of inflection of f.
Solution. f ′(x )=12 −3 x2 = 3(4 − x2) and f ′′(x) =−6x.
From this it is clear that f′ (x) > 0 if x2 < 4 and f′ (x) < 0
if x2 > 4. We also have f′′ (x) > 0 if x < 0 and f′′ (x) < 0 if x > 0.
Thus the function is increasing in the interval (−2 , 2), and
decreasing in intervals (−∞ , −2) and (2, ∞).
It is concave up in the interval (−∞ , 0) and concave down in
the interval (0, −∞ ).
Here is the graph of the function.
Example. For the function f ( x) = 3x4 − 4 x3 + 2 find all intervals
on which f is increasing, all intervals on which f is decreasing,
all open intervals on which f is concave up, all intervals on
which f is concave down, and all points of inflection of f.
Solution.
f ′(x )=12 x3 −12x 2 =12 x 2(x −1) and f ′′(x) = 36x2 − 24x =12 x(3x − 2).
From this it is clear that f′ (x) > 0 if x > 1 and f′ (x) < 0
if x<1. We also have f′′ (x) > 0 if x < 0 and if x > 2/3, while
f′′(x) < 0 if 0 < x < 2/3.
Thus the function is increasing in the intervals (1, ∞) and the
function is decreasing in interval (−∞ , 1) .
It is concave up in the intervals (−∞ , 0) and (2/3, ∞), and it is
concave down in the interval (0, 2/3 ).
2
38
27
1
0
2 1
3
At x equal 0, 2/3, and 1, either the direction or the curvature
changes. We plot these points on the graph, and divide the plane
into regions.
2
38
27
1
0
f ′ (x) < 0
f ′′ (x) > 0
f ′ (x) < 0
f ′′ (x) < 0
2 1
3
f ′ (x) < 0
f ′′ (x) > 0
f ′ (x) > 0
f ′′ (x) > 0
Recall the characterization of the pieces of a smooth curve
1
2
Concave up
increasing – decreasing
1. f ′ > 0, f ′′ > 0
3. f ′ > 0, f ′′ < 0
3
4
Concave Down
increasing – decreasing
2. f ′ < 0, f ′′ > 0
4. f ′ < 0, f ′′ < 0
2
38
27
1
0
f ′ (x) < 0
f ′′ (x) < 0
2 1
3
f ′ (x) < 0
f ′′ (x) > 0
f ′ (x) > 0
f ′′ (x) > 0
2
38
27
1
0
2 1
3
f ′ (x) < 0
f ′′ (x) > 0
f ′ (x) > 0
f ′′ (x) > 0
2
38
27
1
0
2 1
3
f ′ (x) > 0
f ′′ (x) > 0
2
38
27
1
0
2 1
3
Here is the actual graph.