Lecture Note
Increasing and Decreasing Functions, Min and Max, Concavity
-
University:
California State University, Northridge -
Course:
MATH 150A | Calculus I Academic year:
2004
-
Views:
289
Pages:
44
Author:
Janelle Woodard
Increasing and Decreasing
Functions, Min and Max,
Concavity
studying properties of the function
using derivatives Increasing and Decreasing
Functions
characterizing function’s
behaviour Definition: (I = [, ], (, ), [, ), (, ])
f (x) is increasing on I if for each
pair x1, x2 ∈ I
f(x2)
f(x2) > f(x1) if x 2 > x 1
f(x1)
x1
x2 > x1 ⇒ f (x2) > f (x1)
I
x2 Definition: (I = [, ], (, ), [, ), (, ])
f (x) is increasing on I if for each
pair x1, x2 ∈ I
f(x2)
f(x2) > f(x1) if x 2 > x 1
f(x1)
I
x1
x2
x2 > x1 ⇒ f (x2) > f (x1)
Definition:
f (x) is decreasing on I if for
each pair x1, x2 ∈ I
f(x1)
f(x2) < f(x1) if x2 > x1
f(x2)
x1
I
x2 > x1 ⇒ f (x2) < f (x1)
Increasing/decreasing = strict monotonicity
x2 Definition: (I = [, ], (, ), [, ), (, ])
f (x) is non-decreasing on I if for
each pair x1, x2 ∈ I
f(x2) $ f(x1) if x 2 > x1
f(x2)
f(x1)
I
x1
x2
x2 > x1 ⇒ f (x2) ≥ f (x1)
Definition:
f (x) is non-increasing on I if for
each pair x1, x2 ∈ I
f(x1)
f(x2)
f(x2) # f(x1) if x2 > x1
x1
I
x2
x2 > x1 ⇒ f (x2) ≤ f (x1)
Non-decreasing/increasing = non-strict monotonicit y=sin(x)
Example 1.
Function sin(x)
is strictly monotonic on each
interval
- π_
2
π_
2
3π
__
2
[−π/2 + kπ, π/2 + kπ], k = 0, ±1, ±2, ±3, . . . y=sin(x)
Example 1.
Function sin(x)
is strictly monotonic on each
interval
- π_
2
π_
2
3π
__
2
[−π/2 + kπ, π/2 + kπ], k = 0, ±1, ±2, ±3, . . .
It is increasing on
[−π/2 + kπ, π/2 + kπ], k = 0, ±2, ±4, ±6, . . . y=sin(x)
Example 1.
Function sin(x)
is strictly monotonic on each
interval
- π_
2
π_
2
3π
__
2
[−π/2 + kπ, π/2 + kπ], k = 0, ±1, ±2, ±3, . . .
It is increasing on
[−π/2 + kπ, π/2 + kπ], k = 0, ±2, ±4, ±6, . . .
It is decreasing on
[−π/2 + kπ, π/2 + kπ], k = ±1, ±3, ±5, . . . y=tan(x)
__
- 3π
2
- π_
2
π_
2
3π
__
2
Example 2. Function tan(x) is
increasing on each interval
[−π/2 + kπ, π/2 + kπ], k = 0, ±1, ±2, ±3, . . . y=tan(x)
__
- 3π
2
- π_
2
π_
2
3π
__
2
Example 2. Function tan(x) is
increasing on each interval
[−π/2 + kπ, π/2 + kπ], k = 0, ±1, ±2, ±3, . . .
Note, that you still can’t say tan(x) increases
everywhere! y=tan(x)
__
- 3π
2
- π_
2
π_
2
3π
__
2
Example 2. Function tan(x) is
increasing on each interval
[−π/2 + kπ, π/2 + kπ], k = 0, ±1, ±2, ±3, . . .
Note, that you still can’t say tan(x) increases
everywhere!
Indeed, for x1 = π/4 and x2 = 3π/4,
x2 > x1
but
tan(x2) = −1 < tan(x1) = 1 Derivative and monotonicity
What derivative can tell
about the function? Theorem A. If f (x) is increasing on I, and f 0(x)
exists, then f 0(x) ≥ 0 on I. Theorem A. If f (x) is increasing on I, and f 0(x)
exists, then f 0(x) ≥ 0 on I.
Proof. Since f (x) is increasing,
f (t) − f (x)
> 0,
t−x
∀x, t ∈ I
f(t)
f(t)-f(x)>0
t-x<0
f(x)
Therefore,
f (t) − f (x)
f (x) = lim
≥0
t→x
t−x
t-x>0
f(t)-f(x)<0
f(t)
t
x
I
t Theorem A. If f (x) is increasing on I, and f 0(x)
exists, then f 0(x) ≥ 0 on I.
Proof. Since f (x) is increasing,
f (t) − f (x)
> 0,
t−x
∀x, t ∈ I
f(t)
f(t)-f(x)>0
t-x<0
f(x)
Therefore,
f (t) − f (x)
f (x) = lim
≥0
t→x
t−x
t-x>0
f(t)-f(x)<0
f(t)
t
x
I
t
Note that “≥” can not be replaced with “>”!
(f (x) = x3 is increasing everywhere but f 0(0) = 0). Theorem B. If f 0(x) > 0 on I, f (x) increases. Theorem B. If f 0(x) > 0 on I, f (x) increases.
Requires Lagrange’s theorem:
∀x1, x2 ∈ I, there exists value
c between x1 and x2 such that
f(x2)
slope=f '(c)
f(x1)
x1
I
f (x2) − f (x1) = f 0(c)(x2 − x1)
If f 0(c) > 0 on I, x2 > x1⇒f (x2) > f (x1)
c
x2 Theorem B. If f 0(x) > 0 on I, f (x) increases.
Requires Lagrange’s theorem:
∀x1, x2 ∈ I, there exists value
c between x1 and x2 such that
f(x2)
slope=f '(c)
f(x1)
x1
I
f (x2) − f (x1) = f 0(c)(x2 − x1)
If f 0(c) > 0 on I, x2 > x1⇒f (x2) > f (x1)
Note: f (x) is increasing ⇒ f 0(x) ≥ 0
But f 0(x) ≥ 0 6⇒ f (x) is increasing.
c
x2 Theorem A. If f (x) is decreasing on I, and f 0(x)
exists, then f 0(x) ≤ 0 on I.
Theorem B. If f 0(x) < 0 on I, f (x) decreases. EXAMPLES f 0(x) > 0(f 0(x) < 0) ⇒ is increasing (decreasing)
EXAMPLE 3. Find where f (x) = x2 − 5x + 1 is
increasing and where it is decreasing. f 0(x) > 0(f 0(x) < 0) ⇒ is increasing (decreasing)
EXAMPLE 3. Find where f (x) = x2 − 5x + 1 is
increasing and where it is decreasing.
Consider
0
f (x) = 2x − 5
f 0(x) > 0 if x > 5/2, f 0(x) < 0 if x < 5/2. By
Thm. B:
5
f (x) is increasing for x >
2
5
f (x) is decreasing for x <
2 f 0(x) > 0(f 0(x) < 0) ⇒ is increasing (decreasing)
EXAMPLE 4. Find where f (x) = (x2 −3x)/(x+1)
is increasing and where it is decreasing. f 0(x) > 0(f 0(x) < 0) ⇒ is increasing (decreasing)
EXAMPLE 4. Find where f (x) = (x2 −3x)/(x+1)
is increasing and where it is decreasing.
Consider
(x + 3)(x − 1)
0
f (x) =
(x + 1)2
f 0(x) > 0 for x < −3 and x > 1 (increasing)
0
f (x) < 0 for −3 < x < −1 and −1 < x < 1
(decreasing) Concavity.
what the second derivative
can tell about the function? Two way of increasing:
f(x2)
f(x2)
f(x1)
x1
I
f(x1)
x2
x1
I
x2 Two way of increasing:
f(x2)
f(x2)
f(x1)
x1
I
f(x1)
x2
x1
How to distinguish these two cases?
I
x2 f(x2)
Definition f (x) is concave up on
I if f 0(x) increases on I.
f(x1)
x1
I
x2 f(x2)
Definition f (x) is concave up on
I if f 0(x) increases on I.
f(x1)
x1
I
x2
f(x2)
Definition f (x) is concave down
on I if f 0(x) decreases on I.
f(x1)
x1
I
x2 Second derivative and Concavity
00
0
f (x) > 0 ⇒ f (x) is increasing = Concave up
f 00(x) < 0 ⇒ f 0(x) is decreasing = Concave down
Concavity changes = Inflection point Second derivative and Concavity
00
0
f (x) > 0 ⇒ f (x) is increasing = Concave up
f 00(x) < 0 ⇒ f 0(x) is decreasing = Concave down
Concavity changes = Inflection point
Example 5. Where the graph of f (x) = x3 − 1 is
concave up, concave down?
00
00
Consider f (x) = 2x. f (x) < 0 for x < 0,
concave down; f 00(x) > 0 for x > 0, concave up. f 00(x) > 0(f 00(x) < 0) ⇒ concave up(down)
EXAMPLE 6. Find where the graph of f (x) =
x − sin(x) is concave up, concave down? f 00(x) > 0(f 00(x) < 0) ⇒ concave up(down)
EXAMPLE 6. Find where the graph of f (x) =
x − sin(x) is concave up, concave down?
0
f (x) = 1 − cos(x),
00
f (x) = sin(x)
f 00(x) > 0 for x ∈ [kπ, (k + 1)π], k =
0, ±1, ±2, . . . (concave up)
f 00(x) < 0 for x ∈ [(k − 1)π, kπ], k =
0, ±1, ±2, . . . (concave down)
kπ, k = 0, ±1, ±2, . . . inflection points Minima and Maxima.
critical points, first derivative test
second derivative test. Definition. f (c) is a local maximum value of
f (x) if there exists an interval (a, b) containing c
such that ∀x ∈ (a, b), f (c) ≥ f (x).
Definition. f (c) is a local minimum value of f (x)
if there exists an interval (a, b) containing c such
that ∀x ∈ (a, b), f (c) ≤ f (x).
loc max
f '(x) DNE
loc max
loc max
end point
loc max
f '(x)=0 loc max
f '(x) DNE
loc min
f '(x)=0
loc min
loc min
end point
loc min
f '(x)=0
f '(x)=0
loc min
f '(x)=0 Critical Point Theorem. If f (c) is a local min
(max), then c is a critical point, that is
a) an end point
0
b) a stationary point, that is f (c) = 0
0
c) a singular point, that is f (c) does not exists
(a) and c) are proved by examples.)
loc max
f '(x) DNE
loc max
loc max
end point
loc max
f '(x)=0 loc max
f '(x) DNE
loc min
f '(x)=0
loc min
loc min
end point
loc min
f '(x)=0
f '(x)=0
loc min
f '(x)=0 Proof b) If f (c) is max, then
f (t) − f (c)
< 0,
t−c
x > c,
f(c)
f(t)
t-c<0
t
Or,
f (t) − f (c)
> 0,
t−c
f (t) − f (c)
lim
≤ 0,
t→c+
t−c
t-c>0
c
I
x < c,
f (t) − f (c)
lim
≥ 0,
t→c−
t−c
f (t) − f (c)
f (c) = lim
=0
t→c
t−c
0
f(t)-f(c)<0
f(t)-f(c)<0
t First Derivative Test
f 0(x) > 0 to the left, f 0(x) < 0 to the right of c ⇒
increases to the left, decreases to the right of c ⇒
max at x = c.
f 0(x) < 0 to the left, f 0(x) > 0 to the right of c ⇒
decreases to the left, increases to the right of c ⇒
min at x = c.
f(c)
min
f(t)-f(c)<0
f(t)-f(c)>0
max
f(t)
f(t)
t-c<0
t-c<0
t-c>0
t-c>0
f(c)
t
c
I
t
t
c
I
t Second Derivative Test
f 00(c) < 0 and f 0(c) = 0 ⇒
f 0(x) is decreasing near c and passing 0 at c ⇒
f 0(x) > 0 to the left, f 0(x) < 0 to the right of c ⇒
increases to the left, decreases to the right of c ⇒
max at x = c.
00
0
f (c) > 0 and f (c) = 0 ⇒
f 0(x) is increasing near c and passing 0 at c ⇒
f 0(x) < 0 to the left, f 0(x) > 0 to the right of c ⇒
decreases to the left, increases to the right of c ⇒
min at x = c. EXAMPLES EXAMPLE 7.
Find where the graph of
f (x) = x ln(x) is increasing, decreasing, concave
up, concave down, has max, min? EXAMPLE 7.
Find where the graph of
f (x) = x ln(x) is increasing, decreasing, concave
up, concave down, has max, min?
1
f (x) = ln(x) + 1, f (x) =
x
f 0(x) < 0 for 1 < x < 1/e (decreasing), f 0(x) >
0 for x > 1/e (increasing)
0
00
00
f (x) > 0 for x > 0, (concave up), no inflection
pts.
min at x = 1/e (first derivative test) EXAMPLE 8. Find where the graph of f (x) =
1/(x2 + 1) is increasing, decreasing, concave up,
concave down, has max, min? EXAMPLE 8. Find where the graph of f (x) =
1/(x2 + 1) is increasing, decreasing, concave up,
concave down, has max, min?
2x
0
f (x) = − 2
,
(x + 1)2
2
3x
−1
00
f (x) = 2 2
(x + 1)3
f 0(x) > 0 for x < 0 (increasing), f 0(x) < 0 for
x > 0 (decreasing) max at x = 0 (first derivative
test)
√
√
00
f (x) > 0 for x < −1/ 3 √
and x > 1/√3,
(concave up), f 00(x) <
0
for
−1/
3
<
x
<
1/
3,
√
√
(concave down), −1/ 3, 1/ 3 are inflection pts.
Increasing and Decreasing Functions, Min and Max, Concavity
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