Lesson 25
Definition
Procedure
Examples
Section 3.1: Increasing and Decreasing
Functions; Relative Extrema
March 12th, 2014
Lesson 25
Definition
Procedure
Examples
In this lesson we will discuss how to determine where a function
is increasing and where it is decreasing. This will give us a new
kind of application of the derivative which will be useful in real
world problems.
Lesson 25
Definition
Definition
Procedure
A function f (x) is increasing if f (x) increases as x increases.
It is decreasing if f (x) decreases as x increases.
Examples
By sketching a parabola and looking at the tangent lines, we
see that a function f (x) is increasing on an interval (a, b) if
and only if its derivative f 0 (x) is positive on (a, b), and it is
decreasing on (a, b) if and only if its derivative is negative on
(a, b). In summary...
f (x) is increasing on (a, b) ⇐⇒ f 0 (x) > 0 on (a, b)
f (x) is decreasing on (a, b) ⇐⇒ f 0 (x) < 0 on (a, b)
Lesson 25
Definition
Procedure
Examples
Note the derivative of a function changes sign only around
points where it is zero or not continuous. With this in mind, we
give a procedure for finding where a function is increasing or
decreasing.
Procedure
(1) Find all x values such that f 0 (x) = 0 or f 0 (x) is not
continuous, and mark these numbers on a number line.
(2) Choose a test value c from each interval a < x < b
determined in (1) and compute f 0 (c).
f 0 (c) > 0 ⇒ f increasing on a < x < b
f 0 (c) < 0 ⇒ f decreasing on a < x < b
Lesson 25
Example
Definition
Procedure
Find where the given function is increasing or decreasing.
Examples
f (x) = 3x 3 − x 2 + 5
Following the procedure, we first compute f 0 (x).
f 0 (x) = 9x 2 − 2x = x(9x − 2)
We must now find where f 0 (x) is zero or not continous.
Since f 0 (x) is a polynomial, it is continuous everywhere. It
is zero at x = 0, 29 . Now we put these values on a number
line.
Lesson 25
Definition
Procedure
0
2
9
Examples
We now pick test values in the intervals −∞ < x < 0,
0 < x < 29 , and 29 < x < ∞ and plug them into f 0 (x).
f 0 (−1) = −1(−9 − 2) > 0
f 0 ( 91 ) = 19 (1 − 2) < 0
f 0 (1) = 1(9 − 2) > 0
Note that we don’t care about the actual values of f 0 (x). We
only care about the sign. Now we update our number line by
putting a + over the intervals where f 0 (x) is positive and a −
over the intervals where f 0 (x) is negative.
Lesson 25
Definition
Procedure
Examples
−
+
0
+
2
9
Looking at this number line, we clearly see that f (x) is
increasing on −∞ < x < 0 and 92 < x < ∞ and decreasing on
0 < x < 92 .
Lesson 25
g (t) =
Definition
3
16−t 2
Procedure
Examples
g (t) = 3(16 − t 2 )−1 ⇒ g 0 (t) = −3(16 − t 2 )−2 (−2t)
6t
=
(16 − t 2 )2
So g 0 (t) is zero at t = 0. Since g 0 (t) is a rational polynomial,
it is discontinuous where it is undefined, i.e. where the
denominator is zero.
(16 − t 2 )2 = 0 ⇒ 16 − t 2 = 0 ⇒ (4 + t)(4 − t) = 0
Therefore g 0 (t) is not continuous at t = −4, 4.
Lesson 25
−
−
+
+
Definition
−4
Procedure
0
4
Examples
g 0 (−5) =
g 0 (2) =
−30
(16−25)2
12
(16−4)2
< 0,
> 0,
g 0 (−2) =
g (5) =
−12
(16−4)2
30
(16−25)2
<0
>0
So our conclusion is
g (t) increasing: −∞ < t < −4, −4 < t < 0
g (t) decreasing: 0 < t < 4, 4 < t < ∞
Not that we do not say that g (t) is increasing on −∞ < t < 0.
We leave out the points where the derivative is zero or not
continuous.
Lesson 25
F (x) = x 3 −
Definition
Procedure
Examples
1
x3
6
6
3
= 3xx 4+3 = 3(xx 4+1)
x4
3(x 6 + 1) = 0 ⇒ x 6 = −1,
F 0 (x) = 3x 2 +
F 0 (x) = 0:
no solution.
F 0 (x) not continuous: x 4 = 0 ⇒ x = 0
+
+
0
F 0 (−1) =
3(1+1)
1
> 0,
F 0 (1) =
3(1+1)
1
>0
Thus F (x) is increasing on −∞ < x < 0 and 0 < x < ∞.
F (x) is nowhere decreasing.