|
Unlimited Tutoring & Homework Help
|
Let f be a function defined on an interval I and let x1 and x2 be any two points on I.
(i) f is said to be increasing in the interval I,

Example:
Define

f(x) is increasing, because

Since 2 - 1 < 3 - 1
(ii) f is said to strictly increasing in the interval I if
For example,
Let x1 < x2

f(x) is strictly increasing function.
(iii) f(x) is said to be decreasing function if for x1, x2
I


= 0 for 0
x <2
2
is a decreasing function.
(iv) f (x) is said to be strictly decreasing on an interval I if for
For example,
is strictly decreasing function.Theorem 1:
Let f be continuous on [a, b] and differentiable on the open interval (a, b). Then
(a) f is increasing on [a, b] if f '(x) > 0 for each x
(a, b)
(b) f is decreasing on [a, b] if f '(x) < 0 for each x
(a, b)
This theorem is applied in various problems to check whether a function is increasing or decreasing.
Working Rule to Check Whether a Differentable Function is Increasing or Decresing
(1) Let the given function be f (x) on the real number line R.
(2) Differentiate the function f(x) with respect to x and equate it to zero i.e., put f '(x) = 0. Solve for x. These values of x which satisfy f '(x) = 0 are called Critical values of the function(3) Arrange these Critical values in ascending order and partition the domain of f (x) into various intervals, using the Critical values.
(4) Check the sign of f '(x) in each open intervals.(5) If f '(x) > 0 in a particular interval, then the function is increasing in that particular interval.
If f '(x) < 0 in a particular interval, then the function is decreasing in that particular interval.Example:
Find the intervals on which the function
(a) increasing (b) decreasing
Differentiating the function, we have


The critical values in ascending order are -1, 1. We divide the Real numbers into the intervals 

Since f '(x) < 0, the function is decreasing in the interval
.


= - ve
Since f '(x) < 0, f(x) is decreasing in the interval (1,
). 
