Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Monotonicity
Monotonic Function
Non-Monotonic Function
Monotonicity of a Function at a Point
Monotonicity in an Interval
Non-decreasing and Non-increasing functions
Greatest and Least Value of a Function
Monotonicity is an important part of application of derivatives. The monotonicity of a function gives an idea about the behaviour of the function. A function which is either completely non-increasing or completely non-decreasing is said to be monotonic.
A function is said to be monotonic if it is either increasing or decreasing in its entire domain.
eg : f(x) = 2x + 3 is an increasing function while f(x) = -x^{3} is a decreasing function.
If x_{1} < x_{2} and f(x_{1}) < f(x_{2}) then the function is called increasing function or strictly increasing function.
Decreasing Function:
If x_{1} < x_{2} but f(x_{1}) > f(x_{2}) in the entire domain, then the function is said to be a decreasing function or strictly decreasing function.
Functions which are increasing as well as decreasing in their domain are said to be non-monotonic functions.
Eg: f(x) = |x|
f(x) = sin x is non-monotonic but is increasing in the interval [0, π/2].
We can talk of the concept of monotonicity either at a point or in an interval:
A function is said to be monotonically increasing at x = a if f(x) satisfies:
f(a + h) > f(a) and
f(a - h) < f(a) , for a small positive h.
A function is said to be monotonically decreasing at x = a if f(x) satisfies:
f(a + h) < f(a) and
f(a – h) > f(a) , for a small positive h.
Note: We can talk of monotonicity of f(x) at x = a only if x = a lies in the domain of f(x) without any restriction of continuity or differentiability of f(x) at x = a.
For an increasing function in some interval
if Δx > 0 ⇔ Δy > 0 or Δx < 0 ⇔ Δy < 0, then f is said to be monotonic (strictly) increasing in that interval.
In other words, we can say that if dy/dx > 0 in some interval then y is said to be an increasing function in that interval. Also, if the function f(x) is increasing in some interval, then dy/dx > 0 in that interval.
Similarly, if dy/dx < 0 in some inetrval then y is said to be an decreasing function in that interval. Also, if the function f(x) is decreasing in some interval, then dy/dx < 0 in that interval.
Points to Remember:
At some point, we may have dy/dx to be equal to zero, but f(x) may still be increasing at x = a. Such a point is termed as the point of inflexion and these points indicate the change of concavity of the curve.
If f(x) is increasing for x > a and f is also increasing for x < a then f is also increasing at x = a provided f(x) is continuous at x = a.
If the function f(x) is discontinuous at x = a, then it becomes possible to draw the graph wherein x = a is the point of maxima.
If the function f is decreasing for x > a and it is also decreasing for x < a then f is also decreasing for x = a provided f(x) is continuous at x = a.
If a function si not defined at a particular point like suppose f(a) is not defined then monotonicity cannot be indicated at x = a.
A function f(x) is termed to be monotonically increasing for all such interval (a, b) where f’(x) ≥ 0 , where equality will hold only for discrete values of x i.e. f;(x) does not identically become zero for x ∈ (a,b) or any sub-interval.
The function f(x) is said to be monotonically decreasing for all such interval (a, b) where f’(x) ≤ 0 and again the equality may hold only for discrete values of x.
Example: Prove that f(x) = x – sin x is an increasing function.
Solution: f(x) = x – sin x
So, f’(x) = 1- cos x
Now, f’(x) > 0 everywhere except at x = 0, ±2π, ±4π etc. but since all of these points are discrete and do not form an interval, hence, we can conclude that f(x) is monotonically increasing for x ∈ R.
A function f(x) is said to be non-decreasing in the domain, if for every x_{1}, x_{2} ∈ D, x_{1} > x_{2} and f(x_{1}) ≥ f(x_{2}). In other words, it means that the value of the function f(x) would never decrease with an increase in the value of x.
Similarly, f(x) is said to be non-increasing in a domain if for every x_{1}, x_{2} ∈ D*, x_{1} > x_{2} and f(x_{1}) ≤ f(x_{2}). In other words, it means that the value of the function f(x) would never increase with an increase in the value of x.
If a function is monotonic at x = a it cannot have extremum point at x = a and conversely i.e. a point on the curve cannot simultaneously be an extremum as well as monotonic point.
If f is increasing then nothing definite can be said about the function f’(x) as to whether it is increasing or decreasing.
Case 1: If a function y = f(x) is strictly increasing in the closed interval [a,b] then f(a) is the least value and f(b) is the greatest value.
Case 2: If f(x) is decreasing in [a,b] then f(b) is the least and f(a) is the greatest value of f(x) in [a,b].
Case 3: If (x) is non-monotonic in [a,b] and is continuous then the greatest and least value of f(x) in [a,b] are those where f(x) = 0 or f’(x) does not exist or at the extreme values.
Let us now discuss some examples based on this concept:
Example 1: Find the set of values of x for which ln(1 + x) > x/(1 + x)
Solution: The given function is f(x) = ln(1 + x) – x/(1 + x)
= ln(1 + x) + 1/(1 + x) – 1
Domain: x > -1
f’(x) = 1/(1 +x) – 1/(1 + x)^{2} = x/(1 + x)^{2}
f’(x) ≥ 0 ∀ x ≥ 0.
This means that f(x) is increasing
and f’(x) ≤ 0 ∀ x ≤ 0.
This means that f(x) is decreasing.
and f’(0) = 0
Therefore, f(x) > f(0) ∀ x ∈ D_{f} – {0}
Therefore, f(x) > 0 ∀ x ∈ (-1, 0) ∪ (0,∞).
Example 2: Show that the equation x^{5} – 3x – 1 = 0 has a unique root in [1,2].
Solution: Consider the function x^{5} - 3x – 1 = 0, x ∈ [1,2].
and f’(x) = 5x^{4} – 3 > 0∀ x ∈ [1,2].
So, f(x) is strictly increasing in [1,2].
Also, we have f(1) = 1 - 3 - 1 = -3.
f(2) = 32 – 6 – 1 = 25
From the shape of the curve, we can see that the curve y = f(x) will cut the x-axis exactly once in [1,2].
So, f(x) will vanish exactly once in [1,2].
Example 3: Prove that x/(1 + x) < ln(1 + x) < x ∀ x > 0.
Solution: Consider the function f(x) = ln(1 + x) - x/(1 + x), x > 0.
then, f‘(x) = x/(1 + x) – x/(1 + x)^{2} = x/(1 + x)^{2} > 0 ∀ x > 0
So, f(x) is strictly increasing in (0, ∞)
So, f(x) > f(0^{+}) = 0
i.e. ln(1 + x) > x/(1 + x) which proves the first inequality.
now, consider the function g(x) = x – ln(1 + x), x > 0
then g’(x) = 1 – 1/(1+x)
= x/(1 + x) > 0 ∀ x > 0
So, g(x) is strictly increasing in (0, ∞).
This means g(x) > g(0^{+}) = 0
So, x > ln(1+x) which proves the second part of inequality.
Example 4: Find the intervals in which the function f (x) = 2x^{2} – ln |x| is
(i) decreasing (ii) increasing
Solution: f (x) = 2x^{2} – ln |x|
f'(x) = 4x – 1/|x|.|x|/x ⇒ f' (x) = 4x^{2}–1/x
for increasing function f (x), f' (x) > 0
⇒ 4x^{2}–1/x > 0
⇒ x ∈ (–1/2, 0) U (1/2, ∞)
for decreasing function f (x), f' (x) < 0
⇒ 4x^{2} – 1/x < 0
⇒ x ∈ (∞, –1/2) U (0, 1/2)
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
Rate of Change of Quantities Table of Content...
Concept of Local Maximum and Local Minimum Local...
Motion in a Straight Line Table of Content...
Download IIT JEE Solved Examples of Applications...
Mean Value Theorem Table of Content History of...
Introduction of Application of Derivatives Table...
Objective Type Questions 42. Total number of...
Increasing and Decreasing Functions Table of...
Geometrical Meaning of Derivative at Point The...