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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)
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