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When we are solving some equation, the graph either goes up the slope or down the slope. The former situation is called a positive slope and the latter a negative one. Since derivatives and slope are similar we can associate them to the derivatives of a function. When the value of the function escalates as the input value escalates within an interval, we say that the function is increasing on an interval. But when the function value decreases while the input value escalates we say that the function is decreasing over that interval. Some functions may continue to shoot up throughout the graph as we move from the left side of the graph to the right whereas some might change their course of action.
The input value where the function changes from escalating to declining is called local maximum. When the function has more than one single value we call it local maxima. Whereas when the function value becomes increasing from decreasing with an increasing input value we call it local minimum and local minima for more than one value. Local maximum and local minimum together is called local extrema in increasing and decreasing functions.
Increasing /Decreasing Functions are a part of the application of derivatives and carry a weightage of 6 to 7 marks and the whole unit Calculus carries a total of 56 periods.
Keyterms: Derivatives, Functions, Increasing, Decreasing, Intervals, Slope,
Read More: Determinant of a Matrix
What is an increasing decreasing function?
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These are the applications of the derivatives. The increasing and decreasing nature of the functions in the given interval can be found out by finding the derivatives of the given function.
Function: y = f(x)
When the value of y increases with the increase in the value of x, the function is said to be increasing in nature.
When the value of y decreases with the increases in the value of x, the function is said to be decreasing in nature.
Example: Suppose a graph shows the plot of y = x2 -1:
On the left-hand side of the origin, as the value of x increases, the value of y is decreasing (curve downwards), hence the function is decreasing.
Whereas, on the right-hand side of the origin, the value of y increases with the increase in x (curve upwards), hence the function is increasing.
Also Read:
Definitions
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Given function – y = f(x)
- Increasing function:
If a function increases over an interval, the function will be increasing.
Where the conditions are:
x1 < x2
f(x1) ≤ f(x2)
- Decreasing function:
If the function decreases over the given interval, the function is called a decreasing function:
Where,
x1 < x2
f(x1) ≥ f(x2)
- Strictly increasing function:
A function is said to be strictly increasing when for all the values between the given interval (x1, x2) the function is increasing.
Where, x1 < x2.
Therefore – f(x1) < f(x2)
- Strictly decreasing function:
A function is said to be strictly decreasing when for all the values between the given interval (x1, x2) the function is decreasing.
Where, x1 < x2.
Therefore – f(x1) > f(x2)
- Monotonic function:
If a function falls under the category of any one of the following: Increasing, Strictly increasing, decreasing, strictly decreasing then the function is called a monotonic function.
Read More: Applications of Derivatives Ncert Solutions
How are derivatives used to find out the increasing/decreasing nature of the function?
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Using the First derivative test:
In this test, the derivative of the function is used to find out the nature of the function – monotonic or not, increasing/decreasing.
To find out the nature of the function [f(x)] in the domain (a, b)
- If dfdx ≥ 0 for all x in (a, b), the function is said to be increasing in the domain (a, b).
- If dfdx ≤ 0 for all x in (a, b), the function is said to be decreasing in the domain (a, b).
- If dfdx>0, the function is said to be strictly increasing (strict monotonic).
- If dfdx<0, the function is strictly decreasing (strict monotonic).
NOTE:
If the derivative of the given function equals zero, then the function is said to be a constant function.
A mathematically constant function can be represented as:
dfdx = 0
How to calculate the intervals of the function?
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- Firstly, find the derivative of the given function.
- Now, equate the derivative to 0 and find the value of x.
- After this calculate the various open intervals where the values of the derivatives are found.
- If all the values of the function remain positive, then the function is said to increase.
- If the values decrease, the function will be called a decreasing function.
Read More: Integral Calculus
Sample questions
Q: What is the nature of the function sin(x) in the given intervals, (0, π/4) and (π4,π2)? (2 marks)
Ans: Derivative of the function: d(sinx)dx = cos x
We can see that x∈(0,π4) the value is positive, hence the function is increasing in the interval.
Whereas, in the interval (π4,π/2), the value is negative, hence the function is decreasing in the interval.
Q: What does the interval of decrease refer to in the function? (1 mark)
Ans: The interval of decrease for the function refers to the range where the value of the function decreases.
Q: what is the difference between the increasing and strictly increasing function? (1 mark)
Ans: The function which has all the values positive (including 0) is called an increasing function, whereas the function which has all the values positive (NOT 0) are called strictly increasing functions.
Q: How can we find the increasing or decreasing nature without opting for the first derivative test? (2 marks)
Ans: We can find out the increasing/decreasing nature by trial and error method for the function i. e. we can put in some random number from the given interval and check whether the values are increasing/decreasing and then take the final decision.
Q: Why is increasing decreasing function called the application of derivatives? (2 marks)
Ans: It is because the increasing/decreasing nature of the function can be found out by using the derivative (First derivative test). Also, the constant function can be found using this derivative test with one step.
Q: What are some examples of constant functions? (1 mark)
Ans: f(x) = 2, f(x) = π, etc.
Previous Year Questions
Ques. Find the intervals in which the function f(x) = 3x4 - 4x3 - 12x2 + 5 is: (a) Strictly increasing, (b) Strictly decreasing (2018)
Solution: Given: f(x) = 3x4 - 4x3 - 12x2 + 5
Then f(x) = 12x3 - 12x2 - 24x
Then f’(x) = 0
12x3 - 12x2 - 24x = 0
12x (x2 - x - 2) = 0
12x (x2 - 2x + x - 2) = 0
12x[x(x-2) + 1(x-2)] = 0
12x(x+1)(x-2) = 0
x = 0; x = -1; x = 2
So, The points x = – 1, x = 0and x = 2 divide the real line into four disjoint intervals, namely ( – ∞, – 1), ( – 1, 0), (0, 2) and (2, ∞).
| Interval | Sign of f'(x) = 12x (x + 1)(x – 2) | Nature of Function |
|---|---|---|
| ( – ∞, – 1) | ( – )( – )( – ) = – or < o | Strictly Decreasing |
| ( – 1, 0) | ( – )( + )( – ) = + or > 0 | Strictly Decreasing |
| (0, 2) | ( + )( + )( – )= – or < 0 | Strictly Decreasing |
| (2, ∞) | ( + )( + )( + ) = + or > 0 | Strictly Decreasing |
(a) The given function is strictly increasing in the intervals ( – 1,0) U (2,∞).
(b) The given function is strictly decreasing in the intervals ( – ∞, – 1) U (0,2).
Ques. Show that the function f(x) = 4x3 – 18x2 + 27x – 7 is always increasing on R. (2017)
Solution: The given function is f(x) = 4x3 – 18x2 + 27x – 7, On differentiating both sides with respect to x, we get
f’(x) = 12x2 – 36x + 27
⇒ f’(x) = 3(4x2 – 12x + 9)
⇒f’(x) = 3(2x – 3)2
Which is always positive for all x ∈ R.
Since, f’(x) ≥ 0 \(\forall\)x ∈ R, Therefore, f(x) is always increasing on R.
Ques. Find the intervals in which f(x) = sin3x – cos3x, 0<x<π, is strictly increasing or strictly decreasing. (2016)
Solution: If we take the function f’(x) = sin3x - cos3x
f’(x) = 3cos3x + 3sin3x
= 3 √ 2 { sin 3x cos (\(\frac{ \pi}{4}\)) + cos 3x sin (\(\frac{ \pi}{4}\))}
= 3 √ 2 { sin ( 3x + \(\frac{ \pi}{4}\))}
For the increasing interval f’(x) > 0
3 √ 2 { sin( 3x + \(\frac{ \pi}{4}\))} > 0
sin( 3x + \(\frac{ \pi}{4}\)) > 0
⇒ 0 < 3x + \(\frac{ \pi}{4}\)< π
⇒ 0 < 3x < \(\frac{3 \pi}{4}\)
⇒ 0 < x < π / 4
also sin ( 3x + \(\frac{ \pi}{4}\)) > 0
when, 2π < 3x + \(\frac{ \pi}{4}\) < 3π
=> \(\frac{ 7\pi}{4}\) < 3x < \(\frac{ 11\pi}{4}\)
Therefore, intervals in which function is strictly increasing in 0 < x < π/4 and 7π/12 < x < 11π/12
Similarly, for the decreasing interval f’(x) < 0
3 √ 2 { sin (3x + \(\frac{ \pi}{4}\)) } < 0
sin ( 3x + \(\frac{ \pi}{4}\)) < 0
=>π < 3x + \(\frac{ \pi}{4}\) < 2π
=>\(\frac{3 \pi}{4}\) < 3x < \(\frac{ 7\pi}{4}\)
=> π/4 < x < 7π/12
Also, sin ( 3x + \(\frac{ \pi}{4}\)) < 0
When, 3π < 3x + \(\frac{ \pi}{4}\) < 4π
=> \(\frac{ 11\pi}{4}\) < 3x < \(\frac{ 15\pi}{4}\)
=> (11π /12)
The function is strictly decreasing in \(\frac{ \pi}{4}\) and (11π /12)
π4 < x < 7π12 and 11π12 < x < π
Ques. Find the value(s) of x for which y = [x(x – 2)]2 is an increasing function. (2014 AI)
Solution: y = [x(x – 2)]2
y = [x2 – 2x]2
∴ \(\frac{dy}{dx}\) = 2(x2 – 2x)(2x – 2)
or, \(\frac{dy}{dx}\) = 4x(x – 1)(x – 2)
on equating, \(\frac{dy}{dx}\)= 0, 4x(x – 1)(x – 2) = 0 ⇒ x = 0, x = 1, x = 2
∴ Intervals are ( – ∞,0), (0, 1), (1, 2) and (2, ∞).
Since \(\frac{dy}{dx}\)> 0 in (0,1) or (2, ∞)
∴f(x) is increasing in (0,1) U (2, ∞)
Ques. Find the intervals in which the function given by f’(x) = sinx + cosx, 0 ≤ x ≤ 2π is: (a) Increasing, (b) Decreasing.
Solution: We know, f’(x) = sinx + cosx, 0 ≤ x ≤ 2π, f’(x) = cosx - sinx
Since the main points of function over the interval v 0 ≤ x ≤ 2π is provided by
f’ (x) = 0 ⇒ cos x – sin x = 0
⇒ cos x = sin x
⇒ x = \(\frac{ \pi}{4}, \frac{ 5\pi}{4}\)
Possible intervals are (0,\(\frac{ \pi}{4}\)), (\(\frac{ \pi}{4}, \frac{ 5\pi}{4}\)), (\(\frac{ 5\pi}{4}\), 2π)
If 0 < x < \(\frac{ \pi}{4}\), f’ (x) = cos x – sin x > 0 \(\because\) cos x > sin x
⇒ f’(x) > 0
⇒ f’(x) is strictly increasing.
If \(\frac{ \pi}{4}\) < x < \(\frac{ 5\pi}{4}\), f’(x) = cos x – sin x < 0 \(\because\) cos x < sin x
⇒ f’(x) is strictly decreasing.
If \(\frac{ 5\pi}{4}\) < x < 2π, f’(x) = cos x – sin x < 0 \(\because\) cos x > sin x
⇒ f’(x) is again strictly increasing.
The function provided f(x) = sinx+cosx[0,2π] = strictly increasing
\(\forall\) x ∈ (0, \(\frac{ \pi}{4}\)) and (\(\frac{ 5\pi}{4}\), 2π)
While it is strictly decreasing \(\forall\)x ∈ (\(\frac{ \pi}{4}, \frac{ 5\pi}{4}\))
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