Let F Be a Continuous Strictly Decreasing Real Value Dunction Such That

Increasing and Decreasing Functions: Derivative Tests, Applications

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Increasing and Decreasing Functions: Any activity can be represented using functions, like the path of a ball followed when thrown. If you have the position of the ball at various intervals, it is possible to find the rate at which the position of the ball is changing. If we graph the path the ball follows with respect to time, we can see the change in the position at various time intervals. Such studies of derivatives have practical applications in multiple fields. One of the most important applications is to study the monotonicity of the function on its domain. In this article, we will learn about increasing and decreasing functions in detail.

What is a Function?

The relation \(R\) from the non-empty set \(A\) to the non-empty set \(B\) is a subset of the cartesian product \(A×B.\) It is represented as shown below.

Here, the relation \(R\) is the set of ordered pairs that is written as:

\(\{ (1,5),(3,15),(5,25),(7,35)\} \)

A relation f from a set \(A\) to a set \(B\) is a function if every element of set \(A\) has one and only one image in set \(B.\)

If you consider the relation \(R\) as shown above, element \(10\) in set \(A\) does not have an image in set \(B,\) and therefore, \(R\) is not a function.

If we map the element \(10\) in set \(A\) to \(50\) in set \(B,\) the relation \(R'\) given by: \(\{ (1,5),(3,15),(5,25),(7,35),(10,50)\} \)

\(R'\) is a function defined by the equation \(y=5x,\) where \(x\) is the independent variable whose values vary within the elements of the set \(A\) and \(y\) is the dependent variable whose values range within the elements of the set \(B.\)

What is the Derivative of a Function?

The derivative of a real-valued function measures the tendency of a function to change the values with respect to the change in its independent variable.

The derivative of a function at a point, if exists, is the slope of the tangent line to the graph of the function at that point.

If \(f(x)\) is a real-valued function that is differentiable at a point a, and if the domain contains an open interval \(I\) containing \(a\) and the limit \(\frac{{f(a + h) – f(a)}}{h}\) exists, this limit is called the derivative of the function \(f(x)\) at \(a.\)

Application of Derivatives

There are many applications of the derivative of a function. One of the most important uses is to understand the monotonicity of the function. That is, it can help you know whether a function is increasing or decreasing or stays constant at a point or in an interval.

What is an Increasing Function?

As the word suggests, a function is said to be increasing when the value of the dependent variable \(y\) increases with \(x.\)

Example 1: Consider the graph of the function \(y=5x.\)

Observe that, as the value of \(x\) increases, the corresponding \(y\) values also increase. So, \(y\) is an increasing function.

Example 2: Consider the function \(y = {e^x}\) as an increasing function as the \(y-\)values increase with increasing \(x-\)values.

What is a Decreasing Function?

Contrary to the increasing functions, a function is said to be decreasing when the values of the dependent variable \(y\) decrease as \(x\) increases.

Example 1: Consider the function \(y=-2x:\)

\(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)
\(y=f(x) =-2x\)	\(4\)	\(2\)	\(0\)	\(-2\)	\(-4\)	\(-6\)

Observe that, as the value of \(x\) increases from \(-2\) to \(3,\) the corresponding \(y-\)value decreases from \(4\) to \(-6.\)

So, \(y\) is a decreasing function.

Example 2: The function \(y =\, – \log x\) is a decreasing function as the \(y-\)values decrease with increasing \(x-\)values.

Increasing and Decreasing Functions

Some functions may be increasing or decreasing at particular intervals.

Example: Consider a quadratic function \(y = {x^2}.\)

Observe that the dependent variable y decreases as the independent variable \(x\) increases in the interval \((-∞,0,)\) whereas \(y\) increases with \(x\) in the interval \((0, ∞).\)

This function is decreasing in the interval \((-∞,0)\) and increasing in the interval \((0, ∞).\)

Let us now define increasing and decreasing functions systematically.

Increasing and Decreasing Functions: Definition

Let \(I\) be an interval contained in the domain of a real-valued function \(f.\)

Then \(f\) is said to be:

(i) increasing on \(I\) if \({x_1} < {x_2}\) in \(I.\) This implies \(f\left( {{x_1}} \right) \le f\left( {{x_2}} \right)\) for all \({x_1},{x_2} \in I.\)

(ii) strictly increasing on \(I\) if \({x_1} < {x_2}\) in \(I.\) This implies \(f\left( {{x_1}} \right) < f\left( {{x_2}} \right)\) for all \({x_1},{x_2} \in I.\)

(iii) decreasing on \(I\) if \({x_1} < {x_2}\) in \(I.\) This implies \(f\left( {{x_1}} \right) \ge f\left( {{x_2}} \right)\) for all \({x_1},{x_2} \in I.\)

(iv) strictly decreasing on \(I\) if \({x_1} < {x_2}\) in \(I.\) This implies \(f\left( {{x_1}} \right) > f\left( {{x_2}} \right)\) for all \({x_1},{x_2} \in I.\)

(v) constant on \(I,\) if \(f(x)=c\) for all \(x∈I,\) where \(c\) is a constant.

Increasing Function at a Point \({x_0}:\)

A function \(f\) is said to be increasing at a particular point \({x_0},\) if there exists an
open interval \(I\) containing \({x_0}\) such that \(f\) is increasing in \(I.\)

Decreasing Function at a Point \({x_0}:\)

A function \(f\) is said to be decreasing at a particular point \({x_0},\) if there exists an
open interval \(I\) containing \({x_0}\) such that \(f\) is decreasing in \(I.\)

Example: The function \(y = {x^2}\) is decreasing at \(x=-2\) and increasing at \(x=2\)

How to identify the nature of a function at a given point?

Let us look into a theorem related to it.

Increasing and Decreasing Functions Theorem

Let \(f\) be continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a,b).\)

Then:

Case 1: \(f\) is increasing in \([a,b],\) if \(f'(x)>0,\) for each \(x∈(a,b)\)

Case 2: \(f\) is decreasing in \([a,b],\) if \(f'(x)<0,\) for each \(x∈(a,b)\)

Case 3: \(f\) is a constant function in \([a,b],\) if \(f'(x)=0,\) for each \(x∈(a,b)\)

Local Maximum and Minimum of a Function

Let \(f\) be a continuous real-valued function on the closed interval \([a, b]\) and differentiable on the open interval \((a, b).\)

A point \(c∈[a, b]\) where either \(f\) is not differentiable or its first derivative is zero is called a critical point of \(f.\)

A point \(c∈(a, b)\) is called a local maxima with \(f(c)\) as the local maximum value if there exists an \(h>0\) such that \(f(c)>f(x)\) for all the values in the sub-interval \((c-h, c+h).\)

Similarly, a point \(d∈(a, b)\) is called a local minima with \(f(d)\) as the local minimum value if there exists an \(h>0\) such that \(f(d)>f(x)\) for all the values in the sub-interval \((d-h,d+h).\)

How to Find Local Minima or Maxima?

The first and second derivative tests would help you find that.

First Derivative Test

Let \(f\) be a continuous real-valued function on the closed interval a, b and differentiable on the open interval \((a, b).\)

Case 1: As \(x\) increases through point \(c,\) if \(f'(x)\) changes sign from positive to negative, then \(c\) is a local maxima with \(f(c)\) as the local maximum value.

Case 2: As \(x\) increases through point \(c,\) if \(f'(x)\) changes sign from negative to positive, then \(c\) is a local minima with \(f(c)\) as local minimum value.

Case 3: As \(x\) increases through point \(c,\) if \(f'(x)\) does not change its sign, then \(c\) is neither a point of local maxima or local minima. Such points are called points of inflection.

If the first derivative at a point is zero and the function is twice differentiable, the second derivative test can be used.

Second Derivative Test

Let \(f\) be a function defined on an interval \(I\) and \(c∈I.\) Let f be twice differentiable, then:

Case 1: If the first derivative is zero and the second derivative is less than zero at a point \(c,\) then \(c\) is a point of local maxima with \(f(c)\) as local maximum value.

Case 2: If the first derivative is zero and the second derivative is greater than zero at a point \(c,\) then \(c\) is a point of local minima with \(f(c)\) as local minimum value.

The test does not work if the first and the second derivatives are zero at point \(c.\)

Now, if we can identify the critical points, the function's values at these points and the endpoints can be calculated. The maximum and the minimum values among these would be the function's absolute maximum and minimum values, respectively.

Solved Examples – Increasing and Decreasing Functions

Q.1. Show that \(f(x)=4x+9\) is a strictly increasing function on the set of real numbers.
Ans: Let \({x_1}\) and \({x_2}\) be two real numbers such that \({x_1} < {x_2}.\)
Multiplying both sides by \(4,\) we have:
\({x_1} < {x_2}\)
Adding \(9\) to both sides:
\(4{x_1} + 9 < 4{x_2} + 9\)
That is, \({x_1} < {x_2}\) implies \(f\left( {{x_1}} \right) < f\left( {{x_2}} \right)\)
Therefore, the function is strictly increasing.

Q.2. Decided whether the function \(f(x)=2x-7\) is increasing or decreasing on the set of real numbers \(R.\)
Ans: As the first step, find the first derivative of the function \(f(x)=2x-7.\)
\({f^\prime }(x) = \frac{d}{{dx}}(2x – 7)\)
\( = \frac{d}{{dx}}(2x) – \frac{d}{{dx}}(7)\)
\(=2-0\)
\(=2>0\)
Therefore, by the increasing function theorem, the function is increasing.
If you look at the graph of the function, it is clear that it is an increasing function.

Q.3. Find the least value of a for which the function \(g(x) = {x^2} + ax + 5\) is increasing on the interval \([0, 1].\)
Ans: The first derivative of the function \(g(x)\) is:
\(g'\left( x \right) = 2x + a\)
For the extreme values of the interval \([0, 1],\) the first derivatives are:
\(a=0:\)
\(g'(x)=2x+0\)
\(g'(x)=2x\)
\(a=1:\)
\(g'(x)=2x+1\)
For the function to be increasing, these values should be greater than zero.
That is,
\(g'(x)=2x>0⇒x>0\)
\({g^\prime }(x) = 2x + 1 > 0 \Rightarrow x > – \frac{1}{2}\)
Therefore, the least value of a for which the function \(g(x)\) increases in the
interval \([0, 1]\) is \(0.\)

Q.4. Find the interval in which the function f increases or decreases where f is given by \(f(x) = \cos 2x\) for \( – \frac{\pi }{4} \le x \le \frac{\pi }{4}\)
Ans: \({f^\prime }(x) = – \sin (2x) \times 2\)
\( = – 2 \sin 2x\)
Now, \({f^\prime }(x) = 0 \to \sin 2x = 0 \to x = 0\) as \(x \in \left[ { – \frac{\pi }{4},\frac{\pi }{4}} \right]\)
So, the point \(x=0\) divides the interval \(x \in \left[ { – \frac{\pi }{4},\frac{\pi }{4}} \right]\) into two disjoint intervals \(\left[ { – \frac{\pi }{4},0} \right)\) and \(\left( {0,\frac{\pi }{4}} \right]\)
Now, \(f' (x)<0\) for all \(x∈\) as \( \Rightarrow – \frac{\pi }{2} \le 2x < 0.\)
Also, \(f' (x)<0\) for all \(x \in \left( {0,\frac{\pi }{4}} \right]\) as \(0 < x < – \frac{\pi }{4} \Rightarrow 0 < 2x < \frac{\pi }{2}.\)
Therefore, f is increasing in \(\left[ { – \frac{\pi }{4},0} \right)\) and decreasing in \(\left( {0,\frac{\pi }{4}} \right].\)
Now, since the function is continuous at x = 0, by the first derivative test, f is increasing in \(\left[ { – \frac{\pi }{4},0} \right]\) and decreasing in \(\left( {0,\frac{\pi }{4}} \right].\)
You can see this result using the graph of the function:

Q.5. Prove that the function \(f(x)=log (1+x)\) for \(x>-1\) is an increasing function.
Ans: The first derivative of the function can be calculated as:
\({f^\prime }(x) = \frac{d}{{dx}}( \log (1 + x))\)
\( = \frac{1}{{1 + x}}\)
For \(x>-1,\) the value of \(\frac{1}{{1 + x}} > 0.\) Therefore, the function is increasing
for all the values in the domain.

Summary

The article helps you understand the basic concepts of functions right from their definitions. Then the article defines the derivative of a function and applies the concept of derivatives in finding the monotonicity of a function. The concepts of natures of increasing, strictly increasing, decreasing, strictly decreasing, and constant functions are explained with examples. A theorem as an application of derivatives is also discussed in simple language for better understanding. Further, the characteristics of functions such as local maxima, local minima, critical points are well explained with the first and second derivative tests to find these points in a real-valued function. Finally, a few solved examples are worked out to clarify the concepts explained.

Frequently Asked Questions

Q.1. What is meant by increasing and decreasing function?
Ans: A function is "increasing" when the dependent variable \(y\) increase as that of the independent variable \(x\) increase. A function is "decreasing" when the values of the dependent variable \(y\) decrease as that of the independent variable \(x\) increase.

Q.2. What is the difference between increasing and strictly increasing function?
Ans: Let \(I\) be an interval contained in the domain of a real-valued function \(f.\)
Then \(f\) is said to be:
(i) Increasing on \(I\) if \({x_1} < {x_2}\) in \(I.\) This implies \(f\left( {{x_1}} \right) \le f\left( {{x_2}} \right)\) for all \({x_1},{x_2} \in I.\)
(ii) Strictly increasing on \(I\) if \({x_1},{x_2} \in I\) in \(I.\) This implies \(f\left( {{x_1}} \right) < f\left( {{x_2}} \right)\) for all \({x_1},{x_2} \in I.\)

Q.3. What is strictly decreasing?
Ans: Let \(I\) be an interval contained in the domain of a real-valued function \(f.\)
Then \(f\) is said to be strictly decreasing on \(I\) if \({x_1} < {x_2}\) This implies \(f\left( {{x_1}} \right) > f\left( {{x_2}} \right)\) for all \({x_1},{x_2} \in I.\)

Q.4. Which functions are always increasing?
1. Linear
2. Quadratic
3. Absolute Value
4. Square Root
5. Cubic
6. Cube Root
7. Rational
8. Exponential
Ans: Let \(I\) be an interval contained in the domain of a real-valued function \(f.\) Then \(f\) is said to be increasing on \(I\) if \({x_1} < {x_2}\) in \(I.\) This implies \(f\left( {{x_1}} \right) \le f\left( {{x_2}} \right)\) for all \({x_1},{x_2} \in I.\)
Let \(f\) be continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a,b).\) Then, \(f\) is increasing in \([a, b]\) if \(f'(x)>0\) for each \(x∈(a,b).\) Out of the given functions, only the cubic functions always have the positive first derivatives, thus making the functions always increasing.

Q.5. How do you find the increasing and decreasing intervals?
Ans: First, identify the value of the independent variable in the interval for which the first derivative is zero. Now, the function is increasing on the interval where the first derivative is positive, and it is decreasing where the first derivative is negative.

We hope you find this article on 'Increasing and Decreasing Functions' helpful. In case of any queries, you can reach back to us in the comments section, and we will try to solve them.

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