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First Derivative: Definition, Formulas, and Examples

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First Derivative Test is the test in calculus to find whether a function has a maximum or minimum value in the given interval. As the name suggests, the first derivative is used in this test to find the critical point and then further conditions are used to check each critical point for extrema. Many times in our daily life, we come across many situations where we wish to find the maximum and minimum of something, in those cases, we can use the First Derivative Test if we can mathematically define that situation.

In this article, we will learn this important test i.e., “First Derivative Test”, other than this we will also learn how to use this test to find the extrema of the function as well as solved examples of the same. So, let’s start learning about one of the fundamental tests in calculus i.e., the First Derivative Test.

What is First Derivative Test?

The First Derivative Test is a method used to analyze the critical points of a function and determine whether at that point function attains its maximum or minimum value. It is one of the basic tests in calculus which uses the change in slope to determine the extremum value of the function. As we know, either side of each extremum point slope can be either positive or negative.

The first derivative test is a method used in calculus to analyze the critical points of a function and determine whether they correspond to relative maxima or minima. It is a part of the larger topic of calculus known as optimization, which deals with finding the maximum or minimum values of a function.

Note: The maximum and minimum values are also called the extremum values and the point which is either a point of local maxima or local minima is called an extreme point.

Theorem: If f'(x) exists in the interval [a, b] and the function f(x) has a maximum or minimum value at x = x0 ∈ (a, b) then f'(x0) = 0. The point x0 in the domain of a function at which f'(x0) = 0 is called a critical point.

Conditions for First Derivative Test

Let f(x) be a differentiable function defined on an interval I and let x0 ∈ I and f'(x0) = 0. Then

  • x0 is a point of the local maximum of f(x) if
    • f'(x0) changes sign from positive to negative as x increases through x0 i.e., f'(x) > 0.
  • x0 is a point of local minimum of f(x) if
    • f'(x) changes sign from negative to positive as x increases through x i.e., f'(x) < 0 at every point sufficiently close to x0 and the left of x and F (x) > 0 at every point sufficiently close to and to the right of x0.
  •  f'(x) does not change sign as x increases through x, then x0 is neither a point of local maximum nor a point of local minimum. Such a point is called a point of inflection.

Learn more about Inflection Point.

First Derivative Test for Maxima and Minima

Using the first derivative test we can find the local maxima and minima of any given function. Let’s understand maxima and minima first.

Maxima

Let be a real-valued function defined on an interval I. Then f(x) is said to have a maximum value in l at x0 if x0 ∈ l and f(x0) ≥ f (x) for all x in I. The value given for f(x0) is called the maximum value of f(x).

Minima

Let be a real-valued function defined on an interval I. Then f(x) is said to have a minimum value in I at x0 if x0 ∈ I and f(x0)<=F(x) for all x in I. The value given for f(x0) is called the minimum value of f(x).

Learn more about Maxima and Minima.

Inflection Point

For any function f(x), the point at which the second derivative is either 0 or doesn’t exist and the sign of the second derivative changes from either positive to negative or negative to positive is called the inflection point.

Read the article on Inflection Point to learn more about it.

Steps for First Derivative Test

We can use the following steps to perform the first derivative test.

Step 1: Find the first derivative of the given function.

Step 2: Equate the first derivative to 0, and solve for the dependent variable to find the critical points.

Step 3: With the help of critical points and the domain of the function, find the intervals for the positive and negative slopes of the graph of the function.

Step 4: Check the signs of the derivative around the critical point.

Step 5: Finalize the extremum based on the following cases:

  • If the sign of the first derivative changes from positive to negative as you move from left to right around the critical point, the function has a local maximum at that critical point.
  • If the first sign of the derivative changes from negative to positive as you move from left to right around the critical point, the function has a local minimum at that critical point.
  • If the first derivative does not change sign (either positive or negative) at a critical point, the first derivative test fails and needs to use further tests such as the second derivative test.

Read, More

How to Perform First Derivative Test?

To Perform the First Derivative Test, we can use the following steps:

Step 1: Find f'(x) for the function f(x).

Step 2: Put f'(x) = 0 and solve this equation and obtain different values of x say a, b, c….

Step 3: To test the point x = a, determine the sign of f'(x) for values of slightly less than and slightly greater than a, and see which of the following condition holds true:

  • If f'(x) changes sign from positive to negative, then x = a is the point of the local maximum, and f (a) is a local maximum value.
  • If f'(x) changes sign from negative to positive, then x = a is the points of the local minimum, and f (a) is a local minimum value.
  • If f'(x) does not change its sign, then x = a is a point of inflexion.

Step 4: Similarly, repeat the test for the points x = b, x = c, etc.

Let’s consider the following examples for better understanding.

Example of First Derivative Test

Example 1: Find the local maximum and local minimum value of the function f(x) = sin x – cos x; 0 < x < 2π, using the first derivative test.

Answer:

Given: f(x) = sin x – cos x

⇒ f’ (x) = cos x + sin x

For local maximum or local minimum, f'(x) = 0

⇒ cos x + sin x = 0

⇒ cos x =- sin x

⇒ tan x = – 1 

⇒ – tan x = -tan π/4

⇒ tan x = tan( π- π/4)  or tan(2 π- π/4) 

⇒ x= 3 π/4 or 7 π/4

At x=3 π /4,

For values of x slightly less than 3 π/4,f'(x) is +ve.

For values of x slightly greater than 3 π/4, f'(x) is – ve

Thus, f'(x) changes sign from positive to negative as x increases through 3π/4

Therefore, f(x) has a local maximum at x = 3π/4

Local maximum value = f(3π/4)

⇒ Local maximum value = sin3π/4 – cos3π/4

⇒ Local maximum value  = 1/√2+1/√2=√2

Now, at x=7π/4,

For values of x slightly less than 7 π/4,f'(x) is – ve.

For values of x slightly greater than 3 π/4, f'(x) is + ve

Thus, f'(x) changes sign from negative to positive as x increases through 7π/4

Therefore, f(x) has a Local minimum value at x= 7π/4.

Local minimum value = f(7π/4)

⇒ Local minimum value = sin7π/4 – cos7π/4

⇒ Local minimum value = -1/√2-1/√2= -√2 

Example 2: Find the local maximum and local minimum values of the constant function a.

Answer: 

Let f(x) = a

⇒  f'(x) = 0 for all x. [As a is constant a constant function]

Let c be any real number, then f’ (c) = 0

⇒ When x is slightly < c, f'(x) = 0

⇒ When x is slightly > c, f'(x) = 0

As, f'(x) does not change the sign at x = c. 

Thus c is neither a point of the local maximum nor a point of the local minimum.

Hence, f(x) has neither local maximum nor local minimum.

Example 3: Find the point of local maximum and local minimum for the function, f(x) = x2 -3x; using the first derivative test. Also, find the local maximum and local minimum values.

Answer:

Given: f(x) =x3 – 3x

⇒ f’ (x) = 3×2 -3 = 3(x2 – 1) = 3(x- 1)(x + 1)

Now,

⇒  f’ (x) = 0 = 3(x – 1) (x + 1) = 0 = either x = 1 or x = – 1.

Let us test the nature of the function at the points x = 1 and x=- 1.

At x = 1:

Let us take x = 0.9 to the left of x = 1 and x = 1.1 to the right of x = 1 

⇒ f’ (x) at these points.

⇒ f’ (0.9) = 3(0.9 – 1)(0.9 + 1) = – ve

⇒ f’ (1.1) = 3(1.1 – 1)(1.1 + 1) = + ve.

Thus f’ (x) changes sign from negative to positive as x increases through 1 and hence x= 1 is a point of local minimum.

⇒ Local minimum value = f (1) = (1)3 – 3(1) = – 2.

At x = – 1;

Let us take x = – 1.1 to the left of x = – 1 and x = – 0.9 to the right of x = – 1.

⇒ f’ (-1.1) = 3(- 1.1 – 1)(-1.1 + 1) = + ve f’ (-0.9)  ⇒ 3(- 0.9 – 1)(- 0.9 + 1) = – ve

Thus f’ (x) changes sign from positive to negative as × increases through – 1

Hence, x = – 1 is a point of local maximum.

⇒ Local maximum value = f(- 1) = (-1)3 – 3 (-1) = 2.

Example 4: Determine the local maximum and local minimum values for the following functions:

f(x) = x3-3x2-9x-7

Answer:

Given: f(x) = x3 – 3x2 – 9x – 7

⇒ f’ (x) = 3x2 – 6x – 9 

⇒ f’ (x) = 3 (x2 – 2x – 3)

Now,

⇒ f’ (x) = 0

⇒ 3(x2 – 2x -3) =0 

⇒ x2 – 2x -3 = 0 

 ⇒ (x-3)(x+ 1) = 0

⇒ either x = -1 or x = 3

Let us test the nature of the function at the points x = – 1, 3.

At x = – 1:

When x is slightly < – 1, f’ (x) = 3 (- ve) (- ve) 

⇒ f'(x) = + ve

and  ,when x is slightly > – 1, 

f’ (x) = 3 (- ve ) (+ ve) 

⇒ f'(x) = – ve

Thus f’ (x) changes sign from positive to negative as * increases through – 1.

f(x) has a local maximum at x = – 1

⇒  local maximum value = f(-1)

⇒ f(-1) = (-1)3-3 (-1)2– 9(-1) -7

⇒ f(-1) = – 1 – 3 + 9 – 7 =- 2.

At x = 3 :

When x is slightly < 3,  f’ (x)  = 3(- ve)  (+ ve)  ⇒  – ve

when x is slightly > 3,  f’ (x)   =  3 (+ ve) (+ ve) ⇒  + ve

Thus f’ (x) changes sign from negative to positive as x increases through 3.

⇒  f(x) has a local minimum at x = 3

⇒ Local minimum value = f (3) = (3)3 – 3 (3)2-9 (3) – 7

⇒  27 – 27 – 27 – 7 = – 34.

Example 5. Examine the function (x – 2)3 (x – 3)2 for local maximum and local minimum values. Also find the point of inflection, if any.

Answer:

Let f(x) = (x – 2)3 (x – 3)2

⇒ f'(x) = (x – 2)3. 2(x – 3) + (x – 3)2.3(x – 2)2

⇒  f'(x) = (x-2)2 (x – 3) (2x – 4 + 3x – 9) 

⇒  f'(x) = (x – 2)2(x – 3)(5x – 13)

For local maximum or minimum, f'(x) = 0

 ⇒ (x- 2)2 (x- 3) (5x – 13)=0 

 ⇒ x = 2, 3 or 13/5

At x = 2;

When x is slightly < 2, f'(x) = (+)(- )(-) = + ve

When x is slightly > 2, f'(x) = (+)(- )(- ) = + ve

Thus f'(x) does not change sign as x increases through 2.

⇒  Hence x = 2 is a point of inflection.

At x = 3 :

When x is slightly < 3, f(x) = (+ (- )(+ ) = – ve

When x is slightly > 3, f'(x) = (+)(+ )(+ ) = + ve

Thus f(x) changes sign from negative to positive as x increases through 3.

:. f(x) has a local minimum at x = 3

⇒ Local minimum value = f (3) = (3 – 2)3 (3 – 3)2= 0

At x =13/5

When x is slightly < 13/5 :’f(x)=(+)-) -) = +ve

When x is slightly > 13/5, f(x) = (+) (- )( +) = -ve

Thus f(x) changes sign from positive to negative as x increases through 13/5. 

⇒ f(x) has a local maximum at x = 13/5

⇒ Local maximum value = f (13/5) = (13/5-2)2 (13/5-2)3 (13/5-3) 

⇒  27/125*4/25= 108/3125

FAQs on First Derivative Test

1. What is First Derivative Test?

First Derivative Test is the method in calculus which help us analyze the behaviour of any function near the critical point.

2. How Does the First Derivative Test Work?

First Derivative Test uses the change in the sign of the derivative near the critical point such that by analysing the change in the sign of the derivative around the critical point, we can conclude that the function has either relative maxima, relative minima, or points of inflection at the critical point.

3. What are Critical Points?

Critical points are the points on a function where its derivative is either zero or undefined.

4. What happens if the Derivative is Zero at a Critical Point?

If derivative of any function became zero at the critical point, then the first derivative test become inconclusive in determining the relative maximum, relative minimum, or a point of inflection of the function. In such cases, we need further tests such as Second Derivative Test.

5. Can the First Derivative Test determine Absolute Extrema?

No, the First Derivative Test can only determine relative extrema. To find absolute extrema, you need to consider the function’s values at the critical points and endpoints within a specific interval.



Last Updated : 18 Sep, 2023
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