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.

Solution:

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.

Solution:

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.

Solution:

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

Solution:

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.

Solution:

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

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.