The First Derivative Test is a fundamental technique in calculus used to analyze critical points of a function and determine whether the function attains a relative maximum or minimum at those points. It works by examining how the slope of the function changes sign around a critical point. Since the slope on either side of an extremum can be positive or negative, studying this change allows us to identify the nature of the extremum.
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.
1) 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.
2) 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 to the left of x and F (x) > 0 at every point sufficiently close to and to the right of x0.
3) 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.
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).
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.
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 sign of the first 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.
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.
Related Articles
Example of First Derivative Test
Example 1: 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 2: 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 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) = 3x2 -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
Problem practice: First Derivative
- Problem 1: Find the derivative of f(x) = 4x5 - 3x3 + 2x - 7.
- Problem 2: Find the derivative of k(x) = e2x.
- Problem 3: Determine the derivative of g(x) = sin(x)⋅cos(x).
- Problem 4: Find the derivative of
r(x) = x \cdot e^x . - Problem 5: Compute the derivative of h(x) = ln(x2+1).
- Problem 6: Compute the derivative of
q(x) = \sqrt{4 - x^2} . - Problem 7: Determine the derivative of
p(x) = \frac{3x^2 - 5}{x + 2} . - Problem 8: Determine the derivative of
s(x) = \frac{\ln(x)}{x} .