Introduction to First derivative test

In calculus, the first derivative test is a fundamental derivative concept and is important in analyzing a function over some domain. This test calculates the rate of change of a function by finding its critical points. It has many applications in calculus and physics where the rate of change is involved. Let’s understand how the 1st derivative test for extreme points is used. 

Understanding the First Derivative Test

The first derivative test is a method to calculate the local maximum and minimum points of a function. It helps to determine whether the function changes from increasing to decreasing or vice versa. Since this concept is based on a function's rate of change, the first derivative is used. The first derivative of a function is calculated by differentiating the function one time. By definition, the first derivative of a function f(x) is defined as;

$f’(x)=\lim_{h→0}\frac{f(x+h)-f(x)}{h}$

Although a function can be monotonically analyzed without using the first derivative, calculus is a subject that helps to analyze a function easily. A function is differentiated to find its critical points. The first derivative test helps to identify whether these critical points are local maximum, local minimum or saddle points. Before understanding the formula of the first derivative test, let’s discuss what critical points are.

What are critical points?

The critical points are the points on the graph of a function where the function’s derivative is zero or does not exist. These points are important to consider to determine local maximum and minimum value of a function. Critical point of a function is defined as;

“The point x of a function f(x) is called critical point if f’(x) = 0 or f’(x) > 0.”

What are local maximum and minimum values?

The local maximum and minimum values are the points on the graph of a function where the function takes maximum and minimum values. A point x is known as local maximum if f(x)>0 and is a local minimum if f(x) < 0. These points are used in the first derivative and second derivative test.

First derivative test formula

The first derivative test is used to find the behaviour of a function on the points in its domain. If a function f(x) is defined on an open interval I and there is a point c in the domain of f(x) then the first derivative formula is,

• If f’(x) > 0, i.e. f’(x) changes from positive to negative, then f(c) is the local maximum point of f(x).
• If f’(x) < 0 i.e. f’(x) changes from negative to positive, then f(c) is the local minimum point of f(x).
• If f’(x) = 0 i.e. f’(x) does not change. Then f(c) is neither local maximum or minimum.

The above statements are key points for the first derivative test.

How to apply the first derivative test?

The 1st derivative test can be applied by finding the derivative of the given function. You can also follow the first derivative test steps to simplify calculations. These steps are:

1. Write the function and identify the independent variable. 
2. Calculate the first derivative of f(x), i.e., f’(x) and use the relevant rules according to the type of function. For example, you can use the product rule if there is a product of two functions depending on the same variable.
3. Equate f’(x) to zero to identify the critical points.
4. Analyze the interval where the function is increasing or decreasing and determine the local maximum and minimum points. 

Let’s understand the implementation of the first derivative test in the following examples.

First Derivative Test Example 

Find the local maximum and local minimum values by using 1st derivative test for the function, 

$f(x)=3x^4+4x^3-12x^2+12$

Evaluating derivative with respect to x.

$f'(x)=\frac{d}{dx}[3x^4+4x^3 –12x^2+12]$

Since the function involves power functions, so by using the power rule of derivative,

$f’(x) = 12x^3+12x^2 –24x$

Now to find critical points, substitute f’(x) = 0.

$12x^3+12x^2 –24x = 0$

Simplifying,

$12x(x^2+x – 2) = 0$

Factorizing, 

$12x[x^2+2x – x – 2]=0$

Or, 

$12x[x(x+2) – 1(x+2)] = 0$

$12x(x+2)(x – 1) = 0$

The critical points are 0, -2 and 1 which means that f’(x) is 0 at 0, -2 and 1. Now we have to determine the local maxima and minima. For this, first substitute x=-2 in the given function.

$f(-2) = 3(-2)^4 + 4(-2)^3 – 12(-2)^2 + 12$

$f(-2)= 48 – 32 – 48 + 12$

$f(-2)= -20$

$\text{Minimum point} = (-2, -20)$

Now substituting x = 0 in f(x),

$f(0) = 3(0)^4 + 4(0)^3 – 12(0)^2 + 12 = 12$

$\text{Maximum point} = (0, 12)$

Substituting $x = 1$ in f(x),

$f(1) = 3(1)^4 + 4(1)^3 – 12(1)^2 + 1$

$f(1)= 3 + 4 – 12 + 12$

$f(1)= 7$

$\text{Minimum point} = (1, 7)$

Hence, the local extreme points are:

$\text{Local minimum} = (-2, -20)\quad\text{and}\quad(1, 7)$

$\text{Local maxima} = (0, 12)$

Comparison between first derivative and second derivative test

The comparison between the higher derivative and second derivative test can be easily analysed using the following difference table.

Conclusion

The first derivative test is a method to identify the change in a function, whether it is changing from positive to negative or negative to positive. The critical points of the function are used to apply this test. The function then substitutes these points to calculate local maxima and minima. Hence this test has many applications in calculus and physics because it allows us to find the change in a function at a specific point.