Maxima and minima of a function are very important concepts which are applicable both for Mathematics and Science. This poses an obvious question, how to find them?
To start with it, let us see the first method i.e. first derivative test. This method is based on the basic concept of increasing and decreasing functions. From the definition of the function, we can determine the critical points by \(f’ = 0\)
In fig. 1, we can see that point \( P \)
Let us assume a function which has a critical point at \( P(x = a) \)
 If \( f’ \)
is negative towards just left of \( a \) and it is positive towards just right of \( a \) , then the point\( P(x = a)\) is a local minimum for the function \( f\) .  If \( f’ \)
is positive towards just left of \( a \) and it is negative towards just right of \( a \) , then the point \( P(x = a)\) is a local maximum for the function \( f\) .  If \( f’ \)
has same sign towards just left and just right of \( a \) , then the point \( P(x = a)\) is neither a local maximum nor a local minimum for the function \( f\) .
To summarize what we have learned, let us put it in a tabular form.
Table 1: Different cases possible for first derivative method
Pictorial representation of f’ towards left and right of point P(In all cases, f'(a) = 0) 
Nature of f’ Towards just left 
Towards just right 
Whether Local maximum or minimum 
Negative 
Positive 
Local Minimum 

Positive 
Positive 
Neither 

Positive 
Negative 
Local Maximum 
Let us take an example to understand the application of this test. We will try to find out all the critical points for the function and tell whether they are local maxima or minima.
\(f(x) = 0.2 x^5 + 1.25x^4 + 2x^3 + 2015^{2016}\)
Since it is a polynomial function, it is both continuous and differentiable in its entire domain.
\(f(x) = 0.2 x^5 + 1.25x^4 + 2x^3 + 2015^{2016}\)
Differentiating both sides w. r. t. , we get:
\(f'(x) = x^4 + 5x^3 + 6x^2 + 0\)
\(f'(x) = x^2(x^2 + 5x + 6)\)
\(f'(x) = x^2(x+3)(x+2)\)
For getting the critical points, we equate to 0,
\(f'(x) = 0\)
\(x^2(x+3)(x+2) = 0\)
\(x = 0, 2, 3\)
So, our critical points are 0, 2 and 3.
From equation 1,
\(f'(2.5) < 0\)
The sign of f ‘ is changing from negative to positive as we vary from left to right of . This means that is local minimum.
\(f'(3.5) > 0\)
\(f'(2.5) < 0\)
\(f'(1) > 0\)
The sign of \( f’ \)
So, there you go! That was the first derivative test. But as it turns out, you can’t apply this for every function. For example: \(f(x) = x^2 + 12 sin x\)
For those functions, we use a different approach that is called the second derivative test. For more information, visit BYJU’S and learn new concepts every day.
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