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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » Minima of functions

Minima of functions

In this article we will learn about minima of the function, definition of minima of the function, first order and second order derivative test of minima of the function.

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A function is said to have an extreme value, which can be maxima or minima. The maximum value of the function in the given range is known as maxima, and the minimum value of the function in the given range is known as minima of the function. The value of local minima in the function lies in the particular interval.

What is Minima of the Function?

Minima of the function is the minimum value of the function within a given range. It is the value of the function at a point which is greater than the value of the function at that particular point. The minima of any function exists in the form of local minima and absolute( or global) minima. Let us see the definition of minimum value of the function, local minima of the function, and absolute (or global) minima of the function.

Minimum value of the function –  Let a function f, a point C exists in the interval L  such that f(c) ≤ f(x) for all x £ L.

Local minima –  Let f be a function having an interior point c  within the domain of f, such that c will be the local minima if f(c) ≤ f(x) for all x in (c – h, c + h), where h is a very small number which approaches to zero but not equal to zero 

Absolute(or global minima) – There can be only one global minima within the entire range of the domain. For a function f, there exists a point x= c if f(x)  ≥ f(c)  for all x £ L  (L is the domain of the function f(x)) The point c is known as the global minima of the function.

How to find Minima of the Function?

The minima of the function can be found using the derivative test. We can use the first derivative test and second derivative test to find the minima of the function. Let us understand the derivative tests in detail.

First order derivative test 

As we move towards the minimum point of the function, the slope of the function decreases and becomes zero at the minimum point. It increases when we move away from the minimum point. The slope of the function is given by its derivative, and we will find the local minima through the first derivative test. In the first derivative test, first we differentiate the function f(x) to find f’(x) and as we know that at a critical point c, f’( c) = 0. If the sign of f’(x) changes from negative to positive which indicates x increases through point c, then we can say that c is the point of local minima and by finding f(c) we will get the minimum value of the function.

Second order derivative test

In the second order derivative test, we will first find the slope of the function, that is the first derivative of the function and then we will find the second derivative of the function to check whether the function exists in the given range. 

For a function f(x), the slope at a critical point ‘c’ is given by f’(x) ( by putting f’(x) = 0 ). Then we will find the second derivative of f(x) and if f’’(x) > 0 then ‘c’ is the point of local minima and the value of the function corresponding to ‘ x = c’ (say f(c) ) is the minimum value of the function.

Solved Examples 

  • Calculate the minima of the function, f(x) = x3 + x2 using the first derivative test.

Solution. First we will find the  first order derivative of the function which is given by, 

f’(x) = 3×2 + 2x

taking f’(x) = 0

3×2+ 2x = 0 

x(3x+2) = 0

  • x = 0 and x = -2/3

substituting the values in the function, we will get f(0) = 0 

and f(-2/3) = (-23)3 + (-23)2   = 427

hence, the points are  (0,0) and (-2/3,4/27).

Now, we will check the sign for f’(x) at x = 0 and x = 4/27 

At x = 0, the f’(x) sign change from positive to negative, hence, 0 is the point of local minima 

At x = 4/27 the f’(x) sign changes from negative to positive, hence, 4/27 is the point of local maxima.

  • Calculate the minima of the function, f(x) = x2 using the second order  derivative test.

Solution. First, find the first derivative of the function 

f’(x)  = 2x

putting  f’(x) = 0,

2x= 0

=>    x = 0

x = 0 is the critical point 

Now taking second derivative  we will get, f’’(x) = 2

Since f’’(0) = 2 

And f’’(0) > 0

Hence, x = 0 is the point of local minima.

Conclusion

In this article, we have learned about the minima of the function, which is the minimum value of the function. The minima of the function exist in two forms, the absolute(or global) value and the local minima value. The absolute value is the only value that exists in the entire domain, whereas the local minima value is the value of the function at which the slope of the function is equal to zero. We use first and second-order derivative tests to find the local minima of the function. 

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