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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » MAXIMA FUNCTION OF ONE VARIABLE
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MAXIMA FUNCTION OF ONE VARIABLE

In this article we will study the critical point , critical number which is also known as stationary point with examples.

Table of Content
  •  

INTRODUCTION

In mathematics, the maxima and minima of a function is called extrema. The highest point within the given range on the curve is called maxima point.The lowest point within the given range on the curves is called minima point. The extremum is found using the calculus of variation  if an extremum consists of its own function. Maxima and minima can also be seen in sets but for now we will study it only for calculus.

CRITICAL POINT

The  critical points for a function f(x) are those points of the domain where either f’(x) = 0 or doesn’t exist. At a critical point the value of function is a critical value.

For eg: f(x) = x2+8x+1

              f’(x)=2x+8

              F’(x)=0

              2x+8  = 0

              x = -4 are critical point of a function

CRITICAL NUMBERS OR STATIONARY POINT

The critical numbers of a function are those at which its first derivative is equal to 0.

Or we can  call critical number stationary points which also says the same that the stationary points of a function are those points where the derivative of the function) is zero.

For eg: x3 − 6x2 + 9x

F’(x3 − 6x2 + 9x) =0

3x2-12x+9=0

3(x − 3)(x − 1)=0

X=3,1

Critical number or stationary point(3, −2) and (1, 2).

 

MAXIMA POINTS AND MINIMA POINTS

The highest point within the given range on the curve is called maxima point.

The lowest point within the given range on the curves is called minima point.

The local maximum: function changes from positive to zero to negative. The function  is  known as decreasing.

The local minimum function changes from negative, to zero, to positive. The function is increasing.

radio

Let’s take one example

Q1: find relative  min and max for f(x) =  x3-3x2-9x+1?

Ans:f(x) = 3x2 -6x -9

               = 3(x2-2x-3) = 0

                  x2-3x+x-3 = 0

                  (x+1)(x-3) = 0

                   x=-1,3

SECOND ORDER DERIVATIVE TEST FOR LOCAL EXTREMUM

  1. If n is odd , f  has neither a local max. nor local min at x=xo.
  2. If n is even , f has local max if f(xo)<0 , f has local min if f(xo)>0.

Let’s take example of this

Ques1:- find local max and local min?

Ans:-f(x)=2x3-15x2+36x+18

         F’(x)=6x2-30x+36 = 6(x2-5x+6) = 0

              x2-5x+6 = 0

            x(x-2)-3(x-2)= 0

            (x-3)(x-2)= 0

              x= 3,2

         f”(x) = 12x-30

         f”(2) = 24-30 = -6<0

         f”(3) = 36-30 = 6>0

       local max at x=2

       local min at x=3

CONCLUSION

In this article we study the critical point , critical number which is also known as stationary point with examples. We will also talk about the first order derivatives test for relative local extrema and second derivatives test for  relative local extrema.

In mathematics, the maxima and minima of a function is called extrema. The highest point within the given range on the curve is called maxima point.The lowest point within the given range on the curves is called minima point.

The extremum is found using the calculus of variation  if an extremum consists of its own function. Maxima and minima can also be seen in sets but for now we will study it only for calculus.

Further we will do some questions related to maxima and minima.

 

faq

Frequently asked questions

Get answers to the most common queries related to the IIT JEE Examination Preparation.

Find the increasing and decreasing of function ex+e-x?

Ans: f(x) = ex+e-x f’(x) = ex-e-x f’(x) = 0 ex-e...Read full

Find the stationary point of the function y = x2− 2x + 3 and hence determine the nature of this point?

Ans: y = x2 − 2x + 3 dy/dx = 2x − 2 d2y/ dx2 = 2. The function has only ...Read full

Find the local max. and local min. Of function f(x) = x^3-6x^2+9x+15?

Ans: f’(x) = f’(x) = 0 so , = 0 3( X2-4x+3 = 0 x(x-3)-1(x-3) = 0 x = 1,3...Read full

Ans: f(x) = ex+e-x

f’(x) = ex-e-x

f’(x) = 0

ex-e-x = 0

e-x(e2x-1) = 0

e2x-1 = 0

e2x = 1

taking log both sides

2x = log(1)

2x = 0

x =0

increasing at (0,)

decreasing at (,0)

Ans: y = x2 − 2x + 3

dy/dx = 2x − 2

d2y/ dx2 = 2.

The function has only one stationary point when x = 1 (and y = 2). d2y/ dx2 = 2 > 0

a stationary point is a local minimum.

therefore the function y = x2 − 2x + 3 has a local minimum at the point (1, 2).

Ans: f’(x) =

f’(x) = 0

so , = 0

3(

X2-4x+3 = 0

x(x-3)-1(x-3) = 0

x = 1,3

f”(x) = 6x -12

f”(1) = 6-12 =-6<0

f”(3) = 18-12 = 6>0

local min at 3

local max at 1.

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  • Fehling’s solution
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