5.Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(iv) f (x) = ( x-1) ^2 + 3 , x \epsilon [ -3 , 1 ]

Answers (1)

Given function is 
f(x) = (x-1)^2+3
f^{'}(x) =2(x-1) \\ f^{'}(x)= 0\\ 2(x-1)= 0\\ x=1         
Hence, x = 1 is the critical point  of function f(x) = (x-1)^2+3
Now, we need to check the value of function f(x) = (x-1)^2+3  at  x = 1  and at the end points of given range  i.e. at x = -3 and x =  1
f(1) = (1-1)^2+3 = 0^2+3 = 3
            
f(-3) = (-3-1)^2+3= (-4)^2+3 = 16+3= 19
f(1) = (1-1)^2+3 = 0^2+3 = 3
Hence,  absolute maximum value of function  f(x) = (x-1)^2+3  occurs at x = -3 and value is 19
and absolute minimum value of function f(x) = (x-1)^2+3  occurs at x = 1 and value is 3

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Easy Installments)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions