0

Guest

Courses New

find the minimum value of f(x)=(x²-x+1)÷(x²+x+1)

find the minimum value of f(x)=(x²-x+1)÷(x²+x+1)

Grade:12

1 Answers

Sunil Raikwar
askIITians Faculty 45 Points
10 years ago
y=(x^2-x+1)/(x^2+x+1)
yx^2+yx+y=x^2-x+1
(y-1)x^2+(y+1)x+y-1=0
since x belong to R
Therefore, Discriminant>=0
(y+1)^2-4(y-1)(y-1)>=0
y^2+2y+1-4y^2+8y-4>=0
-3y^2+10y-3>=0
3y^2-10y+3<=0
(y-3)(3y-1)<=0
y belong to 1/3 to 3 so minimum value is 1/3

Thanks & Regards
Sunil Raikwar
askIITians faculty



Think You Can Provide A Better Answer ?

Other Related Questions on Integral Calculus

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More