let fx 1x2 x 1 Then the range of f isSolutionfx1x2x1 y1x2

let f(x)= 1/(x^2 + x + 1) . Then the range of f is?

f(x)=1./(x^2+x+1)

y=1/x^2+x+1

multiplying both sides by x^2+x+1

y(x^2+x+1)=1

subtract 1 from both sides

yx^2+yx+y-1=0

It is in quadratic equation

we have to check out the discriminant which is b^2 - 4ac

y^2-4y(y-1)

-3y^2+4y

ANd to find the range we have to set the discriminant have to be greater than or equal to zero

-3y^2+4y>=0

y(-3y+4)>=0

0<=y<=4/3

0<=f(x)<=4/3

And that\'s the range