Let `A={x inR: x >=1/2} and B={x in R: x>=3/4}.` If `f:A->B` is defined as `f(x)=x^2-x=1,` then the solution set of the equation `f(x)=f^-1(x)` is

Let f(x)=2x+1. AA x , then the solution of the equation f(x)=f^(-1)(x) is

Let f(x)=2x+1. AA x in R , then the solution of the equation f(x)=f^(-1)(x) is

Let f(x)-2x-x^(2),x<=1 then the roots of the equation f(x)=f^(-1)(x) is

Let f(x)=x^(2)- x+1, x ge 1//2 , then the solution of the equation f^(-1)(x)=f(x) is

Let f(x)=x^(2)-x+1, AA x ge (1)/(2) , then the solution of the equation f(x)=f^(-1)(x) is

Let f(x)=x^(2)-x+1, AA x ge (1)/(2) , then the solution of the equation f(x)=f^(-1)(x) is

Let A={x in R:x<=1} and f:A rarr A be defined as f(x)=x(2-x). Then ,f^(-1)(x) is

A function f: [3/2, oo) to [ 7/4, oo) defined as, f(x) = x^2 - 3x +4. Then compute f^(-1)(x) and find the solution of the equation, f(x) = f^(-1) (x) .

Let f:RtoR be defined by f(x)=x/(1+x^2),x inR . Then the range of f is