Let `f(x) =(x+1)^2-1,x ge -1` then
`{x: f(x) =f^(-1)(x) }` =

Let f(x)=(x+1)^(2)-1, x ge -1 . Then the set {x:f(x)=f^(-1)(x)} is

Let f(x) = (x+1)^2 -1, x ge -1 ). Then the set {x: f(x) = f^(-1) (x)} is

If f(x)=(x+1)^(2)-1, x ge -1 then the set S={x, f(x)=f^(-1)x} is

Let f(x) = (x+2)^(2) - 4, x ge 2 . Let S = {x : f(x) =f^(-1)(x)} , Then S is equals to

Let f: R rarr R be given by f(x)=(x+1)^(2)-1, x ge-1 . Then f^(-1)(x)=

Let f:[-1,oo] in [-1,oo] be a function given f(x)=(x+1)^(2)-1, x ge -1 Statement-1: The set [x:f(x)=f^(-1)(x)]={0,1} Statement-2: f is a bijection.

Let f:[-1,oo] in [-1,oo] be a function given f(x)=(x+1)^(2)-1, x ge -1 Statement-1: The set [x:f(x)=f^(-1)(x)]={0,1} Statement-2: f is a bijection.

Let f:[-1,oo] in [-1,oo] be a function given f(x)=(x+1)^(2)-1, x ge -1 Statement-1: The set [x:f(x)=f^(-1)(x)]={0,1} Statement-2: f is a bijection.