lf `f: RR rarr RR`, `f(x) = x^2,` then `{x| f(x) = - 1}` equals

If f: RR rarr RR, f(x) = {(1,; "when" x in QQ), (-1,;"when" x notin QQ):} then image set of RR under f is

If f: R rarr R, f(x) ={(1,; "when " x in QQ), (-1,; "when " x !in QQ):} then image set of RR under f is

Let RR be the set of real numbers and the mapping f: RR rarr RR be defined by f(x)=2x^(2) , then f^(-1) (32)=

Let RR be the set of real numbers and the mapping f: RR rarr RR and g: RR rarr RR be defined by f(x)=5-x^(2) and g(x) =3x -4 , state which of the following is the value of (f o g) (-1) ?

Let RR^(+) be the set of positive real numbers and f: RR rarr RR ^(+) be defined by f(x) =e^(x) . Show that, f is bijective and hence find f^(-1)(x)

Let the function f: RR rarr RR be defined by, f(x) =x^(2), . Find (i) f^(-1) (25) , (ii) f^(-1) (5) , (iii) f^(-1) (-5)

Let the functions f: RR rarr RR and g: RR rarr RR be defined by f(x)=x+1 and g(x)=x-1 Prove that , (g o f)=(f o g)=I_(RR)

function f and g are defined as follows: f:RR-{1} rarr RR, where f(x) =(x^(2)-1)/(x-1) and g: RR rarr RR g(x)=x+1, RR being the set of real numbers .Is f=g ? Give reasons for your answer.

Prove that the mapping f: RR rarr RR defined by , f(x)=x^(2)+1 for all x in RR is neither one-one nor onto.

Let RR be the set of real numbers . If the functions f:RR rarr RR and g: RR rarr RR be defined by , f(x)=3x+2 and g(x) =x^(2)+1 , then find ( g o f) and (f o g) .