" (iv) "f:R rarr R:int(x)=e^(x)

If R is the set of real numbers prove that a function f:R rarr R,f(x)=e^(x),x in R is one to one mapping.

Which of the following function is one-one and onto both? f:R rarr R,f(x)=e^(x)

consider f:R rarr R,f(x)=e^(x-1)+x^(3)+x and g be the inverse of f

Which one of the following function is suriective but not injective? (A) f:R rarr R,f(x)=x^(3)+x+1(B)f:R,oo)rarr(0,1],f(x)=e^(|x|)(C)f:R rarr R,f(x)=x^(3)+2x^(2)-x+1(D)f:R rarr R+,f(x)=sqrt(1+x^(2))

Function f:R rarr R;f(x)=(e^(x^(2))-e^(-x^(2)))/(e^(x^(2))+e^(-x^(2))) is :

Which of the following function is surjective but not injective.(a) f:R rarr R,f(x)=x^(4)+2x^(3)-x^(2)+1(b)f:R rarr R,f(x)=x^(2)+x+1(c)f:R rarr R^(+),f(x)=sqrt(x^(2)+1)(d)f:R rarr R,f(x)=x^(3)+2x^(2)-x+1

Discuss the surjectivity of f:R rarr R given by f(x)=x^(3)+2 for all x in R

Let f:R rarr R defined by f(x)=(e^(x^(2))-e^(-x^(2)))/(e^(x^(2))+e^(-x^(2))), then f(x) is

The function f:R rarr R, f(x)=x^(2) is

The function f:R rarr R, f(x)=x^(2) is