`f:[-3pi,2pi]rarr[-1,1]` defined by `f(x)=sin 3x.`

If f:[(pi)/(2),(3 pi)/(2)]rarr[-1,1] and is defined by f(x)=sin x,thenf^(-1)(x) is given by

The function f : [0,2 pi] to 1 [-1,1] defined by f(x) = sin x is

Let f : [-1, -1/2] rarr [-1, 1] is defined by f(x)=4x^(3)-3x , then f^(-1) (x) is

Let f : [-1, -1/2] rarr [-1, 1] is defined by f(x)=4x^(3)-3x , then f^(-1) (x) is

let f:[-pi/3,pi/6]rarr B defined by f(x)=2cos^2x+sqrt3sin2x+1 . Find B such that f^-1 exists. Also find f^-1

let f:[-(pi)/(3),(pi)/(6)]rarr B defined by f(x)=2cos^(2)x+sqrt(3)sin2x+1. Find B such that f^(-1) exists.Also find f^(-1)

Let f:[-1,1]rarr[-(pi)/(2),(pi)/(2)] is defined as f(x)=sin^(-1)x and g:R rarr[-1,1] is defined as g(x)=sin x. if D_(1) denotes the domain of f(x), then which of the following is correct?

Let f : [-(pi)/(2), (pi)/(2)] rarr [3, 11] defined as f(x) = sin^(2)x + 4 sin x + 6 . Show that f is bijective function.