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Inverse Function

Let f : X → Y be a function defined by y = f(x) such that f is both one - one and onto. Then there exists a unique function g : Y → X such that for each y ε Y,

g(y) = x <=> y =  f(x). The function g so defined is called the inverse of f.

Further, if g is the inverse of f, then f is the inverse of g and the two functions f and g are said to be the inverses of each other. For the inverse of a function to exists, the function must be on-one and onto.

Method to Find Inverse of a Function

If f-1 be the inverse of f, then fof-1 = f-1 of = I, where I is an identity function.

fof-1 = I => (fof-1(x)) = I (x) = x.

Apply the formula of f on f-1 (x), we will get an equation in f-1 (x) and x.

Solve it to get f-1 (x).

Note : A function and its inverse are always symmetric with respect to the line y = x.

Let f : R → R defined by f(x) = (ex-e-x)/2 . Find f-1 (x).

Solution: We have f(f-1(x)) = x

        =>  (ef-1(x) - e-f-1(x))/2 = x

        =>  e2f-1(x) - 2xef-1(x)  -1 = 0

        =>  ef-1(x)  = x + √(x2 +1).

But negative sign is not possible because L.H.S. is always positive.

Thus ef-1(x) = x + √(x2 +1) . Hence, f-1(x) = log(x + √(x2 +1)) .

We give below some standard functions along with their inverse functions:

Inverse Function




f:[0,∞)→[0,∞) defined by f(x)=x2

f-1:[0,∞)→[0,∞) defined by f-1(x) = √x


f:[-∏/2,∏/2] →[-1,1] defined by f(x)=sin x

f1 [-1,1]→[-(∏/2),∏/2]  defined by f-1(x)=sin-1x


f:[0,∏]→[-1,1] defined by f(x)=sinx

f1:[-1,1]→[0,∏] defined by f1(x)=cos-1x


f:[-∏/2,∏/2] →(-∞,∞) defined by f(x)=tan x

f1:(-∞,∞)→[-(∏/2),∏/2] defined by f1(x)=tan-1 x


f:(0,∏)→(-∞,∞) defined by f(x) = cot x

f-1:(-∞, ∞)→(0,∏) defined by f-1(x)=cot-1 x


f:[0,∏/2)U(n/2,n]→(-∞, -1]U[1,∞) defined by f(x) = sec x

f-1:(-∞,-1]U[1,∞) →[0,∏/2)U(∏/2,∏]  defined by f-1 (x) = sec-1 x


f:[-(∏/2),0)(0,n/2]→(-∞,-1]U[1,∞) defined by f(x) = cosec x

f-1:(-∞,-1]U[1,∞) →[0,-(∏/2))U(0,∏/2]  defined by f-1 (x) = cosec-1 x


f:R → R+ defined by f(x) = ex

f-1(x):R+ → R defined by f-1 (x) = In x.

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