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.
downloads 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.
Illustration: 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:
FUNCTIONS |
INVERSE FUNCTION |
1. |
f:[0,∞)→[0,∞) defined by f(x)=x2 |
f-1:[0,∞)→[0,∞) defined by f-1(x) = √x |
2. |
f:[-∏/2,∏/2] →[-1,1] defined by f(x)=sin x |
f1 [-1,1]→[-(∏/2),∏/2] defined by f-1(x)=sin-1x |
3. |
f:[0,∏]→[-1,1] defined by f(x)=sinx |
f1:[-1,1]→[0,∏] defined by f1(x)=cos-1x |
4. |
f:[-∏/2,∏/2] →(-∞,∞) defined by f(x)=tan x |
f1:(-∞,∞)→[-(∏/2),∏/2] defined by f1(x)=tan-1 x |
5. |
f:(0,∏)→(-∞,∞) defined by f(x) = cot x |
f-1:(-∞, ∞)→(0,∏) defined by f-1(x)=cot-1 x |
6. |
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 |
7. |
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 |
8. |
f:R → R+ defined by f(x) = ex |
f-1(x):R+ → R defined by f-1 (x) = In x. |
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