Inverse of Function Solver

Introduction to inverse function solver

If  ' f ' is a relation from A to B , then the relation  { (b,a) : (a,b) `in` f } is denoted as f-1 .

If ' f ' is a relation from A to B , then  (f   -1)-1  =  f   .

Theorem : If   f : A `|->` B is one-one , then f -1 is bijection from f(A) to A .

Proof : Let  f : A `|->` B be one-one .

Clearly f  -1  is a relation from   f(A) to A .

Let  b `in` f(A)  . then there exists  a `in` A such that   f(a) = b . Since f is one-one , a is the only element of A such that  f(a) = b . Thus given b `in` f(A) there is a  unique element a in A such that  (a,b) `in` f . Hence given b `in` f(A) there is unique element a `in` A such that  (b,a) `in` f-1 . Hence f-1 is function  from f(A) to A  and  f-1 (b) = a if and only  if f(a) = b . Clearly f-1 is an onto function . If  b1 , b2 `in` f(A) and f-1(b1) = f-1(b2) = a  say , then b1 = f(a) = b2 . Thus f-1 is one to one . Therfore  f-1 : f(A) `|->` A is bijection .

Definition : If  f : A `|->` B  is a bijection then the relation f -1 : { (b,a) : (a,b) `in` f } is a function from B to A and is called the inverse function of f .

Let us see some problems in inverse function solver.

Examples on Inverse Function Solver

Problem: Find the inverse function of   f(x) = 4x +7 ?

Solution :   Given  f(x)  =  4x  + 7

Let   y  =  f(x)    `rArr`   x  =  f-1(y) .

`rArr`    y   =  4x   +   7  
`rArr`     x   =  `(y-7)/(4)`
`rArr`     f-1(y)   =  `(y-7)/(4)`

`:.`   f-1(x)  =  `(x-7)/(4)`

Answer :  f-1  =  `(x-7)/(4)` .

Problem: Find the inverse function of   f(x) = e4x+7  ?

Solution : Given  f(x) = e4x+7

Let    y  =  f(x)   `rArr`   x  =  f-1(y)

`rArr`    y   =  e4x+7  
`rArr`    log y =  4x + 7
`rArr`     x  =  `(logy-7)/(4)`    =  f-1(y)

f-1(x)  =  `(logx-7)/(4)` 

Answer : f-1(x)   =  `(logx-7)/(4)`

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Solved Problems on Inverse Function Solver

1) If A = { 1 , 2 , 3 } , B = { a , b , c } and  f = { (1,a) , (2,b) , (3,c) } , then find the inverse function of f ?

Solution : Given    A = { 1 , 2 , 3 } , B = { a , b , c }

Also given the function  f = { (1,a) , (2,b) , (3,c) }

Now , inverse of f  i.e. f-1  =  { (a,1) , (b,2) , (c,3) }

and f-1 : B `|->` A is also a bisection .

Answer :   f-1  =  { (a,1) , (b,2) , (c,3) }

2) If A = { 1 , 2 , 3 } , B = { a , b , c, d }  and g = { (1,b) , (2,a) , (3,c) } . Find the inverse function ?

Solution : Given        A = { 1 , 2 , 3 } , B = { a , b , c, d }  and g = { (1,b) , (2,a) , (3,c) }

Here g-1 does not exist .

g : A `|->` B  is one to one but not onto .  g-1 is bijection from { a, b c }  to A .