Tuesday, October 6, 2015



Hello friends
So far we have been learning about function transformations in class, and about inverse of a function. 
This post is about how to find inverse of a function algebraically. Let us consider an example.


Example:Let f:RR    be defined by f(x)=3x+1  . Find the inverse function f 1   .


Solution: For any input x  , the function machine corresponding to f  spits out the value y=f(x)=3x+1  . We want to find the function f 1   that takes the value y  as an input and spits out x  as the output. In other words, y=f(x)  gives y  as a function of x  , and we want to find x=f 1 (y)  that will give us x  as a function of y  .
To calculate x  as a function of y  , we just take the expression y=3x+1  for y  as a function of x  and solve for x  .
yy1y13  =3x+1=3x=x  
Therefore, we found out that x=y/31/3  , so we can write the inverse function as
f 1 (y)=y3 13 . 

In the definition of the function f 1   , there's nothing special about using the variable y  . We could use any other variable, and write the answer as f 1 (x)=x/31/3  or f 1 ()=/31/3  . The placeholder variable used in the formula for a function doesn't matter.
To verify that f 1   is really the inverse of f  , we should show that the composition of f  and f 1   doesn't do anything to the input. In this case, the order shouldn't matter, and the functions ff 1   and f 1   f$ should both do nothing. Let's check this.
First, we apply f  followed by f 1   .
(f 1 f)(x) =f 1 (f(x))=f 1 (3x+1)=(3x+1)/31/3=x+1/31/3=x  
Second, we apply f 1   followed by f  .
(ff 1 )(x) =f(f 1 (x))=f(x/31/3)=3(x/31/3)+1=x1+1=x  
In both cases, applying both f  and f 1   to x  gave us back x  . Indeed, f 1 (x)=x/31/3  .


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