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INVERSE FUNCTION

Before we know about the definition of inverse function,we will suppose that a function F(x,y)=0 where function y=f(x) as the Vertical Line T (VLT) and also x=g(y) as the Horizontal Line T (HLT) are have the relation.

And we can solve the function by using function y=f(x) where x=g(x). If we looking for the function of y=x² and then x=g(y) is a horizontal line which intersects the graph in two points.

Let write y= 2x-1

and look at the graph of that function

So, in the x-intersect we find point of (½,0).

look at the line of y=x and we can substitute y=x into the function y = 2x - 1, so : x = 2x - 1

1 + x = 2x

1 = x

And , we have intersection point between line y = 2x – 1 and y = x, of course in point of (1,1).

Now, from these relation we want to solve the equation

2x – 1 = y

2x = y + 1

x = ½(y + 1)

x = ½y + ½

then exchange x into y,and y into x ,so obtain :

y = ½x + ½

We looking back on the graph then we get the other line. Let we graph a line containing of point (1,1) and (0,-1).

We have f(x) = 2x – 1 and g(x) = ½x + ½, so

f(g(x)) = 2(g(x)) – 1

= 2 (½x + ½) – 1

= x + 1 – 1

= x

On the other hand

g(f(x)) = ½(f(x)) + ½

= ½(2x – 1) + ½

= x - ½ + ½

= x

So the important problem of two functions is g = f^-1

f(g(x)) = f(f^-1(x))

= x

g(f(x)) = f^-1(f(x))

= x

The example :

Write the function of y = (x-1 )/(x+2)f

The x-intercept is gonna be equal to 1, and y-intercept is gonna be -1.

The solution is :

y = (x-1 )/(x+2)

y ( x + 2 ) = x - 1

yx + 2y = x – 1

yx – x = -1 – 2y

(y - 1) x = -1 – 2y

x^-1 = (-1-2y) / (y-1) ;

Then exchange x into y,and y into x;so we get:

y^-1 = (-1-2x) / (x-1)

Let; x = 0, y = -1

y = 0, x = -½

There are vertical asymtot x = 1 and horizontal asymtot y = -2

We can see from the figure of function y = (x-1 )/(x+1) and y^-1 = (-1-2x) / (x-1)that the two functions are reflected each other. So, the favorite function to look for the inverses are two function .