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Video 4
INVERSE FUNCTION

INVERSE FUNTION
Before we start study the Inverse Function, better we review the definition of a function.
We have relation of F(x,y), this represent to function y=f(x) if we can solve for y uniquely in term of x as 1-1 function if also we can solve x uniquely in term of y.
If we see y=x2, it is a function but not 1-1 function, because if we choose a value of y, we have two value of x. 1-1 function only mapping a x into a y.
So lets start at y = 2x-1
Let’s look at the graph function
y intercepts -1 , x intersects ½, the slope =m= 2
We have a point (1/2,0) and (0, -1) . To draw this line, find (1/2,0) and (0, -1) first, then draw a line that across the points.
Then attend it
y = x intersects y = 2x-1
x = 2x-1 (add 1 in both of side)
1+x = 2x (move the x into right side)
1 = x
Substitute x = 1 in the equation y = x and y = 2x-1, we get y =1 (x =1 satisfying the both equations because x is intersect of y= x and y = 2x-1
Of course, the point of intersection of y= x and y = 2x-1 is (1,1)
We want to solve for x in term of y
2x-1 = y
2x = y+1
x = ½ (y+1)
x = ½ y + ½
Then in this equation, I want to give exchange rule of x and y, so I’m going to do now is rewrite x into y, and y into x,
We get y = ½ x + ½ ……………..(**)
The x intercepts in -1, and y intercepts in ½
We have point (0,1/2) and (-1,0)
So the inverse function to given function passes through that point (1,1)
We have f(x) = 2x-1
We have g(x) = ½ x + ½
Then take the value of g and plug in for f
We want to compute f(g(x))
f(g(x)) = 2 […] – 1 , then fill the bracket with ½ x + ½
= 2 [½ x + ½] – 1= x + 1 -1= x
in the other hand, we take g(f(x))
g(f(x)) = ½ […..] + ½ = ½ [2x-1] +½ = x-½ +½ = x
Then we write
g = f -1
f(g(x)) = f(f -1 (x)) = x and
g(f(x)) = f -1 (f(x)) = x
Other example. Let’s take the function y =(x-1)/(x+2). This function has a vertical asymptote as line x = -2 and also has horizontal asymptote as a line y = 1
x intercept is approximately 1and y intercept is approximately -½
The function hyperbole
Then let’s take the equation y =(x-1)/(x+2) and solve for x in term of y
y(x+2) = (x-1)
yx-2y = x-1 (take 2y into the right side, move x to the left side)
yx-x =-1-2y
(y-1)x =-1-2y
X = (-1-2y)/(y-1)
we are going to exchange the rule of y and x
y= (-1-2x)/(x-1)
at x = 0 y = -1
at y = 0 -1-2x = 0
-1 = 2x or x =-½
y= (-1-2x)/(x-1) has vertical asymptote as a line x = 1 and also horizontal asymptote as a line y = -2
We have
f(x) = (x-1)/(x+2)
f -1(x) = (-1-2x)/(x-1)
Then
f(f -1 (x)) = (((-1-2x)/(x-1))-1)/(( (-1-2x)/(x-1))+2)
= ((((-1-2x)-(x-1))/(x-1)))/((((-1-2x)+2(x-1)/(x-1)))
= (-1-2x-x+1)/(-1-2x+2x-2)
= -3x/-3
= x
Then find f -1(f(x))
f -1(f(x)) =((-1-2((x-1)/(x+2)))/((x-1)/(x+2) – 1))
= (-1(x+2)-2((x-1)/(x+2))/((x-1)- 1(x+2)/(x+2))
= ((-1x-2-2x+2)/(x+2))/((x-1- x-2)/(x+2)))
= (-1x-2-2x+2)/(x-1- x-2)
= -3x/-3
= x
We find that
f(g(x)) = x
g(f(x)) = x
Range of g is a domain of f, and the range of f is a domain of g
The two function are inverses function.
we just write g = f -1