Question 481128: The one-to-one function is defined by
f(x) = x/(8-9x)
Find f^-1, the inverse of f . Then, give the domain and range of f^-1 using interval notation.
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The one-to-one function is defined by
f(x) = x/(8-9x)
Find f^-1, the inverse of f .
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1st: Interchange x and y to get:
x = y/(8-9y)
2nd: Solve for "y":
8x-9xy = y
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y+9xy = 8x
y(9x+1) = 8x
f^-1(x) = 8x/(9x+1)
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Domain: All Real Numbers except x = -1/9
Range: All Real Numbers except y = 8/9
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Cheers,
Stan H.
Then, give the domain and range of f^-1 using interval notation.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! let y = f(x) to get:
y = x/(8-9x)
make y = x and x = y to get:
x = y/(8-9y)
multiply both sides of this equation by (8-9y) to get:
x(8-9y) = y
perform the indicated operation to get:
8x - 9xy = y
add 9xy to both sides of this equation to get:
8x = 9xy + y
factor out the y to get:
8x = y(9x+1)
divide both sides of this equation by (9x+1) to get:
8x/(9x+1) = y
commute this equation to get:
y = 8x/(9x+1)
your original equation is:
y = x/(8-9x)
your inverse equation is:
y = 8x/(9x+1)
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if y = 8x/(1+9x) is an inverse equation of y = x/(8-9x), then the coordinate pairs of your original equation will be reversed from the coordinate pairs of your inverse equation.
this means that:
(x,y) = (a,b) in the original equation will become (x,y) = (b,a) in the inverse equation.
an example:
let x = 5 in your original equation.
solve the original equation to get y = -5/37.
your coordinate pairs from the original equation are (x,y) = (5,-5/37).
now let x = -5/37 in your inverse equation.
solve the inverse equation to get y = 5.
your coordinate pairs from the inverse equation are (x,y) = (-5/37,5).
since x,y = (a,b) in your original equation and x,y = (b,a) in you inverse equation, then this is a confirmation that your inverse equation has been calculated correctly.
graph both equations.
if you graph your original equation and your inverse equation, they should look like a reflection about the line y = x.
the following graph shows both equations and also shows the equation of y = x.
you can see that the graph of the original equation and the graph of the inverse equation look like reflections about the line y = x.
note that the original equation appears to have an asymptote at somewhere around x = 1 and that the inverse equation appears to have an asymptote at somewhere around y = 1.
the asymptote for the original equation is really at x = 8/9.
when that happens, the original equation of y = x/(8-9x) gets a denominator of 0 which means that the value of y is undefined which means there is a vertical asymptote at x = 8/9.
the vertical asymptote for the original equation should become the horizontal asymptote for the inverse equation.
this means that the inverse equation should have a horizontal asymptote at y = (8/9).
the easiest way to find the horizontal asymptote is to solve the inverse equation for x.
when you do that, you get back to the original equation, with the exception that x = y and y = x.
you wind up with x = y/(8-9y) which leads to an asymptote of y = 8/9.
the domain of your original equation is equal to all real values of x except x = 8/9.
the range of your original equation is all real values of y.
the domain of your inverse equation is equal to all real values of x.
the range of your inverse equation is equal to all real values of y except y = 8/9.
in the next graph, a vertical line was drawn at x = 8/9 (as best i can produce it using the algebra.com graphing software) and a horizontal line was drawn at y = 8/9, to show you where the vertical and horizontal asymptotes are.
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