So you want to find f^(-1)(x) if
f(x) = (5x - 1) / (7x + 3)
Your first step is to make f(x) into y. That is
y = (5x - 1) / (7x + 3)
Now, swap the x and y terms, and then solve for y.
x = (5y - 1) / (7y + 3)
Multiply both sides by (7y + 3),
x(7y + 3) = 5y - 1
Expand the left hand side,
7xy + 3x = 5y - 1
Now, move everything with a y in it to the left hand side; everything else goes to the right hand side.
7xy - 5y = -3x - 1
Factor out a y on the left hand side to obtain
y(7x - 5) = -3x - 1
Now, divide both sides by (7x - 5).
y = (-3x - 1) / (7x - 5)
Just to clean this up a bit, those two negative signs on top look ugly. Let's factor out a (-1) on the top, and factor out a (-1) on the bottom. This will swap the terms on the bottom.
y = [(-1)(3x + 1)] / [(-1)(5 - 7x)]
y = (3x + 1) / (5 - 7x)
Now that you've solved for y, this will be your inverse, and you state:
f^(-1)(x) = (3x + 1) / (5 - 7x)
2007-02-06 16:04:59
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answer #1
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answered by Puggy 7
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If you have a function f(x), then the inverse of the function is the function g(x) where f(g(x)) = g(f(x)). One way to find the inverse of a function f(x) is to set f(x) = y, then solve for x in terms of y.
In this case, I don't know if "5x - 1 / 7x + 3" is suppose to mean "(5x - 1) / (7x + 3)", or "5x - 1/(7x) + 3", or "5x - (1/7)x + 3". So whatever it is, set it equal to y and solve for x. This will give you an expression where x = a function of y. This should be the inverse of f(x) too.
2007-02-06 16:05:02
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answer #2
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answered by Anonymous
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f(x)=y=5x-x/7+3 (I'm assuming you mean x/7, not 1/7x)
The algebraic method is simply to switch the x and the y, and solve for y. You get
x=5y-y/7+3
x-3=5y-y/7
x-3=y(5-1/7)
x-3=y(34/7)
y=7(x-3)/34
The graphical method is to graph f(x) your original function and reflect it over the line y=x. The resulting function will be f-1(x), or the inverse.
2007-02-06 16:08:13
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answer #3
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answered by Anonymous
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let 5x-1/7x+3=y
multiply 7x+3 to both sides collect term containig x and find its value in terms of y replace y by x and x by y you will get the inverse
2007-02-06 16:06:21
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answer #4
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answered by tarundeep300 3
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