Posted by Rachal on Wednesday, February 2, 2011 at 8:24pm.
The onetoone function f is defined by f(x)=(4x1)/(x+7).
Find f^1, the inverse of f. Then, give the domain and range of f^1 using interval notation.
f^1(x)=
Domain (f^1)=
Range (f^1)=
Any help is greatly appreciated.
Algebra  helper, Wednesday, February 2, 2011 at 7:05pm
f(x)=(4x1)/(x+7)
y = (4x1)/(x+7)
Rewrite as:
y = (4x)/(x+7) 1/(x+7)
Multiply both sides by x+7:
(x + 7)y = 4x  1
Expand out terms of the left hand side:
xy + 7y = 4x  1
xy  4x = 7y  1
x(y  4) = 7y  1
Divide both sides by y  4:
x = (7y  1)/(y  4)
f^1 = (7x  1)/(x  4)
Can you do the domain and range now?
Algebra  Rachal, Wednesday, February 2, 2011 at 7:11pm
I don't know if this is right but this is what I came up with.
f^1=(7x+1)/(x4)
domain f(^1)=(inf,7)U(7,inf)
range f(^1)=(inf,4)U(4,inf)
Let me know if it looks right. Thanks

Math  inverse  MathMate, Thursday, February 3, 2011 at 12:11am
The domain and range suggested apply to f(x). You will see that the vertical asymptote is at x=7 when the denominator becomes zero.
f(x)=(4x1)/(x+7)
domain f(^1)=(inf,7)U(7,inf)
range f(^1)=(inf,4)U(4,inf)
The domain and range of f^{1}(x) is equal to the range and domain respectively of f(x). Double check by evaluating the denominator at the singular points.
Post again if you need confirmation.
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