Question

The one-to-one function f is defined by f(x)=(4x-1)/(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)=(4x-1)/(x+7)
y = (4x-1)/(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)/(x-4)
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)=(4x-1)/(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.

I'm still not getting it

Answer 1

original:
y = (4x-1)/(x+7)

inverse: interchange x and y variables

x = (4y-1)/(y+7)
cross-multiply
xy + 7x = 4y - 1
xy - 4y = -7x - 1
y(x-4) = -7x - 1
y = (7x+1)/(4-x)

domain: any real x , except x?4

Answer 2

You have corrected stated:
f(x)=(4x-1)/(x+7)
domain f(^-1)=(-inf,-7)U(-7,inf)
range f(^-1)=(-inf,4)U(4,inf)

and Reiny had worked out the inverse:
f^-1 = (7x+1)/(4-x)

The numerator is a polynomial, so domain is ℝ (always true for polynomials), but denominator is (4-x), which becomes zero and creates a vertical asymptote at x=4.

The domain is the combination of the two, with all the limitations, so it is ℝ-{4}, or in interval notation, (-∞4)∪(4,∞).

Since there is a horizontal asymptote at y=-7, so the range is limited to (-∞7)∪(7,∞).

You will note that the domain of f(x) is the same as the range of f^-1(x), and the range of f(x) is the same as the domain of f^-1(x). This property is generally true for inverse functions, if an inverse exists.

If you study the graph of f(x) and f^-1(x), it will be much easier to understand.

See:
http://img834.imageshack.us/img834/7039/1296725887.png

Answer 3

I want to thank all of you for taking the time to help me understand the problem