# Algebra-I'm still stuck

posted by on .

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

• Algebra-I'm still stuck - ,

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

• Algebra-I'm still stuck - ,

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

• Algebra-I'm still stuck - ,

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