Posted by **Rachal** on Wednesday, February 2, 2011 at 8:24pm.

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.

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