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
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
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
I want to thank all of you for taking the time to help me understand the problem
To find the inverse of a function, you need to switch the x and y variables and solve for y. Let's go through the steps again to find the inverse of f(x).
Step 1: Start with f(x) = (4x-1)/(x+7).
Step 2: Replace f(x) with y, so the equation becomes y = (4x-1)/(x+7).
Step 3: Switch the x and y variables, so the equation becomes x = (4y-1)/(y+7).
Step 4: Solve for y. Multiply both sides by y+7 to get rid of the denominator: x(y+7) = 4y-1.
Step 5: Expand and simplify: xy + 7x = 4y - 1.
Step 6: Move the terms involving y to one side: xy - 4y = -7x - 1.
Step 7: Factor out the y: y(x - 4) = -7x - 1.
Step 8: Divide both sides by (x-4): y = (-7x - 1)/(x - 4).
So, the inverse of f(x) is f^(-1)(x) = (-7x - 1)/(x - 4).
Now let's determine the domain and range of the inverse function using interval notation.
For the domain of f^(-1)(x), we need to consider the values of x that make the denominator (x - 4) non-zero. So, the domain is all real numbers except x = 4. In interval notation, this can be written as: (-∞, 4) U (4, ∞).
For the range of f^(-1)(x), we need to find the possible values of y. Since the numerator (-7x - 1) and denominator (x - 4) can take on any real value, the range of the inverse function is all real numbers. In interval notation, this is written as: (-∞, ∞).
Therefore, the domain of f^(-1)(x) is (-∞, 4) U (4, ∞) and the range is (-∞, ∞).