# Thread: inverse & rational functions

1. ## inverse & rational functions

I have the solutions to them, I am just trying to understand how they were reached.
Find the inverse of the given functions and find the domain for f(x) and f^-1(x)
so I understand that you just flip the x and y
but what does it mean domain?
heres a couple
f(x)=x-2 over x+1
I know dom f(x)= [-2, infinity) but how is dom f^-1(x) = x+2 over 1-x??
maybe that explanation will help me with the other ones

rational functions:
heres a couple, I understand how to do most of them but I get stuck at little aspects of it.
find domain, intercepts, and asymptotes when appropriate:
y=(x-2)^2 + 1
why is there no x intercept? I got it as being x= square root of 5

y= 1 over x-1 +2
(the 2 is separate from the fraction)
How is that =0 solved? for some reason I got x=0, I guess its because I don't know how to solve that equation since the x is in denominator. but the solution says x=1/2

Final is tomorrow:/
Thanks so much for all your help!

2. If $f^{-1}(x)$ is the inverse function of $f(x)$, then the domain of $f^{-1}(x)$ is the range of $f(x)$ and the range of $f^{-1}(x)$ is the domain of $f(x)$. Like the variables $x$ and $y$ are 'flipped' to find the inverse function, the domain and range are also 'flipped'.

The graph of $y = (x - 2)^2 + 1$ crosses the x-axis if and only if $y = 0$. However, $y$ never equals 0 because the discriminant of $y$ is negative. To review, the discriminant of $ax^2 + bx + c$ is given by $b^2 - 4ac$.

You may recall the discriminant reveals the number of real solutions to a quadratic equation. If the disciminant is positive, there are 2 real solutions and the graph crosses the x-axis twice. If the discriminant is zero, there is one real solution and the graph crosses the x-axis once. If the discriminant is negative, there are no real solutions and the graph does not cross the x-axis.

Before you can find the discriminant of $y$, you have to expand $y$. For example, $y = (x - 2)^2 + 1 = x^2 - 4x + 4 + 1 = x^2 - 4x + 5$.

To find the x-intercept(s) of $y = \frac{1}{x - 1} + 2$, substitute $y = 0$ and solve for $x$.

$0 = \frac{1}{x - 1} + 2$

Multiply both sides of the equation by $x - 1$.

$(x - 1) \times 0 = (x - 1) \times \frac{1}{x - 1} + 2$

Anything times 0 is still 0.

$0 = (x - 1) \times (\frac{1}{x - 1} + 2)$

Distribute $(x - 1)$ into $(\frac{1}{x - 1} + 2)$.

$0 = (x - 1) \times \frac{1}{x - 1} + (x - 1) \times 2$

$(x - 1)$ and $\frac{1}{x - 1}$ cancel, producing 1.

$0 = 1 + 2(x - 1)$

Subtract 1 from both sides of the equation.

$-1 = 2(x - 1)$

Divide by 2 on both sides of the equation.

$\frac{-1}{2} = x - 1$

Add 1 to both sides of the equation.

$\frac{-1}{2} + 1 = x$

Find the least common denominator of $\frac{-1}{2}$ and $1$.

$\frac{-1}{2} + \frac{2}{2} = x$

Write the left side of the equation as only one fraction.

$\frac{-1 + 2}{2} = x$

Simplify the numerator.

$\frac{1}{2} = x$

3. im sorry, I'm still confused. Is there anyway you (or anyone) could walk me through how to solve the whole problem?
Here is another problem if this would be easier=
f(x)= the square root of (x+2)

4. If by solve $f(x) = \sqrt{x + 2}$ you mean find the x-intercept(s), then set $f(x)$ to $0$ and solve for $x$.

$f(x) = 0$

Substitute $f(x) = \sqrt{x + 2}$.

$\sqrt{x + 2} = 0$

Square both sides of the equation.

$(\sqrt{x + 2})^2 = 0^2$

On the left side of the equation, the square root and the square cancel. On the right side of the equation, 0 squared is still 0.

$x + 2 = 0$

Subtract two from both sides of the equation.

$x = -2$

By the way, what confused you in my previous reply? I may be able to clarify it.

5. O im sorry! when I saw the first post I didn't see the second part, I meant I am still confused about the inverse functions so that is what that problem is for. but I just read over the 2nd part of that post and that clears rational functions up! thank you soooo much

6. I'll try to clarify what I wrote on inverse functions.

You already know how to find the inverse of a function: flip the variables.

To find the domain and range of an inverse function, flip the domain and range of the original function.

For example, if the domain of a function is (-3,3) and the range of the function is (0,10), then the domain of the inverse is (0,10) and the range of the inverse is (-3,3). Do you see how the domain and range flipped? It's as easy as finding the domain and range of the original function and flipping them.

7. I understand that concept, but how would I find the range for lets say
(1/x)-1
b/c the domain is all real numbers correct?