# Domain/range of functions inverse

1. Jul 17, 2011

### Nelo

1. The problem statement, all variables and given/known data

For the following function find the inverse of f(x) graph the inverse and f(x) and determine the domain and range of its inverse
2. Relevant equations

f(x) = (x+1)^2

f(x)inverse = (plusminus) [sqrt] x-1

3. The attempt at a solution

I know the range for f(x)
but how is the domain for the inverse x is greater than or equal to 0? It passes to the negetive side...
I have both graphs, but i dont understand how to get the domain and range.... How can

2. Jul 17, 2011

### vela

Staff Emeritus
What is "it"?

3. Jul 17, 2011

### GreenPrint

To find range for rational functions:
1. Find the derivative of the function
2. Set the derivative equal to zero and solve
3. Plug values back into original equation
4. Omit complex results if your working in the set of real numbers

Last edited: Jul 17, 2011
4. Jul 17, 2011

### LCKurtz

It makes no sense to me. Try it on f(x) = x3.

5. Jul 17, 2011

### GreenPrint

*adds in "for rational functions" and takes out note*

Last edited: Jul 17, 2011
6. Jul 17, 2011

### LCKurtz

But your "inverse" isn't a function since it gives two values. Technically, your given function doesn't have an inverse because it isn't 1-1.

7. Jul 17, 2011

### LCKurtz

So how does your method apply to f(x) = 1/x?

8. Jul 17, 2011

### eumyang

You realize that since f(x) is not one-to-one, f-1(x) is NOT a function, right?

You're looking at the y-values (assuming you graphed correctly.) Yes, some y-values of the graph of f-1(x) are negative, but you need to look at the x-values when dealing with the domain.

I don't know about you, but I had to learn about domains and ranges a year before I learned how to take derivatives. So I don't think this method is going to help.

9. Jul 17, 2011

### 1MileCrash

It will work for continuous ranges. If the derivative is 0 then at that point the function is not growing or falling. The range will always be from one point where f'(x) = 0 to another point where f'(x) = 0. Your example grows without bound, so it's from some number to infinity.

Derivative is 3x^2

3x^2 = 0
x = 0

Since that's the only solution, it's the only real domain limit. Since solving for and x greater than 0 yields a greater y, x = 0 is the lower range, 0 cubed is 0, domain is 0, infinity.

Or another example:
y = 5x^2 + 7x + 5
y' = 10x + 7

10x + 7 = 0
x = -7/10

x = -7/10 is the only real solution.

5(-7/10)^2 + 7(-7/10) + 5
= 2.55

Range is [2.55, infinity)

More useful for examples that don't just grow without bound.

$y=\frac{4x+1}{x^2+x+1}$
Derivative:
$-\frac{4x^2+2x-3}{(x^2+x+1)^2}$

0 = 4x^2 + 2x - 3

x =
$\frac{-2+\sqrt{52}}{8}$
$\frac{-2-\sqrt{52}}{8}$

Solving for those x in original equation:
Range is [-3.070, 1.737]

10. Jul 17, 2011

### GreenPrint

moved to calculus and beyond

Last edited: Jul 18, 2011
11. Jul 17, 2011

### LCKurtz

This looks to me like an ad hoc method that sometimes, maybe usually, works. But there are probably unstated exceptions or additions to the method. If

$$f(x) = \frac{x^2}{x^2+4}$$

the derivative is 0 when x = 0 and only there. How does your method give the range (without adding more cases to the "method")?

12. Jul 17, 2011

### 1MileCrash

That's not a range where the function will actually equal the limiting range at a point. This function approaches 1.

Derivative method will find 0 as the beginning of the range because the function actually equals zero at a point.

13. Jul 17, 2011

### LCKurtz

That's my point. You are suggesting a method that supposedly works on rational functions with continuous range but it is apparently an incomplete method. It is just a tool that might be helpful on some functions, but maybe not on others, even if they are rational functions.

14. Jul 17, 2011

### 1MileCrash

No, I am just explaining to you a method that someone else posted. You remarked that it made no sense, so I showed the sense of it.

Finding where derivative = 0 shows maximum and minimum values, it's not a leap of logic that it helps with finding the ranges of a function as well.

It will work, for any continuous function, that has a range that would be in brackets on both ends. I'm pretty sure that's correct, anyway. For any noncontinuous function, I'd think breaking it up into intervals would work fine.

15. Jul 17, 2011

### LCKurtz

Point taken. It's late and I haven't kept track of who posted what first.

Yes, I agree that can be a useful tool in some cases. GreenPrint posted it at the beginning of this thread presented it as though it was "the way" to do the problem. There's more to it, that's all I'm saying. Anyway, time to hit the sack here.