Domain/range of functions inverse

[Nelo]

Homework Statement

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

Homework Equations

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

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

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 don't understand how to get the domain and range... How can

 

[vela]

but how is the domain for the inverse x is greater than or equal to 0? It passes to the negetive side...

What is "it"?

 

[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

 

[LCKurtz]

To find range:
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

If the above makes no sense at all then I assume all your expected to do is just think about it to get your answer. I hate that method though.

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

 

[GreenPrint]

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

 

[LCKurtz]

Homework Statement

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

Homework Equations

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

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

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 don't understand how to get the domain and range... How can

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.

 

[LCKurtz]

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

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

 

[eumyang]

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

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

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

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

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...

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.

To find range for rational functions:
1. Find the derivative of the function
2. Set the derivative equal to zero and solve
...

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.

 

[1MileCrash]

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

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]

 

[GreenPrint]

moved to calculus and beyond

 

[LCKurtz]

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...

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")?

 

[1MileCrash]

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")?

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.

 

[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")?

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.

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.

 

[1MileCrash]

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.

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.

 

[LCKurtz]

No, I am just explaining to you a method that someone else posted.

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

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.

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.