Domains, Ranges and Inverse Functions

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August 13th 2012, 11:53 PM
ineedhelplz

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Domains, Ranges and Inverse Functions

In the attachment, regarding question 7, I had some confusion with whether or not to use the modulus function and why only the positive (1/x) is taken for 7b (or maybe it's only the negative, -1/x).

It would be fantastic if you could shed some light on this and clearly explain the solutions

Thanking you in advance
August 14th 2012, 01:05 AM
Prove It

Re: Domains, Ranges and Inverse Functions

What set are you using S to represent?
August 14th 2012, 02:44 AM
ineedhelplz

Re: Domains, Ranges and Inverse Functions

Is that for question 7? What do you mean?
August 14th 2012, 03:06 AM
emakarov

Re: Domains, Ranges and Inverse Functions

Quote:

Originally Posted by ineedhelplz

In the attachment, regarding question 7, I had some confusion with whether or not to use the modulus function

"Modulus" is usually called "absolute value" in English.

Quote:

Originally Posted by ineedhelplz

and why only the positive (1/x) is taken for 7b (or maybe it's only the negative, -1/x).

I don't see the word "positive" in 7b. Do you mean 8b? But then I don't see 1/x anywhere in these problems.
August 14th 2012, 03:42 AM
ineedhelplz

Re: Domains, Ranges and Inverse Functions

Sorry. When does the absolute value of 'x' replace just 'x' when squaring and square rooting?

Regarding question 7b, the f^-1(x) is 1/x^2

Then f(f^-1) = -1/sqrt(1/x^2) which equals 'x'. That means the sqrt taken was the negative 1/x not the positive. Why is this so?

Is this because of certain domain and range rules with inverse functions and functions of functions? What are they?

Thanks again
August 14th 2012, 03:45 AM
tom@ballooncalculus

Re: Domains, Ranges and Inverse Functions

The OP is rightly alarmed that the usual swap-and-rearrange procedure results in

$f^{-1}(x) = y = \frac{1}{x^2}$

and hence

$f(f^{-1}(x)) = f(y) = \frac{-1}{\sqrt{\frac{1}{x^2}}} = -x$

Which is a problem. So we are only allowed a negative in the expression of f inverse if it's inside the square, e.g.

whoops, maybe I'm confused, sorry if so... drawing a pic...

Ok, for my benefit anyway...

http://www.ballooncalculus.org/draw/graph/domains.png
August 14th 2012, 04:24 AM
ineedhelplz

Re: Domains, Ranges and Inverse Functions

Thanks tom, I'm glad you understand; would you be able to explain your drawing please? I sort of understand but don't at the same time.

Does the absolute function come in to play at all in this question? (If so why is it? If not when does it?)
August 14th 2012, 04:30 AM
tom@ballooncalculus

Re: Domains, Ranges and Inverse Functions

Quote:

Originally Posted by ineedhelplz

I sort of understand but don't at the same time.

You and me both! Someone will put us out of our misery soon, no doubt.

I'm sure we can take "square root" as conventionally short for "absolute value of square root", since we are assured f is a function.
August 14th 2012, 04:31 AM
emakarov

Re: Domains, Ranges and Inverse Functions

Quote:

Originally Posted by ineedhelplz

When does the absolute value of 'x' replace just 'x' when squaring and square rooting?

We have $\left(\sqrt{x}\right)^2=x$ , but only for x ≥ 0, and $\sqrt{x^2}=|x|$ for all real x.

Quote:

Originally Posted by ineedhelplz

Regarding question 7b, the f^-1(x) is 1/x^2

Yes.

Quote:

Originally Posted by ineedhelplz

Then f(f^-1) = -1/sqrt(1/x^2) which equals 'x'. That means the sqrt taken was the negative 1/x not the positive. Why is this so?

The left-hand side, f(f⁻¹), is not a complete expression: it lacks an x. Also, I don't see where -1 comes from in the right-hand side. We have $f(f^{-1}(x))=\frac{1}{\sqrt{1/x^2}}=\frac{1}{1/|x|}=|x|$ . But if we consider the domain of f⁻¹ to be ℝ⁺ = {x ∈ ℝ | x > 0}, then f(f⁻¹(x)) = x.

This raises an important question: what is the domain of f⁻¹: ℝ - {0} or ℝ⁺? This depends on the definition of the inverse function, which I encourage you to double-check in your textbook. One option is to say that the domain of f⁻¹ is the range (i.e., image) of f. In this case, the range of f is ℝ⁺, which is a proper subset of the codomain ℝ. Then, as we've seen, f(f⁻¹(x)) = x for all x in the domain of f⁻¹.

In higher mathematics, though, the domain of the inverse function is usually defined as the codomain of the original function, i.e., ℝ in this case. Then f does not have an inverse because 1 / x² is not defined on 0. Even if we restrict the domain of f⁻¹ to ℝ - {0}, we would have f(f⁻¹(x)) = |x|, not x. According to this definition, in order to have an inverse, f must have its codomain declared as ℝ⁺, not ℝ.
August 14th 2012, 04:41 AM
ineedhelplz

Re: Domains, Ranges and Inverse Functions

Quote:

Originally Posted by emakarov

Also, I don't see where -1 comes from in the right-hand side.

The -1 is from the original definition of f(x) = -1/sqrt(x)

How does your explanation differ now?

Thanks so much, what you've said makes a lot of sense
August 14th 2012, 04:56 AM
tom@ballooncalculus

Re: Domains, Ranges and Inverse Functions

Quote:

Originally Posted by ineedhelplz

The -1 is from the original definition of f(x) = -1/sqrt(x)

Yes.

Incidentally, if we read $\sqrt{x}$ as $-|\sqrt{x}|$ (which is unconventional but just as plausible in theory as a way of rendering $\sqrt{x}$ as a function), then we don't run into the same problem with the natural algebraic formulation of the inverse.

Edit:

Quote:

Originally Posted by emakarov

We have $\frac{-1}{\sqrt{1/x^2}}=\frac{-1}{1/|x|}=-|x|=x$ because |x| = -x for x < 0.

So no problem at all, of course. (Bow)
August 14th 2012, 04:57 AM
emakarov

Re: Domains, Ranges and Inverse Functions

Quote:

Originally Posted by ineedhelplz

The -1 is from the original definition of f(x) = -1/sqrt(x)

Wow, I thought that this was an accidental pen stroke. This is bad typesetting. It should look like this: $\frac{-1}{\sqrt{x}}$ , or, better, like this: $-\frac{1}{\sqrt{x}}$ .

Quote:

Originally Posted by ineedhelplz

Then f(f^-1) = -1/sqrt(1/x^2) which equals 'x'. That means the sqrt taken was the negative 1/x not the positive. Why is this so?

The inverse function is still $f^{-1}(x)=1/x^2$ , but its domain, as well as the range of f, is now ℝ⁻ = {x ∈ ℝ | x < 0}. This does not mean that we take the negative square root. We have $\frac{-1}{\sqrt{1/x^2}}=\frac{-1}{1/|x|}=-|x|=x$ because |x| = -x for x < 0.