MHB Find the Range of a Rational Function.

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To find the range of the rational function y = 1/x algebraically, one must first determine its inverse, which is also y = 1/x. The domain of this inverse is all real numbers except zero, indicating that the range of the original function is also all real numbers except zero. This means that y can take any value in the real numbers, excluding zero. The discussion also briefly touches on a similar function, g = 1/x^2, which has the same domain but a different range. The overall conclusion is that the range of y = 1/x is all real numbers except zero.
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Find the range of y = 1/x algebraically.

Steps

1. Find the inverse y = 1/x.

2. Find domain of inverse of y = 1/x.

3. The domain of the inverse is the range of the original function given.

Correct?
 
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RTCNTC said:
Find the range of y = 1/x algebraically.

To be a bit more precise (perhaps with an eye towards calculus and analysis), let's talk about the function $f : x \mapsto y = \frac{1}{x}$ with domain $\mathbb{R} \setminus \{0\}$ and co-domain $\mathbb{R}$.

This means that $f$ assigns to every non-zero real number $x$ the real number $y = \frac{1}{x}$.

RTCNTC said:
Steps

1. Find the inverse y = 1/x.

Yes, here that works, because $f$ is indeed invertible. In general, you need to determine those real numbers $y$ for which the equation $f(x) = y$ has at least one nonzero solution $x$, i.e. those $y \in \mathbb{R}$ for which $\frac{1}{x} = y$ has at least one solution $x \in \mathbb{R} \setminus \{0\}$.

RTCNTC said:
2. Find domain of inverse of y = 1/x.

Yes, for this particular $f$ this is equivalent to what I wrote above.

RTCNTC said:
3. The domain of the inverse is the range of the original function given.

Correct?

Yes, with the remarks above.

For example, can you do the same question for $g : x \mapsto y = \frac{1}{x^2}$, again with domain $\mathbb{R} \setminus \{0\}$ and co-domain $\mathbb{R}$?
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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