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 $\mathbb{R}$?
 
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