Question
How do you find the range of (3x-4)/(6x-1) without using a graph?
Answers
GPT-4o mini
To find the range of the function \( y = \frac{3x-4}{6x-1} \), we can manipulate the equation to express \( x \) in terms of \( y \).
Starting from the equation:
\[
y = \frac{3x - 4}{6x - 1}
\]
We can multiply both sides by the denominator \( 6x - 1 \) (assuming \( 6x - 1 \neq 0 \)):
\[
y(6x - 1) = 3x - 4
\]
Expanding the left side gives:
\[
6yx - y = 3x - 4
\]
Now, we can rearrange the equation to isolate terms involving \( x \):
\[
6yx - 3x = y - 4
\]
Factoring out \( x \) from the left-hand side:
\[
x(6y - 3) = y - 4
\]
Assuming \( 6y - 3 \neq 0 \), we can solve for \( x \):
\[
x = \frac{y - 4}{6y - 3}
\]
Next, we need to establish the conditions under which this expression is valid. Specifically, we must avoid division by zero, so we need to find when \( 6y - 3 = 0 \):
\[
6y - 3 = 0 \implies y = \frac{3}{6} = \frac{1}{2}
\]
Therefore, \( y \) cannot be \( \frac{1}{2} \).
Now let's consider the limit of \( y \) as \( x \) approaches certain values. To understand the behavior of \( y \), we analyze the limits at \( x \to \infty \) and \( x \to -\infty \):
\[
\lim_{x \to \infty} \frac{3x - 4}{6x - 1} = \lim_{x \to \infty} \frac{3 - \frac{4}{x}}{6 - \frac{1}{x}} = \frac{3}{6} = \frac{1}{2}
\]
\[
\lim_{x \to -\infty} \frac{3x - 4}{6x - 1} = \lim_{x \to -\infty} \frac{3 - \frac{4}{x}}{6 - \frac{1}{x}} = \frac{3}{6} = \frac{1}{2}
\]
Since both limits approach \( \frac{1}{2} \), we see that the function approaches this value but never actually reaches it.
Now we should determine the behavior as \( x \) tends towards other values. Since \( y = \frac{3x-4}{6x-1} \) is a rational function and is continuous for all \( x \) except where \( 6x - 1 = 0 \) (which occurs at \( x = \frac{1}{6} \)), it can take all real values except for \( \frac{1}{2} \). Thus, \( y \) can reach all values \( (-\infty, \frac{1}{2}) \) and \( (\frac{1}{2}, +\infty) \).
Thus, we conclude that the range of \( y = \frac{3x - 4}{6x - 1} \) is:
\[
\boxed{(-\infty, \frac{1}{2}) \cup (\frac{1}{2}, +\infty)}
\]
Starting from the equation:
\[
y = \frac{3x - 4}{6x - 1}
\]
We can multiply both sides by the denominator \( 6x - 1 \) (assuming \( 6x - 1 \neq 0 \)):
\[
y(6x - 1) = 3x - 4
\]
Expanding the left side gives:
\[
6yx - y = 3x - 4
\]
Now, we can rearrange the equation to isolate terms involving \( x \):
\[
6yx - 3x = y - 4
\]
Factoring out \( x \) from the left-hand side:
\[
x(6y - 3) = y - 4
\]
Assuming \( 6y - 3 \neq 0 \), we can solve for \( x \):
\[
x = \frac{y - 4}{6y - 3}
\]
Next, we need to establish the conditions under which this expression is valid. Specifically, we must avoid division by zero, so we need to find when \( 6y - 3 = 0 \):
\[
6y - 3 = 0 \implies y = \frac{3}{6} = \frac{1}{2}
\]
Therefore, \( y \) cannot be \( \frac{1}{2} \).
Now let's consider the limit of \( y \) as \( x \) approaches certain values. To understand the behavior of \( y \), we analyze the limits at \( x \to \infty \) and \( x \to -\infty \):
\[
\lim_{x \to \infty} \frac{3x - 4}{6x - 1} = \lim_{x \to \infty} \frac{3 - \frac{4}{x}}{6 - \frac{1}{x}} = \frac{3}{6} = \frac{1}{2}
\]
\[
\lim_{x \to -\infty} \frac{3x - 4}{6x - 1} = \lim_{x \to -\infty} \frac{3 - \frac{4}{x}}{6 - \frac{1}{x}} = \frac{3}{6} = \frac{1}{2}
\]
Since both limits approach \( \frac{1}{2} \), we see that the function approaches this value but never actually reaches it.
Now we should determine the behavior as \( x \) tends towards other values. Since \( y = \frac{3x-4}{6x-1} \) is a rational function and is continuous for all \( x \) except where \( 6x - 1 = 0 \) (which occurs at \( x = \frac{1}{6} \)), it can take all real values except for \( \frac{1}{2} \). Thus, \( y \) can reach all values \( (-\infty, \frac{1}{2}) \) and \( (\frac{1}{2}, +\infty) \).
Thus, we conclude that the range of \( y = \frac{3x - 4}{6x - 1} \) is:
\[
\boxed{(-\infty, \frac{1}{2}) \cup (\frac{1}{2}, +\infty)}
\]