Function Range
- Thread starter Ematrix
- Start date
Hello, Ematrix!
The Range is a bit tricky to determine.
As \(\displaystyle x\,\to\,-7^+,\;f(x)\,\to\,\infty\) . . . As \(\displaystyle x\,\to-7^-,\;f(x)\,\to\,\)-\(\displaystyle \infty\)
It looks like the range is all real \(\displaystyle y.\)
But consider the limit of \(\displaystyle f(x)\) as \(\displaystyle x\,\to\,\infty:\L\;\;\lim_{x\to\infty}\frac{2x\,-\,5}{x\,+\,7}\)
Divide top and bottom by \(\displaystyle x:\L\;\;\lim_{x\to\infty}\frac{2\,-\,\frac{5}{x}}{1\,+\,\frac{7}{x}}\;=\;\frac{2\,-\,0}{1\,+\,0}\;=\;2\)
The graph has a horizontal asymptote: \(\displaystyle \,y\:=\:2\)
But a graph can cross a horizontal asymptote
\(\displaystyle \;\;\)"before" it approaches 2 at the extreme right and extreme left.
So we must ask: Can \(\displaystyle f(x) = 2\,?\)
\(\displaystyle \;\;\)We would have: \(\displaystyle \,\frac{2x\,-\,5}{x\,+\,7}\:=\:2\;\;\Rightarrow\;\;2x\,-\,5\:=\:2x\,+\,14\;\;\Rightarrow\;\;0\,=\,19\;\) ??
\(\displaystyle \;\;\)The answer is: no . . . \(\displaystyle f(x)\) never equals \(\displaystyle 2.\)
Range: all real \(\displaystyle y\,\neq\,2.\)
Domain: all real \(\displaystyle x\,\neq\,-7\)
The Range is a bit tricky to determine.
As \(\displaystyle x\,\to\,-7^+,\;f(x)\,\to\,\infty\) . . . As \(\displaystyle x\,\to-7^-,\;f(x)\,\to\,\)-\(\displaystyle \infty\)
It looks like the range is all real \(\displaystyle y.\)
But consider the limit of \(\displaystyle f(x)\) as \(\displaystyle x\,\to\,\infty:\L\;\;\lim_{x\to\infty}\frac{2x\,-\,5}{x\,+\,7}\)
Divide top and bottom by \(\displaystyle x:\L\;\;\lim_{x\to\infty}\frac{2\,-\,\frac{5}{x}}{1\,+\,\frac{7}{x}}\;=\;\frac{2\,-\,0}{1\,+\,0}\;=\;2\)
The graph has a horizontal asymptote: \(\displaystyle \,y\:=\:2\)
But a graph can cross a horizontal asymptote
\(\displaystyle \;\;\)"before" it approaches 2 at the extreme right and extreme left.
So we must ask: Can \(\displaystyle f(x) = 2\,?\)
\(\displaystyle \;\;\)We would have: \(\displaystyle \,\frac{2x\,-\,5}{x\,+\,7}\:=\:2\;\;\Rightarrow\;\;2x\,-\,5\:=\:2x\,+\,14\;\;\Rightarrow\;\;0\,=\,19\;\) ??
\(\displaystyle \;\;\)The answer is: no . . . \(\displaystyle f(x)\) never equals \(\displaystyle 2.\)
Range: all real \(\displaystyle y\,\neq\,2.\)
skeeter
Elite Member
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- Dec 15, 2005
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f(x) = (2x - 5)/(x + 7)
domain is all real numbers except x = -7
one method to find the range of a function is to find the domain of its inverse ...
y = (2x - 5)/(x + 7)
swap "x" and "y" ...
x = (2y - 5)/(y + 7)
solve for "y" ...
x(y + 7) = 2y - 5
xy + 7x = 2y - 5
xy - 2y = -7x - 5
y(x - 2) = -(7x + 5)
y = -(7x + 5)/(x - 2) ... this is the inverse of f(x), f<sup>-1</sup>(x). note that its domain is all real values except x = 2 ... so, the range of the original function, f(x), is all real values except y = 2.
domain is all real numbers except x = -7
one method to find the range of a function is to find the domain of its inverse ...
y = (2x - 5)/(x + 7)
swap "x" and "y" ...
x = (2y - 5)/(y + 7)
solve for "y" ...
x(y + 7) = 2y - 5
xy + 7x = 2y - 5
xy - 2y = -7x - 5
y(x - 2) = -(7x + 5)
y = -(7x + 5)/(x - 2) ... this is the inverse of f(x), f<sup>-1</sup>(x). note that its domain is all real values except x = 2 ... so, the range of the original function, f(x), is all real values except y = 2.