If \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\rm{x}}}{{{\rm{x}} - 1}},\) then what is \(\frac{{{\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{f}}\left( {{\rm{a}} + 1} \right)}}\) equal to?
Understanding the Function and the Problem
The problem asks us to evaluate a specific ratio involving a given function \({\rm{f}}\left( {\rm{x}} \right)\) at two different points, \({\rm{a}}\) and \({\rm{a}} + 1\). The function is defined as \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\rm{x}}}{{{\rm{x}} - 1}}\). We need to find the value of \(\frac{{{\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{f}}\left( {{\rm{a}} + 1} \right)}}\) and see which of the given options it matches.
Step-by-Step Calculation of the Ratio
To find the ratio, we first need to calculate the values of \({\rm{f}}\left( {\rm{a}} \right)\) and \({\rm{f}}\left( {{\rm{a}} + 1} \right)\) using the given definition of \({\rm{f}}\left( {\rm{x}} \right)\).
Calculating f(a)
Substitute \({\rm{x}} = {\rm{a}}\) into the function \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\rm{x}}}{{{\rm{x}} - 1}}\):
\({\rm{f}}\left( {\rm{a}} \right) = \frac{{\rm{a}}}{{{\rm{a}} - 1}}\)
Calculating f(a + 1)
Substitute \({\rm{x}} = {\rm{a}} + 1\) into the function \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\rm{x}}}{{{\rm{x}} - 1}}\):
\({\rm{f}}\left( {{\rm{a}} + 1} \right) = \frac{{{\rm{a}} + 1}}{{\left( {{\rm{a}} + 1} \right) - 1}} = \frac{{{\rm{a}} + 1}}{{\rm{a}}}\)
Calculating the Ratio \(\frac{{{\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{f}}\left( {{\rm{a}} + 1} \right)}}\)
Now we can compute the ratio:
\(\frac{{{\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{f}}\left( {{\rm{a}} + 1} \right)}} = \frac{{\frac{{\rm{a}}}{{{\rm{a}} - 1}}}}{{\frac{{{\rm{a}} + 1}}{{\rm{a}}}}}\)
To simplify this complex fraction, we multiply the numerator by the reciprocal of the denominator:
\(\frac{{{\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{f}}\left( {{\rm{a}} + 1} \right)}} = \frac{{\rm{a}}}{{{\rm{a}} - 1}} \times \frac{{\rm{a}}}{{{\rm{a}} + 1}}\)
Multiply the numerators and the denominators:
\(\frac{{{\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{f}}\left( {{\rm{a}} + 1} \right)}} = \frac{{{\rm{a}} \times {\rm{a}}}}{{\left( {{\rm{a}} - 1} \right) \times \left( {{\rm{a}} + 1} \right)}}\)
Using the difference of squares formula \(({\rm{a}} - {\rm{b}})({\rm{a}} + {\rm{b}}) = {{\rm{a}}^2} - {{\rm{b}}^2}\) in the denominator:
\(\frac{{{\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{f}}\left( {{\rm{a}} + 1} \right)}} = \frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} - {{1}^2}}} = \frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} - 1}}\)
So, the simplified ratio is \(\frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} - 1}}\).
Evaluating the Options
Now we need to check which of the given options, when evaluated using the function \({\rm{f}}\left( {\rm{x}} \right)\), yields the expression \(\frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} - 1}}\).
Option 1: \({\rm{f}}\left( { - \frac{{\rm{a}}}{{{\rm{a}} + 1}}} \right)\)
Substitute \({\rm{x}} = - \frac{{\rm{a}}}{{{\rm{a}} + 1}}\) into \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\rm{x}}}{{{\rm{x}} - 1}}\):
\({\rm{f}}\left( { - \frac{{\rm{a}}}{{{\rm{a}} + 1}}} \right) = \frac{{ - \frac{{\rm{a}}}{{{\rm{a}} + 1}}}}{{ - \frac{{\rm{a}}}{{{\rm{a}} + 1}} - 1}} = \frac{{ - \frac{{\rm{a}}}{{{\rm{a}} + 1}}}}{{\frac{{ - {\rm{a}} - \left( {{\rm{a}} + 1} \right)}}{{{\rm{a}} + 1}}}} = \frac{{ - \frac{{\rm{a}}}{{{\rm{a}} + 1}}}}{{\frac{{ - {\rm{a}} - {\rm{a}} - 1}}{{{\rm{a}} + 1}}}} = \frac{{ - \frac{{\rm{a}}}{{{\rm{a}} + 1}}}}{{\frac{{ - 2{\rm{a}} - 1}}{{{\rm{a}} + 1}}}}\)
Multiply by the reciprocal of the denominator:
\(= \frac{{ - {\rm{a}}}}{{{\rm{a}} + 1}} \times \frac{{{\rm{a}} + 1}}{{ - 2{\rm{a}} - 1}} = \frac{{ - {\rm{a}}}}{{ - 2{\rm{a}} - 1}} = \frac{{\rm{a}}}{{2{\rm{a}} + 1}}\)
This does not match \(\frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} - 1}}\).
Option 2: \({\rm{f}}\left( {{{\rm{a}}^2}} \right)\)
Substitute \({\rm{x}} = {{\rm{a}}^2}\) into \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\rm{x}}}{{{\rm{x}} - 1}}\):
\({\rm{f}}\left( {{{\rm{a}}^2}} \right) = \frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} - 1}}\)
This matches the simplified ratio \(\frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} - 1}}\).
Option 3: \({\rm{f}}\left( {\frac{1}{{\rm{a}}}} \right)\)
Substitute \({\rm{x}} = \frac{1}{{\rm{a}}}\) into \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\rm{x}}}{{{\rm{x}} - 1}}\):
\({\rm{f}}\left( {\frac{1}{{\rm{a}}}} \right) = \frac{{\frac{1}{{\rm{a}}}}}{{\frac{1}{{\rm{a}}} - 1}} = \frac{{\frac{1}{{\rm{a}}}}}{{\frac{{1 - {\rm{a}}}}{{\rm{a}}}}}\)
Multiply by the reciprocal of the denominator:
\(= \frac{1}{{\rm{a}}} \times \frac{{\rm{a}}}{{1 - {\rm{a}}}} = \frac{1}{{1 - {\rm{a}}}}\)
This does not match \(\frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} - 1}}\).
Option 4: f(-a)
Substitute \({\rm{x}} = - {\rm{a}}\) into \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\rm{x}}}{{{\rm{x}} - 1}}\):
\({\rm{f}}\left( { - {\rm{a}}} \right) = \frac{{ - {\rm{a}}}}{{ - {\rm{a}} - 1}} = \frac{{ - {\rm{a}}}}{{ - \left( {{\rm{a}} + 1} \right)}} = \frac{{\rm{a}}}{{{\rm{a}} + 1}}\)
This does not match \(\frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} - 1}}\).
Conclusion
Comparing the calculated ratio with the evaluated options, we find that \(\frac{{{\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{f}}\left( {{\rm{a}} + 1} \right)}}\) is equal to \({\rm{f}}\left( {{{\rm{a}}^2}} \right)\).
Revision Table: Function Evaluation Summary
| Expression | Calculation | Result |
|---|---|---|
| \({\rm{f}}\left( {\rm{x}} \right)\) | Given definition | \(\frac{{\rm{x}}}{{{\rm{x}} - 1}}\) |
| \({\rm{f}}\left( {\rm{a}} \right)\) | Substitute \({\rm{x}} = {\rm{a}}\) | \(\frac{{\rm{a}}}{{{\rm{a}} - 1}}\) |
| \({\rm{f}}\left( {{\rm{a}} + 1} \right)\) | Substitute \({\rm{x}} = {\rm{a}} + 1\) | \(\frac{{{\rm{a}} + 1}}{{\rm{a}}}\) |
| \(\frac{{{\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{f}}\left( {{\rm{a}} + 1} \right)}}\) | Ratio calculation | \(\frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} - 1}}\) |
| \({\rm{f}}\left( {{{\rm{a}}^2}} \right)\) | Substitute \({\rm{x}} = {{\rm{a}}^2}\) | \(\frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} - 1}}\) |
Additional Information: Properties of Functions
This problem demonstrates evaluating a function at specific points and simplifying algebraic expressions involving those evaluations. Understanding function notation and basic algebraic manipulation is key. The function \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\rm{x}}}{{{\rm{x}} - 1}}\) is a rational function. It is undefined when the denominator is zero, i.e., when \({\rm{x}} - 1 = 0\), which means \({\rm{x}} = 1\). In the context of this problem, we assume that the values of \({\rm{a}}\), \({\rm{a}} + 1\), \({\rm{a}} - \frac{{\rm{a}}}{{{\rm{a}} + 1}}\), \({\rm{a}}^2\), \(\frac{1}{{\rm{a}}}\), and \(-{\rm{a}}\) do not cause the denominator of \({\rm{f}}\left( {\rm{x}} \right)\) or any intermediate expression to become zero.
For instance, \({\rm{f}}\left( {\rm{a}} \right)\) requires \({\rm{a}} - 1 \ne 0\), so \({\rm{a}} \ne 1\). \({\rm{f}}\left( {{\rm{a}} + 1} \right)\) requires \(({\rm{a}} + 1) - 1 \ne 0\), so \({\rm{a}} \ne 0\). The ratio calculation \(\frac{{{\rm{f}}\left( {\rm{a}} \right)}}{{{\rm{f}}\left( {{\rm{a}} + 1} \right)}}\) also requires \({\rm{f}}\left( {{\rm{a}} + 1} \right) \ne 0\). Since \({\rm{f}}\left( {{\rm{a}} + 1} \right) = \frac{{{\rm{a}} + 1}}{{\rm{a}}}\), this means \({\rm{a}} + 1 \ne 0\), so \({\rm{a}} \ne -1\). Also, the intermediate step \(\frac{{\rm{a}}}{{{\rm{a}} - 1}} \times \frac{{\rm{a}}}{{{\rm{a}} + 1}}\) requires \({\rm{a}} + 1 \ne 0\).
The simplified ratio \(\frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} - 1}}\) is undefined if \({\rm{a}}^2 - 1 = 0\), which means \({\rm{a}} = 1\) or \({\rm{a}} = -1\). The option \({\rm{f}}\left( {{{\rm{a}}^2}} \right)\) is undefined if \({\rm{a}}^2 - 1 = 0\), which is consistent.
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