Understanding the Function and the Problem

The problem asks us to take a given function \(f(x) = \dfrac{x}{x-1}\) and express \(f(3x)\) in terms of \(f(x)\). This involves understanding function evaluation and algebraic manipulation to isolate variables and substitute expressions.

First, let's find the expression for \(f(3x)\). To do this, we replace every instance of \(x\) in the definition of \(f(x)\) with \(3x\).

Given: \(f(x) = \dfrac{x}{x-1}\)

Replacing \(x\) with \(3x\), we get:

\(f(3x) = \dfrac{3x}{3x-1}\)

Now, we need to express this result, \(f(3x)\), using \(f(x)\). This means we need to find a way to replace the \(x\) terms in the expression for \(f(3x)\) with an expression involving \(f(x)\).

Expressing x in terms of f(x)

We start with the original definition of the function:

\(f(x) = \dfrac{x}{x-1}\)

Let's use \(y\) to represent \(f(x)\) for simplicity while we solve for \(x\):

\(y = \dfrac{x}{x-1}\)

To isolate \(x\), we can multiply both sides by \((x-1)\):

\(y(x-1) = x\)

Distribute \(y\) on the left side:

\(yx - y = x\)

Now, gather all terms containing \(x\) on one side and terms without \(x\) on the other:

\(yx - x = y\)

Factor out \(x\) from the terms on the left side:

\(x(y-1) = y\)

Finally, divide by \((y-1)\) to get \(x\) by itself:

\(x = \dfrac{y}{y-1}\)

Since \(y = f(x)\), we can write \(x\) in terms of \(f(x)\):

\(x = \dfrac{f(x)}{f(x)-1}\)

Substituting x into f(3x)

We have the expression for \(f(3x)\):

\(f(3x) = \dfrac{3x}{3x-1}\)

And we have expressed \(x\) in terms of \(f(x)\):

\(x = \dfrac{f(x)}{f(x)-1}\)

Now, substitute the expression for \(x\) into the equation for \(f(3x)\):

\(f(3x) = \dfrac{3 \left(\dfrac{f(x)}{f(x)-1}\right)}{3 \left(\dfrac{f(x)}{f(x)-1}\right) - 1}\)

Simplifying the Expression

Let's simplify the complex fraction. First, simplify the numerator and the denominator separately.

Numerator: \(3 \left(\dfrac{f(x)}{f(x)-1}\right) = \dfrac{3f(x)}{f(x)-1}\)

Denominator: \(3 \left(\dfrac{f(x)}{f(x)-1}\right) - 1 = \dfrac{3f(x)}{f(x)-1} - 1\)

To combine the terms in the denominator, we find a common denominator, which is \((f(x)-1)\):

\(\dfrac{3f(x)}{f(x)-1} - 1 = \dfrac{3f(x)}{f(x)-1} - \dfrac{f(x)-1}{f(x)-1}\)

Now, subtract the numerators:

\(\dfrac{3f(x) - (f(x)-1)}{f(x)-1} = \dfrac{3f(x) - f(x) + 1}{f(x)-1} = \dfrac{2f(x) + 1}{f(x)-1}\)

So, the expression for \(f(3x)\) becomes:

\(f(3x) = \dfrac{\dfrac{3f(x)}{f(x)-1}}{\dfrac{2f(x)+1}{f(x)-1}}\)

To divide fractions, we multiply the numerator by the reciprocal of the denominator:

\(f(3x) = \dfrac{3f(x)}{f(x)-1} \times \dfrac{f(x)-1}{2f(x)+1}\)

Assuming \(f(x)-1 \neq 0\), we can cancel out the \((f(x)-1)\) terms:

\(f(3x) = \dfrac{3f(x)}{2f(x)+1}\)

This is the expression for \(f(3x)\) in terms of \(f(x)\).

Comparing with Options

Let's compare our simplified expression with the given options:

1. \(\dfrac{3f(x)}{2f(x)+1}\)
2. \(\dfrac{3f(x)}{3f(x)-1}\)
3. \(3f(x) - 1\)
4. \(\dfrac{3f(x)}{3f(x)-3}\)

Our derived expression \(\dfrac{3f(x)}{2f(x)+1}\) matches Option 1.

Conclusion

By first finding the expression for \(f(3x)\) and then solving the original function equation for \(x\) in terms of \(f(x)\), we were able to substitute and simplify to find the relationship between \(f(3x)\) and \(f(x)\). The result is \(\dfrac{3f(x)}{2f(x)+1}\).

Revision Table: Key Steps

Additional Information: Function Transformations

This problem is an example of exploring the relationship between different evaluations of a function, specifically \(f(ax)\) in terms of \(f(x)\). Understanding function transformations is key in algebra and calculus. When we evaluate \(f(3x)\), we are essentially looking at how scaling the input by a factor of 3 affects the output of the function. The resulting expression shows a specific relationship between the original function's value at \(x\) and the transformed function's value at \(3x\).

For a function \(f(x)\), expressing \(f(ax+b)\) or \(f(g(x))\) in terms of \(f(x)\) often requires similar steps: evaluate the transformed function, solve the original function for \(x\) in terms of \(f(x)\), and substitute.