What is the product?
$\frac{4n}{4n-4} \cdot \frac{n-1}{n+1}$
The given expression is the product of two rational functions: \[\frac{4n}{4n-4} \cdot \frac{n-1}{n+1}.\] To simplify this, we need to factor the expressions in the numerator and denominator where possible and then cancel out common factors. We assume \(n eq 1\) and \(n eq -1\) so that the denominators are not zero.
First, we look at the first fraction \(\frac{4n}{4n-4}\). Notice that the denominator \(4n-4\) contains a common factor of 4. Factoring out this 4, we get: \[4n-4 = 4(n-1).\] Therefore, the first fraction becomes: \[\frac{4n}{4(n-1)}.\] We can now cancel the common factor of 4 from both the numerator and the denominator: \[\frac{4n}{4(n-1)} = \frac{n}{n-1}.\]
Now we substitute the simplified first fraction back into the original product: \[\frac{n}{n-1} \cdot \frac{n-1}{n+1}.\] Next, we multiply these two fractions. Since there is a factor of \((n-1)\) in the denominator of the first fraction and a factor of \((n-1)\) in the numerator of the second fraction, we can cancel these common terms (assuming \(n eq 1\)): \[\frac{n}{\cancel{n-1}} \cdot \frac{\cancel{n-1}}{n+1} = \frac{n}{n+1}.\]
Comparing our result, \(\frac{n}{n+1}\), with the provided options, we see it matches the second option. Thus, the simplified product is \(\frac{n}{n+1}\). The final answer is \(\boxed{\frac{n}{n+1}}\).
In real-world applications, simplifying complex rational expressions is common in fields like electrical engineering or physics. For example, if you are analyzing a circuit's impedance that is described by such a rational function, simplifying the formula allows you to quickly determine how the impedance behaves as the variable \(n\) (which might represent frequency) changes. This simplifies calculations and provides immediate insight into the behavior of the system.