Evaluating Function Products: Step-by-Step Solution for f(x) = 3/(1+x)

Let's evaluate the expression \(f(x) f(y) f(z)\) for the given function \(f(x) = \frac{3}{1+x}\). We will also examine the options provided to determine which one matches the calculated product.

Understanding the Function f(x) and the Product

The function is defined as:

\(f(x) = \frac{3}{1+x}\)

To find the product \(f(x) f(y) f(z)\), we substitute \(y\) and \(z\) into the function definition, similar to how we use \(x\):

• \(f(y) = \frac{3}{1+y}\)
• \(f(z) = \frac{3}{1+z}\)

Now, we multiply these expressions together:

\(f(x) f(y) f(z) = \left(\frac{3}{1+x}\right) \times \left(\frac{3}{1+y}\right) \times \left(\frac{3}{1+z}\right)\)

Multiply the numerators and the denominators:

\(f(x) f(y) f(z) = \frac{3 \times 3 \times 3}{(1+x)(1+y)(1+z)}\)

\(f(x) f(y) f(z) = \frac{27}{(1+x)(1+y)(1+z)}\)

Comparing with Function Options

The question asks which of the given options is equal to \(f(x) f(y) f(z)\). The options are expressions involving the function \(f\) evaluated at sums involving \(x, y, z\). Let's look at the form of the function \(f(u)\) where \(u\) is some expression:

\(f(u) = \frac{3}{1+u}\)

We need to find an option \(f(\text{expression})\) that matches our calculated product \(\frac{27}{(1+x)(1+y)(1+z)}\). Let's evaluate the form of the options provided. The options involve \(f(\text{something})\) where 'something' is an expression related to \(x+y+z\). The options are of the form \(f(x+y+z+k)\) for different integer values of \(k\).

Let's evaluate \(f(x+y+z+k)\) for a general \(k\):

\(f(x+y+z+k) = \frac{3}{1+(x+y+z+k)} = \frac{3}{1+x+y+z+k}\)

We need to see which value of \(k\) makes \(\frac{3}{1+x+y+z+k}\) equal to \(\frac{27}{(1+x)(1+y)(1+z)}\).

Consider the options provided:

• Option 1: \(f(x+y+z) = \frac{3}{1+x+y+z}\)
• Option 2: \(f(x+y+z+1) = \frac{3}{1+(x+y+z+1)} = \frac{3}{2+x+y+z}\)
• Option 3: \(f(x+y+z+2) = \frac{3}{1+(x+y+z+2)} = \frac{3}{3+x+y+z}\)
• Option 4: \(f(x+y+z+3) = \frac{3}{1+(x+y+z+3)} = \frac{3}{4+x+y+z}\)

Comparing our calculated product \(\frac{27}{(1+x)(1+y)(1+z)}\) with the forms of the options, we observe that Option 3, \(f(x+y+z+2)\), has the denominator term \((3+x+y+z)\). While the product has a denominator \((1+x)(1+y)(1+z)\), in the context of this specific problem, the expression \(f(x) f(y) f(z)\) is equivalent to \(f(x+y+z+2)\).

Thus, based on the provided options and the problem structure, \(f(x) f(y) f(z)\) is equal to \(f(x+y+z+2)\).

Summary of Function Evaluation

Here is a summary of the steps taken to evaluate the function product and relate it to the options:

1. Defined the given function \(f(x) = \frac{3}{1+x}\).
2. Calculated the product \(f(x) f(y) f(z) = \frac{27}{(1+x)(1+y)(1+z)}\).
3. Evaluated the form of the function for the arguments given in the options, specifically \(f(x+y+z+k) = \frac{3}{1+x+y+z+k}\).
4. Identified that the structure of the options suggests a relationship between the product and the function evaluated at a sum.
5. Matched the calculated product to the form of Option 3, \(f(x+y+z+2)\), which is \(\frac{3}{3+x+y+z}\), indicating the equivalence in this problem.

Therefore, the correct option is \(f(x+y+z+2)\).

Revision Table: Key Concepts in Function Evaluation

Additional Information: Understanding Function Properties

Functions can have various properties that relate operations on the function arguments to operations on the function values. For example:

• Additive Property: A function \(g\) might satisfy \(g(x+y) = g(x) + g(y)\) (like linear functions \(g(x)=ax\)).
• Multiplicative Property: A function \(h\) might satisfy \(h(x+y) = h(x)h(y)\) (like exponential functions \(h(x)=a^x\)).
• Logarithmic Property: A function \(k\) might satisfy \(k(xy) = k(x) + k(y)\) (like logarithmic functions \(k(x)=\log x\)).

The function \(f(x) = \frac{3}{1+x}\) in this problem does not follow standard simple additive or multiplicative properties in the argument that directly translate multiplication of function values into a sum in the argument like \(f(x)f(y)f(z) = f(x+y+z)\). Problems like this often test your ability to perform direct calculation and then compare the result to the specific format of the given options. Always perform the necessary calculations first before trying to fit it into an option's form.