Solving the Functional Equation Problem

The question asks us to find the sum of function values from f(2) to f(6), given a functional equation and a specific value of the function at x=1.

The given functional equation is \(f(x-y) = \frac{f(x)}{f(y)}\), where \(f(y) \ne 0\). We are also given that \(f(1) = 0.5\).

Determining the Form of the Function \(f(x)\)

Let's use the given property to find the nature of the function \(f(x)\). A common type of function that satisfies such a property is an exponential function.

Consider \(f(x) = a^x\) for some base 'a'.

Let's test this form in the given equation:

LHS: \(f(x-y) = a^{x-y}\)

RHS: \(\frac{f(x)}{f(y)} = \frac{a^x}{a^y} = a^{x-y}\)

Since LHS = RHS, the function \(f(x) = a^x\) satisfies the given functional equation. We also need \(f(y) \ne 0\), which is true for \(a^y\) if \(a \ne 0\).

Now we use the given condition \(f(1) = 0.5\) to find the value of 'a'.

\(f(1) = a^1 = a\)

We are given \(f(1) = 0.5\), so \(a = 0.5 = \frac{1}{2}\).

Thus, the function is \(f(x) = \left(\frac{1}{2}\right)^x\).

Calculating the Required Function Values

We need to find the values of \(f(2), f(3), f(4), f(5), f(6)\) using the function \(f(x) = \left(\frac{1}{2}\right)^x\).

• \(f(2) = \left(\frac{1}{2}\right)^2 = \frac{1^2}{2^2} = \frac{1}{4}\)
• \(f(3) = \left(\frac{1}{2}\right)^3 = \frac{1^3}{2^3} = \frac{1}{8}\)
• \(f(4) = \left(\frac{1}{2}\right)^4 = \frac{1^4}{2^4} = \frac{1}{16}\)
• \(f(5) = \left(\frac{1}{2}\right)^5 = \frac{1^5}{2^5} = \frac{1}{32}\)
• \(f(6) = \left(\frac{1}{2}\right)^6 = \frac{1^6}{2^6} = \frac{1}{64}\)

We can summarize these values in a table:

Calculating the Sum \(f(2) + f(3) + f(4) + f(5) + f(6)\)

Now we add the calculated values:

Sum = \(f(2) + f(3) + f(4) + f(5) + f(6)\)

Sum = \(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64}\)

To add these fractions, we find a common denominator, which is 64.

• \(\frac{1}{4} = \frac{1 \times 16}{4 \times 16} = \frac{16}{64}\)
• \(\frac{1}{8} = \frac{1 \times 8}{8 \times 8} = \frac{8}{64}\)
• \(\frac{1}{16} = \frac{1 \times 4}{16 \times 4} = \frac{4}{64}\)
• \(\frac{1}{32} = \frac{1 \times 2}{32 \times 2} = \frac{2}{64}\)
• \(\frac{1}{64} = \frac{1}{64}\)

Now, add the fractions with the common denominator:

Sum = \(\frac{16}{64} + \frac{8}{64} + \frac{4}{64} + \frac{2}{64} + \frac{1}{64}\)

Sum = \(\frac{16 + 8 + 4 + 2 + 1}{64}\)

Sum = \(\frac{31}{64}\)

The sum \(f(2) + f(3) + f(4) + f(5) + f(6)\) is \(\frac{31}{64}\).

Revision Table: Key Concepts

Additional Information: Properties of \(f(x) = a^x\)

The function \(f(x) = a^x\) has several useful properties related to its base 'a':

• Base a > 1: The function is increasing. As x increases, \(a^x\) increases.
• Base 0 < a < 1: The function is decreasing. As x increases, \(a^x\) decreases (which we saw with \(f(x) = (1/2)^x\)).
• f(0): \(f(0) = a^0 = 1\) (assuming \(a \ne 0\)). This was consistent in our derivation.
• f(x+y): \(f(x+y) = a^{x+y} = a^x \cdot a^y = f(x)f(y)\). This is another common functional equation for exponential functions.
• f(x-y): \(f(x-y) = a^{x-y} = \frac{a^x}{a^y} = \frac{f(x)}{f(y)}\), as used in this problem.

Identifying the function type from the functional equation is a key step in solving such problems. The form \(f(x-y) = f(x)/f(y)\) strongly suggests an exponential function.