Let `f:R rarr R` be a continuous function such that
`f(x)-2f(x/2)+f(x/4)=x^(2)`.
f(3) is equal to

Options

A. f(0)
B. 4+f(0)
C. 9+f(0)
D. 16+f(0)

Answer

To solve the problem, we need to find the value of \( f(3) \) given the functional equation:

\[ f(x) - 2f\left(\frac{x}{2}\right) + f\left(\frac{x}{4}\right) = x^2 \]

### Step 1: Substitute \( x \) with \( \frac{x}{2} \)

We first replace \( x \) with \( \frac{x}{2} \) in the original equation:

\[ f\left(\frac{x}{2}\right) - 2f\left(\frac{x}{4}\right) + f\left(\frac{x}{8}\right) = \left(\frac{x}{2}\right)^2 = \frac{x^2}{4} \]

### Step 2: Substitute \( x \) with \( \frac{x}{4} \)

Next, we replace \( x \) with \( \frac{x}{4} \):

\[ f\left(\frac{x}{4}\right) - 2f\left(\frac{x}{8}\right) + f\left(\frac{x}{16}\right) = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \]

### Step 3: Substitute \( x \) with \( \frac{x}{8} \)

Now we replace \( x \) with \( \frac{x}{8} \):

\[ f\left(\frac{x}{8}\right) - 2f\left(\frac{x}{16}\right) + f\left(\frac{x}{32}\right) = \left(\frac{x}{8}\right)^2 = \frac{x^2}{64} \]

### Step 4: Generalize the pattern

Continuing this process, we can generalize the equation for \( n \):

\[ f\left(\frac{x}{2^n}\right) - 2f\left(\frac{x}{2^{n+1}}\right) + f\left(\frac{x}{2^{n+2}}\right) = \frac{x^2}{4^n} \]

### Step 5: Sum the equations

Now we sum all these equations from \( n = 0 \) to \( n \to \infty \):

1. The left side will simplify to:
\[ f(x) - f\left(\frac{x}{2}\right) + \left(-2f\left(\frac{x}{2}\right) + f\left(\frac{x}{4}\right)\right) + \ldots \]
Most terms will cancel out, leaving:
\[ f(x) - f(0) \]

2. The right side will be a geometric series:
\[ x^2 \left(1 + \frac{1}{4} + \frac{1}{16} + \ldots\right) = x^2 \cdot \frac{1}{1 - \frac{1}{4}} = \frac{4}{3} x^2 \]

### Step 6: Final equation

Thus, we have:
\[ f(x) - f(0) = \frac{4}{3} x^2 \]

Rearranging gives:
\[ f(x) = f(0) + \frac{4}{3} x^2 \]

### Step 7: Find \( f(3) \)

Now we substitute \( x = 3 \):
\[ f(3) = f(0) + \frac{4}{3} \cdot 3^2 = f(0) + \frac{4}{3} \cdot 9 = f(0) + 12 \]

### Conclusion

Thus, the value of \( f(3) \) is:
\[ f(3) = f(0) + 12 \]