It is given that \[ f(x)=\left\{\begin{array}{ll}7-x^{2} & \text { for } 0

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To evaluate $$ f(27) + f(45) $$, we first note that the function $$ f(x) $$ is periodic with period 4, meaning $$ f(x + 4) = f(x) $$ for all real numbers $$ x $$.

**Step 1: Reduce $$ x $$ modulo 4**

1. **For $$ f(27) $$:**
   $$
   27 \mod 4 = 3
   $$
   So, $$ f(27) = f(3) $$.

2. **For $$ f(45) $$:**
   $$
   45 \mod 4 = 1
   $$
   So, $$ f(45) = f(1) $$.

**Step 2: Determine which piece of the function to use**

1. **Evaluating $$ f(3) $$:**
   Since $$ 2 < 3 \leq 4 $$, we use the second piece of the function:
   $$
   f(3) = 2(3) - 1 = 6 - 1 = 5
   $$

2. **Evaluating $$ f(1) $$:**
   Since $$ 0 < 1 \leq 2 $$, we use the first piece of the function:
   $$
   f(1) = 7 - (1)^2 = 7 - 1 = 6
   $$

**Step 3: Add the results**

$$
f(27) + f(45) = 5 + 6 = 11
$$

**Answer:** 11
$$ f(27) + f(45) = 11 $$

It is given that

and that for all real values of .
(i) Evaluate .

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