If 
f
(
x
)
f(x) is a function that satisfies 
(
3
−
f
(
x
+
1
)
)
(
6
+
f
(
x
)
)
=
18
(3−f(x+1))(6+f(x))=18 and 
f
(
0
)
=
3
f(0)=3, what is the value of 
∑
i
=
0
x
1
f
(
i
)
∑ 
i=0
x
​
  
f(i)
1
​
 ?

We are given the functional equation:

\[
(3 - f(x+1))(6 + f(x)) = 18
\]

and the initial condition:

\[
f(0) = 3.
\]

We are tasked with finding the value of the sum \( \sum_{i=0}^{x} \frac{1}{f(i)} \).

### Step 1: Analyze the functional equation

Let's expand the equation:

\[
(3 - f(x+1))(6 + f(x)) = 18
\]

Expanding this:

\[
(3)(6) + (3)(f(x)) - (f(x+1))(6) - (f(x+1))(f(x)) = 18.
\]

This simplifies to:

\[
18 + 3f(x) - 6f(x+1) - f(x+1)f(x) = 18.
\]

Canceling out the 18 on both sides:

\[
3f(x) - 6f(x+1) - f(x+1)f(x) = 0.
\]

### Step 2: Solve for \( f(x+1) \)

We can now solve for \( f(x+1) \) by isolating it. Rearrange the equation:

\[
3f(x) = 6f(x+1) + f(x+1)f(x).
\]

Factor the right-hand side:

\[
3f(x) = f(x+1)(6 + f(x)).
\]

Divide by \( 6 + f(x) \):

\[
f(x+1) = \frac{3f(x)}{6 + f(x)}.
\]

### Step 3: Apply the initial condition \( f(0) = 3 \)

Substitute \( f(0) = 3 \) into the recursive formula to find \( f(1) \):

\[
f(1) = \frac{3f(0)}{6 + f(0)} = \frac{3 \times 3}{6 + 3} = \frac{9}{9} = 1.
\]

Now, use this result to find \( f(2) \):

\[
f(2) = \frac{3f(1)}{6 + f(1)} = \frac{3 \times 1}{6 + 1} = \frac{3}{7}.
\]

### Step 4: Calculate the sum \( \sum_{i=0}^{2} \frac{1}{f(i)} \)

Now we can compute the sum:

\[
\sum_{i=0}^{2} \frac{1}{f(i)} = \frac{1}{f(0)} + \frac{1}{f(1)} + \frac{1}{f(2)}.
\]

Substitute the known values of \( f(0) \), \( f(1) \), and \( f(2) \):

\[
\sum_{i=0}^{2} \frac{1}{f(i)} = \frac{1}{3} + 1 + \frac{7}{3}.
\]

Combine the terms:

\[
\frac{1}{3} + 1 + \frac{7}{3} = \frac{1}{3} + \frac{3}{3} + \frac{7}{3} = \frac{11}{3}.
\]

Thus, the value of \( \sum_{i=0}^{2} \frac{1}{f(i)} \) is \( \frac{11}{3} \).

Would you like more details or have any questions?

Here are 5 related questions:
1. How would the value change if \( f(0) \) were different?
2. What happens if the functional equation changes to \( (2 - f(x+1))(4 + f(x)) = 12 \)?
3. How can we solve for \( f(x+1) \) in a more complex functional equation?
4. What are the properties of recursive functions like this one?
5. How can we approximate this sum for larger values of \( x \)?

**Tip**: Recursive functions often involve finding patterns. Solving for initial values helps identify how future values will behave.

If f(x) is a function that satisfies (3 − f(x+1))(6 + f(x)) = 18 and f(0) = 3, what is the value of ∑i=0^x (1/f(i))?

Explore how to solve the recursive function (3 − f(x+1))(6 + f(x)) = 18 with the initial condition f(0) = 3. Learn how to calculate the summation of the inverse function values step by step. Suitable for high school students in Grades 10-12.

Math Problem Statement

If f ( x ) f(x) is a function that satisfies ( 3 − f ( x + 1 ) ) ( 6 + f ( x ) )

18 (3−f(x+1))(6+f(x))=18 and f ( 0 )

3 f(0)=3, what is the value of ∑ i

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