Goal: Complete the table below.
| \(x\) | \(f(x)\) |
|---|---|
| 1 | 3 |
| 2 | 8 |
| 3 | 15 |
| 4 | 24 |
| 5 | 35 |
| 100 | 10,200 |
Subgoal: Create a polynomial function by looking at differences and differences of differences.
| 2nd differences | differences | \( ax^2 + bx + c \) | \(x\) | \( f(x)\) | differences | 2nd differences | |
|---|---|---|---|---|---|---|---|
| \( a + b + c \) | 1 | 3 | |||||
| \( 3a + b \) | 5 | ||||||
| \( 2a \) | \( a \cdot 4 + b \cdot 2+ c \) | 2 | 8 | 2 | |||
| \( 5a + b \) | 7 | ||||||
| \( 2a \) | \( a \cdot 9 + b \cdot 3 + c \) | 3 | 15 | 2 | |||
| \( 7a + b \) | 9 | ||||||
| \( 2a \) | \( a \cdot 16 + b \cdot 4 + c \) | 4 | 24 | 2 | |||
| \( 9a + b \) | 11 | ||||||
| \( a \cdot 25 + b \cdot 5 + c \) | 5 | 35 | |||||
Notice that the 2nd differences are repeating. This is why a quadratic equation \(a x^2 + b x + c\) works in this case.
Substitution into \( a \cdot x^2 + b \cdot x + c = f(x) \)
\( a \cdot 1^2 + b \cdot 1 + c = 3 = f(1)\)
\( a \cdot 2^2 + b \cdot 2 + c = 8 = f(2) \)
\( a \cdot 2^3 + b \cdot 3 + c = 15 = f(3) \)
Subtraction: \(f(3) - f(2)\)
\(\begin{align} 9a + 3b + c =& 15 \\ -4a - 2b - c =& {-8} \\ \hline 5a + b =& 7 \end{align} \)
Subtraction: \(f(2) - f(1)\)
\(\begin{align*} 4a + 2b + c =& 8 \\ -a - b - c =& {-3} \\ \hline 3a + b =& 5 \end{align*} \)
Subtraction: difference1 - difference2
\(\begin{align*} 5a + b =& 7 \\ -3a - b =& {-5} \\ \hline 2a =& 2 \end{align*} \)
Solve for \(a\)
\(\begin{align*} 2a =& 2 \\ a =& 1 \end{align*} \)
Solve for \(b\)
\(\begin{align*} 3 a + b =& 5 \\ 3 \cdot 1 + b =& 5 \\ 3 + b =& 5 \\ b =& 2 \end{align*} \)
Solve for \(c\)
\(\begin{align*} a + b + c =& 3 \\ 1 + 2 + c =& 3 \\ 3 + c =& 3 \\ c =& 0 \end{align*} \)
Formula: \( f(x) = 1 x^2 + 2 x + 0 = x^2 + 2x \)
Answer: \( f(100) = 100^2 + 2 \cdot 100 = 10{,}200 \)
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