1. **Problem 3:** Find the cubic function passing through points $(-1,3)$, $(0,2)$, and $(1,1)$. 2. A cubic function has the form $$f(x) = ax^3 + bx^2 + cx + d.$$ 3. Use the given points to create equations: - For $(-1,3)$: $$a(-1)^3 + b(-1)^2 + c(-1) + d = 3 \Rightarrow -a + b - c + d = 3.$$ - For $(0,2)$: $$a(0)^3 + b(0)^2 + c(0) + d = 2 \Rightarrow d = 2.$$ - For $(1,1)$: $$a(1)^3 + b(1)^2 + c(1) + d = 1 \Rightarrow a + b + c + d = 1.$$ 4. Substitute $d=2$ into the other equations: - $-a + b - c + 2 = 3 \Rightarrow -a + b - c = 1.$ - $a + b + c + 2 = 1 \Rightarrow a + b + c = -1.$ 5. Add the two equations: $$(-a + b - c) + (a + b + c) = 1 + (-1) \Rightarrow 2b = 0 \Rightarrow b = 0.$$ 6. Substitute $b=0$ back: - $-a - c = 1$ - $a + c = -1$ 7. Add these two: $$(-a - c) + (a + c) = 1 + (-1) \Rightarrow 0 = 0,$$ which is true but does not help. Subtract the second from the first: $$(-a - c) - (a + c) = 1 - (-1) \Rightarrow -2a - 2c = 2 \Rightarrow -a - c = 1.$$ 8. From step 6, $-a - c = 1$ is consistent. Let’s express $c$ in terms of $a$: $$c = -a - 1.$$ 9. The cubic function is: $$f(x) = ax^3 + 0 imes x^2 + (-a - 1)x + 2 = ax^3 - (a + 1)x + 2.$$ 10. Since we have only three points, infinitely many cubic functions fit. To find a unique solution, more points or conditions are needed. However, the problem likely expects the simplest cubic passing through these points. Let’s try $a=0$: $$f(x) = -1 imes x + 2 = -x + 2.$$ Check if it fits all points: - At $x=-1$: $f(-1) = -(-1) + 2 = 1 + 2 = 3$ ✓ - At $x=0$: $f(0) = 0 + 2 = 2$ ✓ - At $x=1$: $f(1) = -1 + 2 = 1$ ✓ This is a linear function, not cubic. Since the problem asks for cubic, we must accept the general form with $a$ arbitrary. **Final answer for problem 3:** $$f(x) = ax^3 - (a + 1)x + 2, \text{ where } a \text{ is any real number}.$$