Key Concepts and Formulas

• Derivatives and Critical Points: A critical point of a function $f(x)$ occurs where $f'(x) = 0$ or is undefined. At a critical point $x=c$, if $f''(c) > 0$, then $f(x)$ has a local minimum at $x=c$.
• Polynomial Functions: A polynomial of degree 3 has the general form $f(x) = ax^3 + bx^2 + cx + d$, where $a \neq 0$.
• System of Equations: Solving a system of linear equations allows us to determine the coefficients of the polynomial.

Step-by-Step Solution

Step 1: Define the polynomial and its derivatives.

Let the cubic polynomial be $f(x) = ax^3 + bx^2 + cx + d$, where $a \neq 0$. We need to find its first and second derivatives.

$f'(x) = 3ax^2 + 2bx + c$ $f''(x) = 6ax + 2b$

Step 2: Use the given conditions to form equations.

We have four conditions: $f(-1) = 10$, $f(1) = -6$, $f'(-1) = 0$, and $f''(1) = 0$. Let's translate these into equations.

• $f(-1) = 10$: $a(-1)^3 + b(-1)^2 + c(-1) + d = 10$ $-a + b - c + d = 10 \quad (1)$

• $f(1) = -6$: $a(1)^3 + b(1)^2 + c(1) + d = -6$ $a + b + c + d = -6 \quad (2)$

• $f'(-1) = 0$: $3a(-1)^2 + 2b(-1) + c = 0$ $3a - 2b + c = 0 \quad (3)$

• $f''(1) = 0$: $6a(1) + 2b = 0$ $6a + 2b = 0 \quad (4)$

Step 3: Solve the system of equations.

From equation (4), we have: $2b = -6a$ $b = -3a \quad (5)$

Substitute (5) into (3): $3a - 2(-3a) + c = 0$ $3a + 6a + c = 0$ $9a + c = 0$ $c = -9a \quad (6)$

Add equations (1) and (2): $(-a + b - c + d) + (a + b + c + d) = 10 + (-6)$ $2b + 2d = 4$ $b + d = 2 \quad (7)$

Substitute (5) into (7): $-3a + d = 2$ $d = 2 + 3a \quad (8)$

Now substitute (5), (6), and (8) into (1): $-a + (-3a) - (-9a) + (2 + 3a) = 10$ $-a - 3a + 9a + 2 + 3a = 10$ $8a + 2 = 10$ $8a = 8$ $a = 1$

Now we can find $b, c, d$: $b = -3a = -3(1) = -3$ $c = -9a = -9(1) = -9$ $d = 2 + 3a = 2 + 3(1) = 5$

Thus, $f(x) = x^3 - 3x^2 - 9x + 5$.

Step 4: Find the first and second derivatives with the solved coefficients.

$f(x) = x^3 - 3x^2 - 9x + 5$ $f'(x) = 3x^2 - 6x - 9$ $f''(x) = 6x - 6$

Step 5: Find the critical points of f(x).

Set $f'(x) = 0$: $3x^2 - 6x - 9 = 0$ $x^2 - 2x - 3 = 0$ $(x - 3)(x + 1) = 0$ $x = 3, x = -1$ So, the critical points are $x = 3$ and $x = -1$.

Step 6: Use the second derivative test to determine local minima.

We need to check the sign of $f''(x)$ at the critical points.

• At $x = -1$: $f''(-1) = 6(-1) - 6 = -12 < 0$ So, $f(x)$ has a local maximum at $x = -1$.

• At $x = 3$: $f''(3) = 6(3) - 6 = 18 - 6 = 12 > 0$ So, $f(x)$ has a local minimum at $x = 3$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when substituting values into equations. A single sign error can propagate through the entire solution.
• Derivative Errors: Double-check your derivatives. Incorrect derivatives will lead to incorrect critical points and incorrect application of the second derivative test.
• Systematic Solving: Solve the system of equations systematically. Use substitution to reduce the number of variables and avoid getting lost in the algebra.

Summary

We started by defining the cubic polynomial and its derivatives. Then, we used the given conditions to form a system of four linear equations. Solving this system, we found the coefficients of the polynomial. Finally, we used the first and second derivatives to find the critical points and determine that the polynomial has a local minimum at $x = 3$.

Final Answer

The final answer is \boxed{3}.