Key Concepts and Formulas:

• Local Extrema: A function $f(x)$ has a local minimum at $x=c$ if $f'(c) = 0$ and $f''(c) > 0$.
• Inflection Point: A point where $f''(x) = 0$ and $f''(x)$ changes sign. If $f'(x)$ has a local minimum at $x=c$, then $f''(c) = 0$.
• Polynomial Representation: A cubic polynomial can be represented as $f(x) = ax^3 + bx^2 + cx + d$.

Step-by-Step Solution:

Step 1: Define the cubic polynomial and its derivatives.

We represent the cubic polynomial as: $f(x) = ax^3 + bx^2 + cx + d$ Why this form? This is the general form of a cubic polynomial, where $a, b, c,$ and $d$ are coefficients we need to determine using the given conditions.

We find the first and second derivatives: $f'(x) = 3ax^2 + 2bx + c$ $f''(x) = 6ax + 2b$

Step 2: Apply the condition "f'(x) has a local minima at x = -1".

Since $f'(x)$ has a local minimum at $x = -1$, we have $f''(-1) = 0$. Why this step? A local extremum of $f'(x)$ implies that its derivative $f''(x)$ is zero at that point. This gives us an equation relating $a$ and $b$. $f''(-1) = 6a(-1) + 2b = 0$ $-6a + 2b = 0$ $b = 3a \quad \text{(Equation 1)}$

Self-check/Tip: For $f'(x)$ to have a local minimum at $x=-1$, we also need $f'''(-1) > 0$. Since $f'''(x) = 6a$, this implies $a > 0$. We'll keep this in mind.

Step 3: Apply the condition "f(x) has a local minima at x = 1".

Since $f(x)$ has a local minimum at $x = 1$, we have $f'(1) = 0$. Why this step? A local extremum of $f(x)$ implies that its derivative $f'(x)$ is zero at that point. This gives us another equation relating $a, b,$ and $c$. $f'(1) = 3a(1)^2 + 2b(1) + c = 0$ $3a + 2b + c = 0$ Substitute $b = 3a$ from Equation 1: $3a + 2(3a) + c = 0$ $9a + c = 0$ $c = -9a \quad \text{(Equation 2)}$

Self-check/Tip: For $f(x)$ to have a local minimum at $x=1$, we also need $f''(1) > 0$. $f''(1) = 6a(1) + 2b = 6a + 6a = 12a > 0$, which implies $a > 0$. This is consistent with our earlier finding.

Step 4: Use the given function values f(1) = -10 and f(-1) = 6.

We use these values to create two more equations involving $a, b, c,$ and $d$. Why this step? We now have expressions for $b$ and $c$ in terms of $a$. We need two more equations to determine $a$ and $d$.

First, use $f(1) = -10$: $f(1) = a(1)^3 + b(1)^2 + c(1) + d = -10$ $a + b + c + d = -10$ Substitute $b = 3a$ and $c = -9a$: $a + 3a - 9a + d = -10$ $-5a + d = -10 \quad \text{(Equation 3)}$

Next, use $f(-1) = 6$: $f(-1) = a(-1)^3 + b(-1)^2 + c(-1) + d = 6$ $-a + b - c + d = 6$ Substitute $b = 3a$ and $c = -9a$: $-a + 3a - (-9a) + d = 6$ $11a + d = 6 \quad \text{(Equation 4)}$

Step 5: Solve the system of equations for a, b, c, d.

We now have two equations with two variables, $a$ and $d$. Why this step? Solving this system will give us the values of $a$ and $d$, allowing us to find $b$ and $c$.

Subtract Equation 3 from Equation 4: $(11a + d) - (-5a + d) = 6 - (-10)$ $16a = 16$ $a = 1$

Substitute $a = 1$ into Equation 3: $-5(1) + d = -10$ $d = -5$

Now find $b$ and $c$: $b = 3a = 3(1) = 3$ $c = -9a = -9(1) = -9$

So we have $a = 1, b = 3, c = -9, d = -5$. Self-check: As predicted, $a=1 > 0$, satisfying the conditions for local minima.

Step 6: Construct f(x) and calculate f(3).

Now we can write the complete polynomial and evaluate it at $x = 3$. Why this step? With all the coefficients determined, we can find the required value of $f(3)$.

$f(x) = x^3 + 3x^2 - 9x - 5$ $f(3) = (3)^3 + 3(3)^2 - 9(3) - 5$ $f(3) = 27 + 27 - 27 - 5$ $f(3) = 22$

The final answer is 22.

Tips and Common Mistakes:

1. Derivative Errors: Double-check the derivatives. Incorrect derivatives will lead to wrong equations and an incorrect solution.
2. Understanding Local Minima of Derivatives: Remember that if $f'(x)$ has a local minimum, then $f''(x) = 0$ at that point.
3. Sign Errors: Be careful with signs when solving the system of equations. A small sign error can lead to a completely different result.

Summary:

We found the cubic polynomial $f(x)$ by using the given conditions about its local minima and the local minima of its derivative, along with the function values at $x=1$ and $x=-1$. We set up a system of equations to solve for the coefficients of the polynomial and then evaluated $f(3)$. The final answer is 22.

Final Answer: The final answer is $\boxed{22}$.