Key Concepts and Formulas

• Critical Points: A critical point of a function $f(x)$ is a point $c$ where $f'(c) = 0$ or $f'(c)$ is undefined.
• Second Derivative Test: If $f'(c) = 0$ and $f''(c) > 0$, then $f(x)$ has a local minimum at $x = c$. If $f'(c) = 0$ and $f''(c) < 0$, then $f(x)$ has a local maximum at $x = c$.
• Local Minimum Value: The local minimum value of a function $f(x)$ at $x=c$ is $f(c)$, where $x=c$ is a local minimum point.

Step-by-Step Solution

Step 1: Analyze the Given Function and Derive its Derivatives

The function is given by: $f(x) = {x^3} - 3{x^2} - {{3f''(2)} \over 2}x + f''(1) \quad \text{(Equation 1)}$ We need to find the first and second derivatives of $f(x)$ to determine the constants $f''(2)$ and $f''(1)$.

Differentiate $f(x)$ with respect to $x$ to find $f'(x)$: $f'(x) = \frac{d}{dx}\left({x^3} - 3{x^2} - {{3f''(2)} \over 2}x + f''(1)\right)$ $f'(x) = 3{x^2} - 6x - {{3f''(2)} \over 2} \quad \text{(Equation 2)}$ Explanation: We apply the power rule of differentiation. The term $- {{3f''(2)} \over 2}x$ differentiates to $- {{3f''(2)} \over 2}$ because $f''(2)$ is a constant. The term $f''(1)$ is also a constant, so its derivative is 0.

Now, differentiate $f'(x)$ with respect to $x$ to find $f''(x)$: $f''(x) = \frac{d}{dx}\left(3{x^2} - 6x - {{3f''(2)} \over 2}\right)$ $f''(x) = 6x - 6 \quad \text{(Equation 3)}$ Explanation: Again, we apply the power rule. The term $- {{3f''(2)} \over 2}$ is a constant, so its derivative is 0.

Step 2: Determine the Values of the Unknown Constants, $f''(2)$ and $f''(1)$

We have $f''(x) = 6x - 6$. We can now use this to find the specific constant values:

1. Calculate $f''(2)$: Substitute $x=2$ into Equation 3: $f''(2) = 6(2) - 6$ $f''(2) = 12 - 6$ $f''(2) = 6$ Explanation: We substitute $x=2$ into the expression for $f''(x)$ because the problem defines a term involving $f''(2)$. This gives us the numerical value of that constant.

2. Calculate $f''(1)$: Substitute $x=1$ into Equation 3: $f''(1) = 6(1) - 6$ $f''(1) = 6 - 6$ $f''(1) = 0$ Explanation: Similarly, we substitute $x=1$ into $f''(x)$ to find the value of the constant term $f''(1)$ present in the original function definition.

Step 3: Rewrite the Function and Its First Derivative with Known Constants

Now that we have $f''(2) = 6$ and $f''(1) = 0$, we can substitute these values back into Equation 1 and Equation 2.

1. Rewrite $f(x)$ (Equation 1): $f(x) = {x^3} - 3{x^2} - {{3(6)} \over 2}x + 0$ $f(x) = {x^3} - 3{x^2} - 9x \quad \text{(Simplified Function)}$ Explanation: We substitute the calculated values of $f''(2)$ and $f''(1)$ into the original function definition. This gives us the complete and simplified form of the function $f(x)$ without any unknown constants.

2. Rewrite $f'(x)$ (Equation 2): $f'(x) = 3{x^2} - 6x - {{3(6)} \over 2}$ $f'(x) = 3{x^2} - 6x - 9 \quad \text{(Simplified First Derivative)}$ Explanation: We substitute the calculated value of $f''(2)$ into the first derivative expression. This simplified form of $f'(x)$ will be used to find the critical points.

Step 4: Find the Critical Points

To find the critical points, we set the first derivative $f'(x)$ to zero: $f'(x) = 0$ $3{x^2} - 6x - 9 = 0$ Explanation: Critical points are where the tangent to the curve is horizontal, meaning the rate of change is zero. These are potential locations for local maxima or minima.

We can simplify the quadratic equation by dividing by 3: $x^2 - 2x - 3 = 0$ Now, we factorize the quadratic equation: $(x - 3)(x + 1) = 0$ This gives us two critical points: $x = 3 \quad \text{or} \quad x = -1$ Explanation: We solve the quadratic equation to find the $x$-values where the slope of the tangent is zero. These are our candidates for local extrema.

Step 5: Apply the Second Derivative Test to Classify Critical Points

We use Equation 3, $f''(x) = 6x - 6$, to classify our critical points:

1. For $x = -1$: Substitute $x = -1$ into $f''(x)$: $f''(-1) = 6(-1) - 6$ $f''(-1) = -6 - 6$ $f''(-1) = -12$ Since $f''(-1) = -12 < 0$, the function has a local maximum at $x = -1$. Explanation: A negative second derivative at a critical point indicates concavity downwards, which corresponds to a local maximum.

2. For $x = 3$: Substitute $x = 3$ into $f''(x)$: $f''(3) = 6(3) - 6$ $f''(3) = 18 - 6$ $f''(3) = 12$ Since $f''(3) = 12 > 0$, the function has a local minimum at $x = 3$. Explanation: A positive second derivative at a critical point indicates concavity upwards, which corresponds to a local minimum.

The problem asks for the sum of all local minimum values. In this case, there is only one local minimum, which occurs at $x=3$.

Step 6: Calculate the Local Minimum Value

To find the local minimum value, we substitute the $x$-coordinate of the local minimum ($x=3$) into the simplified function $f(x)$ (from Step 3): $f(x) = {x^3} - 3{x^2} - 9x$ $f(3) = (3)^3 - 3(3)^2 - 9(3)$ $f(3) = 27 - 3(9) - 27$ $f(3) = 27 - 27 - 27$ $f(3) = -27$ Explanation: The question asks for the value of the local minimum, which is the $y$-coordinate of the point where the local minimum occurs. We find this by evaluating the function at the identified $x$-value.

Common Mistakes & Tips

• Don't forget to find the constants first: When a function is defined in terms of its derivatives at specific points, calculate those constants before proceeding.
• Remember the difference between point and value: The local minimum point is the x-coordinate where the minimum occurs, but the local minimum value is the y-coordinate at that point (i.e., the function value).
• Double-check your calculations: Errors in differentiation or substitution can lead to incorrect answers.

Summary

We found the first and second derivatives of the given function, determined the unknown constants, and then found the critical points by setting the first derivative to zero. Using the second derivative test, we classified these critical points as local maxima or minima. Finally, we calculated the function value at the local minimum point to find the local minimum value, which is -27.

The final answer is $\boxed{-27}$, which corresponds to option (A).