Key Concepts and Formulas

• Local Extrema: A function $f(x)$ has a local maximum or minimum at a point $x=c$ if $f'(c)=0$. These points are called critical points.
• First Derivative Test: Critical points are found by solving $f'(x) = 0$.
• Second Derivative Test: If $f'(c) = 0$ and $f''(c) < 0$, then $f(x)$ has a local maximum at $x=c$. If $f'(c) = 0$ and $f''(c) > 0$, then $f(x)$ has a local minimum at $x=c$.

Step 1: Find the First Derivative and Critical Points

We are given the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$. We need to find where the local maximum and minimum occur, which means finding the critical points. This is done by finding the first derivative, $f'(x)$, and setting it equal to zero.

$f'(x) = \frac{d}{dx}(2x^3 - 9ax^2 + 12a^2x + 1)$ Using the power rule, $\frac{d}{dx}(x^n) = nx^{n-1}$, we get: $f'(x) = 6x^2 - 18ax + 12a^2$ Now, set $f'(x) = 0$ to find the critical points: $6x^2 - 18ax + 12a^2 = 0$ Divide by 6: $x^2 - 3ax + 2a^2 = 0$ Factor the quadratic: $(x - a)(x - 2a) = 0$ Thus, the critical points are $x = a$ and $x = 2a$.

Step 2: Classify the Critical Points Using the Second Derivative Test

To determine whether the critical points $x=a$ and $x=2a$ are local maxima or minima, we need to find the second derivative, $f''(x)$.

$f''(x) = \frac{d}{dx}(6x^2 - 18ax + 12a^2)$ $f''(x) = 12x - 18a$

Now, evaluate $f''(x)$ at each critical point:

• For $x = a$: $f''(a) = 12(a) - 18a = -6a$ Since $a > 0$, $f''(a) < 0$, which means $x = a$ is a local maximum. Therefore, $p = a$.

• For $x = 2a$: $f''(2a) = 12(2a) - 18a = 24a - 18a = 6a$ Since $a > 0$, $f''(2a) > 0$, which means $x = 2a$ is a local minimum. Therefore, $q = 2a$.

Step 3: Use the Given Condition $p^2 = q$ to Solve for $a$

We are given that $p^2 = q$. We found that $p = a$ and $q = 2a$. Substitute these into the given equation: $a^2 = 2a$ $a^2 - 2a = 0$ $a(a - 2) = 0$ This gives us two possible solutions for $a$: $a = 0$ or $a = 2$. However, we are given that $a > 0$, so we must have $a = 2$.

Step 4: Calculate $f(3)$

We are asked to find $f(3)$. We know that $a = 2$, so we can write the function as: $f(x) = 2x^3 - 9(2)x^2 + 12(2)^2x + 1$ $f(x) = 2x^3 - 18x^2 + 48x + 1$ Now, substitute $x = 3$ into the function: $f(3) = 2(3)^3 - 18(3)^2 + 48(3) + 1$ $f(3) = 2(27) - 18(9) + 48(3) + 1$ $f(3) = 54 - 162 + 144 + 1$ $f(3) = 37$

Common Mistakes & Tips

• Remember to factor the first derivative completely to find all critical points.
• Be careful with the signs when applying the second derivative test. A negative second derivative indicates a local maximum, while a positive second derivative indicates a local minimum.
• Always check if the solution for $a$ satisfies the given condition $a>0$.

Summary

We found the critical points of the function by setting the first derivative equal to zero. We then used the second derivative test to classify these points as local maxima or minima. Using the given relationship between the location of the local maximum and minimum, $p^2 = q$, we solved for the value of $a$. Finally, we substituted the value of $a$ back into the original function and evaluated it at $x = 3$ to find $f(3)$.

The final answer is \boxed{37}, which corresponds to option (B).