Key Concepts and Formulas

• Local Extrema: A function $f(x)$ has a local maximum or minimum at a point $c$ where $f'(c) = 0$. The second derivative test can be used to classify these points: if $f''(c) > 0$, then $c$ is a local minimum; if $f''(c) < 0$, then $c$ is a local maximum.
• Area Under a Curve: The area under the curve $y=f(x)$ between $x=a$ and $x=b$ is given by $\int_a^b |f(x)| dx$. If $f(x)$ changes sign on the interval $[a, b]$, the integral must be split into subintervals where $f(x)$ has a constant sign.
• Definite Integral: $\int_a^b f(x) dx = F(b) - F(a)$, where $F(x)$ is the antiderivative of $f(x)$.

Step-by-Step Solution

Step 1: Find the first derivative of f(x)

We are given $f(x) = 2x^3 - 3x^2 - 12x$. To find the local extrema, we first need to find the critical points by calculating the first derivative and setting it equal to zero. $f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x) = 6x^2 - 6x - 12$ The first derivative represents the slope of the tangent line to the function at any point $x$.

Step 2: Find the critical points by setting f'(x) = 0

To find the critical points, we set the first derivative equal to zero and solve for $x$. $6x^2 - 6x - 12 = 0$ Divide by 6: $x^2 - x - 2 = 0$ Factor the quadratic: $(x - 2)(x + 1) = 0$ This gives us the critical points $x = 2$ and $x = -1$.

Step 3: Find the second derivative of f(x)

To classify the critical points, we need to find the second derivative of $f(x)$. $f''(x) = \frac{d}{dx}(6x^2 - 6x - 12) = 12x - 6$ The second derivative tells us about the concavity of the function.

Step 4: Classify the critical points using the second derivative test

We evaluate the second derivative at each critical point:

• At $x = -1$: $f''(-1) = 12(-1) - 6 = -12 - 6 = -18$ Since $f''(-1) < 0$, $x = -1$ is a local maximum. Thus, $a = -1$.
• At $x = 2$: $f''(2) = 12(2) - 6 = 24 - 6 = 18$ Since $f''(2) > 0$, $x = 2$ is a local minimum. Thus, $b = 2$.

Step 5: Find the x-intercepts of f(x) within the interval [-1, 2]

We need to find the points where $f(x) = 0$ in the interval $[-1, 2]$. $2x^3 - 3x^2 - 12x = 0$ Factor out $x$: $x(2x^2 - 3x - 12) = 0$ So, one x-intercept is $x = 0$. The quadratic $2x^2 - 3x - 12 = 0$ has roots at $x = \frac{3 \pm \sqrt{9 + 96}}{4} = \frac{3 \pm \sqrt{105}}{4}$. The roots are approximately $x \approx -1.81$ and $x \approx 3.31$. The only x-intercept in the interval $[-1, 2]$ is $x = 0$.

Step 6: Set up the integral for the area

Since $f(x)$ changes sign at $x = 0$ within the interval $[-1, 2]$, we need to split the integral into two parts: $A = \int_{-1}^{2} |f(x)| \, dx = \int_{-1}^{0} f(x) \, dx + \int_{0}^{2} -f(x) \, dx$ On the interval $[-1, 0]$, $f(x) \ge 0$, and on the interval $[0, 2]$, $f(x) \le 0$.

Step 7: Find the antiderivative of f(x)

The antiderivative of $f(x) = 2x^3 - 3x^2 - 12x$ is $F(x) = \int (2x^3 - 3x^2 - 12x) \, dx = \frac{1}{2}x^4 - x^3 - 6x^2 + C$

Step 8: Evaluate the definite integrals

• $\int_{-1}^{0} f(x) \, dx = F(0) - F(-1) = (0) - (\frac{1}{2}(-1)^4 - (-1)^3 - 6(-1)^2) = 0 - (\frac{1}{2} + 1 - 6) = 0 - (-\frac{9}{2}) = \frac{9}{2}$
• $\int_{0}^{2} -f(x) \, dx = -(F(2) - F(0)) = -((\frac{1}{2}(2)^4 - (2)^3 - 6(2)^2) - 0) = -(\frac{1}{2}(16) - 8 - 6(4)) = -(8 - 8 - 24) = -(-24) = 24$

Step 9: Calculate the total area A

$A = \frac{9}{2} + 24 = \frac{9}{2} + \frac{48}{2} = \frac{57}{2}$

Step 10: Calculate 4A

$4A = 4 \cdot \frac{57}{2} = 2 \cdot 57 = 114$

Common Mistakes & Tips

• Remember to use the absolute value when integrating to find the area. If the function crosses the x-axis, you must split the integral into multiple parts and take the absolute value of each part.
• Be careful with signs when evaluating the antiderivative at the limits of integration.
• Double-check your calculations, especially when dealing with fractions.

Summary

We found the local maximum and minimum of the function $f(x) = 2x^3 - 3x^2 - 12x$ by finding the critical points and using the second derivative test. We then found the x-intercepts of the function and set up the integral to find the area between the curve, the x-axis, and the vertical lines $x = a$ and $x = b$. After evaluating the integral, we multiplied the area by 4 to get the final answer. The value of 4A is 114.

Final Answer

The final answer is \boxed{114}.