To find the local maximums and minimums, as well as the intervals of increase and decrease, we will first need to find the critical points of the function. Critical points occur where the derivative is either zero or undefined.
Let's find the critical points by setting the numerator of the derivative equal to zero:
2x^2 - 5x - 12 = 0
This equation can be factored as:
(2x + 3)(x - 4) = 0
Setting each factor equal to zero gives us two possible critical points:
2x + 3 = 0 --> x = -3/2
x - 4 = 0 --> x = 4
Now let's find the values of x where the denominator of the derivative is equal to zero:
3x^2 - 7x + 2 = 0
Again, we can factor this equation:
(3x - 1)(x - 2) = 0
Setting each factor equal to zero gives us two more possible critical points:
3x - 1 = 0 --> x = 1/3
x - 2 = 0 --> x = 2
So, we have obtained four possible critical points: x = -3/2, 1/3, 2, and 4.
To construct the sign diagram, we will analyze the sign of the derivative at different intervals between these critical points. We'll use these intervals to determine where the function is increasing or decreasing, as well as any local maximum or minimum points.
Interval 1: (-∞, -3/2)
Choose x = -4 as a test value. Substituting it into the derivative gives a positive value:
f'(-4) = (2(-4)^2 - 5(-4) - 12) / (3(-4)^2 - 7(-4) + 2) = 42/26 > 0
Since the derivative is positive in this interval, f(x) is increasing.
Interval 2: (-3/2, 1/3)
Choose x = 0 as a test value. Substituting it into the derivative gives a negative value:
f'(0) = (2(0)^2 - 5(0) - 12) / (3(0)^2 - 7(0) + 2) = -12/2 < 0
Since the derivative is negative in this interval, f(x) is decreasing.
Interval 3: (1/3, 2)
Choose x = 1 as a test value. Substituting it into the derivative gives a positive value:
f'(1) = (2(1)^2 - 5(1) - 12) / (3(1)^2 - 7(1) + 2) = -3/4 > 0
Since the derivative is positive in this interval, f(x) is increasing.
Interval 4: (2, 4)
Choose x = 3 as a test value. Substituting it into the derivative gives a negative value:
f'(3) = (2(3)^2 - 5(3) - 12) / (3(3)^2 - 7(3) + 2) = -3/32 < 0
Since the derivative is negative in this interval, f(x) is decreasing.
Interval 5: (4, +∞)
Choose x = 5 as a test value. Substituting it into the derivative gives a positive value:
f'(5) = (2(5)^2 - 5(5) - 12) / (3(5)^2 - 7(5) + 2) = 53/96 > 0
Since the derivative is positive in this interval, f(x) is increasing.
Taking all this information into account, we can summarize the results:
Local maximums occur at x = -3/2 and x = 2.
Local minimums occur at x = 1/3 and x = 4.
The function is increasing on the intervals (-∞, -3/2), (1/3, 2), and (4, +∞).
The function is decreasing on the intervals (-3/2, 1/3) and (2, 4).
For part b, we can graph a simple example function that satisfies these conditions. A quadratic function of the form f(x) = ax^2 + bx + c, where a, b, and c are constants, would work. We can choose appropriate values for a, b, and c to create a graph that reflects the characteristics we found in part a.