Question:

Let f(x) be a differentiable function defined for all real numbers. The graph of f(x) intersects the x-axis at the points (-3, 0), (1, 0), and (4, 0). The derivative of f(x) is given by f'(x) = 3x^2 - 6x + 2.

(a) Determine the interval(s) where f(x) is increasing or decreasing and the x-coordinate(s) of any local maximum or minimum points of f(x).

(b) Find the x-coordinate(s) of any inflection point(s) of f(x).

(c) Find the area of the region bounded by the graph of f(x), the x-axis, and the vertical lines x = -3 and x = 4.

Answer:

(a)

To determine the intervals where f(x) is increasing or decreasing, we need to analyze the sign of the derivative, f'(x).

Setting f'(x) greater than zero gives us:

3x^2 - 6x + 2 > 0

We can simplify this inequality by factoring the quadratic expression:

(3x - 1)(x - 2) > 0

To find the sign of each factor, we can use a sign chart:

Interval	(3x - 1)	(x - 2)
x < 1/3	negative	negative
1/3 < x < 2	positive	negative
x > 2	positive	positive

From the sign chart, we can see that (3x - 1) and (x - 2) have opposite signs when 1/3 < x < 2. This means that f'(x) > 0 when 1/3 < x < 2. Therefore, f(x) is increasing on the interval (1/3, 2).

Next, we check for any local maximum or minimum points by analyzing the critical points of f(x). Critical points occur when the derivative is equal to zero or undefined. In this case, f'(x) is defined for all real numbers, so we need to find where f'(x) = 0:

3x^2 - 6x + 2 = 0

Using the quadratic formula, we get:

x = (6 ± √(36 - 24))/6 = (6 ± √12)/6 = 1 ± √3/3

So, we have two critical points: x = 1 - √3/3 and x = 1 + √3/3.

To determine whether these critical points correspond to a local maximum or minimum, we can analyze the sign of the derivative to the left and right of each point. Substituting test values into f'(x), we get:

f'(0) = 3(0)^2 - 6(0) + 2 = 2 (positive) f'(2) = 3(2)^2 - 6(2) + 2 = 2 (positive)

Since the sign of f'(x) does not change around the critical points, there are no local extrema.

Therefore, the interval where f(x) is increasing is (1/3, 2), and there are no local maximum or minimum points.

(b)

Inflection points occur when the concavity of the function changes, which corresponds to the sign change of the second derivative, f''(x). To find inflection points, we need to find where f''(x) = 0 or undefined.

Since f'(x) = 3x^2 - 6x + 2, the second derivative is obtained by differentiating f'(x):

f''(x) = (3x^2 - 6x + 2)' = 6x - 6

Setting f''(x) equal to zero gives us:

6x - 6 = 0

x = 1

So, x = 1 is the x-coordinate of the potential inflection point.

To determine whether it is a true inflection point, we can analyze the concavity to the left and right of the point by substituting test values into f''(x):

f''(0) = 6(0) - 6 = -6 (negative) f''(2) = 6(2) - 6 = 6 (positive)

Since the sign of f''(x) changes around x = 1, it is indeed an inflection point.

Therefore, the function f(x) has an inflection point at x = 1.

(c)

To find the area of the region bounded by the graph of f(x), the x-axis, and the vertical lines x = -3 and x = 4, we need to calculate the definite integral of f(x) over the interval [-3, 4]:

∫[from -3 to 4] f(x) dx

Using the antiderivative of f'(x), which is f(x), we have:

∫[from -3 to 4] f(x) dx = F(4) - F(-3)

To find F(x), we integrate f(x):

F(x) = ∫[from -3 to x] f(x) dx = ∫[from -3 to x] ( 3t^2 - 6t + 2) dt

Evaluating the integral, we get:

F(x) = t^3 - 3t^2 + 2t (from -3 to x)

F(x) = (x^3 - 3x^2 + 2x) - (-3^3 - 3(-3)^2 + 2(-3))

F(x) = (x^3 - 3x^2 + 2x) - (-27 + 27 - 6)

F(x) = x^3 - 3x^2 + 2x + 15

Now, substituting the limits of integration, we get:

F(4) - F(-3) = (4^3 - 3(4)^2 + 2(4) + 15) - (-3^3 - 3(-3)^2 + 2(-3) + 15)

= (64 - 48 + 8 + 15) - (-27 + 27 - 6 + 15)

= 39

Therefore, the area of the region bounded by the graph of f(x), the x-axis, and the vertical lines x = -3 and x = 4 is 39 square units.