AP Calculus AB Exam Question:

Consider the following functions defined on the interval [0, 4]:

f(x) = x^3 - 4x^2 + 3x + 2

g(x) = x^2 - 2x + 1

(a) Find the x-coordinates of all points of intersection between the graphs of f(x) and g(x).

(b) Find the area of the region enclosed between the graphs of f(x) and g(x) on the interval [0, 4].

Step-by-Step Solution:

(a) To find the x-coordinates of the points of intersection, we need to solve the equation f(x) = g(x). This can be done by setting the two equations equal to each other:

x^3 - 4x^2 + 3x + 2 = x^2 - 2x + 1

Rearranging the equation, we get:

x^3 - 5x^2 + 5x + 1 = 0

To solve this equation, we can use synthetic division or graphing calculator. By using synthetic division, we find that x = 1 is a solution. Performing synthetic division:

1  |  1  -5   5   1
   |     1  -4   1
---------------
     1  -4   1   2

The remainder is 2, which is not equal to zero. Therefore, x = 1 is not a root.

To find the other two roots, we can either factor the cubic equation or use numerical methods. In this case, we can observe that the graph of the cubic function intersects the x-axis at approximately x = 0.4046 and x = 4.5954 (using a graphing calculator or software).

Hence, the x-coordinates of all points of intersection between the graphs of f(x) and g(x) are approximately x = 0.4046 and x = 4.5954.

(b) To find the area between the curves, we integrate the absolute difference of the two functions over the interval [0, 4]. The area between the curves can be calculated using the formula:

Area = ∫[0, 4] |f(x) - g(x)| dx

Since f(x) is greater than or equal to g(x) for all x in the interval [0, 4], we can rewrite the integral as:

Area = ∫[0, 4] (f(x) - g(x)) dx

Substituting the given functions, we have:

Area = ∫[0, 4] [(x^3 - 4x^2 + 3x + 2) - (x^2 - 2x + 1)] dx

Simplifying the expression inside the integral:

Area = ∫[0, 4] (x^3 - 4x^2 + 3x + 2 - x^2 + 2x - 1) dx

Area = ∫[0, 4] (x^3 - 5x^2 + 5x + 1) dx

Integrating term by term, we get:

Area = [1/4 x^4 - (5/3) x^3 + (5/2) x^2 + x] evaluated from 0 to 4

Plugging in the upper limit of integration, we have:

Area = [1/4 (4)^4 - (5/3) (4)^3 + (5/2) (4)^2 + 4]

Area = [64/4 - 320/3 + 80 + 4]

Area = 16 - 320/3 + 80 + 4

Area = 36.6667

Therefore, the area of the region enclosed between the graphs of f(x) and g(x) on the interval [0, 4] is approximately 36.6667 square units.