Question:

Let f(x) = x^2 - 3x + 2 and g(x) = 2x - 4 be two functions defined on the closed interval [0, 3].

1. Find the x-coordinate(s) of the point(s) of intersection of f(x) and g(x).

2. Determine the area enclosed between the graphs of f(x) and g(x) on the interval [0, 3].

Answer:

1. To find the x-coordinate(s) of the point(s) of intersection, we need to set f(x) = g(x) and solve for x.

   Setting x^2 - 3x + 2 = 2x - 4, we have:

   x^2 - 5x + 6 = 0
   

   Factoring the quadratic equation, we get:

   (x - 2)(x - 3) = 0
   

   Therefore, x - 2 = 0 or x - 3 = 0.

   Solving each equation, we find that x = 2 or x = 3.

   Thus, the point(s) of intersection are (2, f(2)) and (3, f(3)).

2. To find the area enclosed between the graphs of f(x) and g(x), we need to find the definite integral of |f(x) - g(x)| over the interval [0, 2] and add it to the definite integral of |g(x) - f(x)| over the interval [2, 3].

   The absolute difference between f(x) and g(x) is given by |f(x) - g(x)|:

   |f(x) - g(x)| = |(x^2 - 3x + 2) - (2x - 4)|
                 = |x^2 - 3x + 2 - 2x + 4|
                 = |x^2 - 5x + 6|
   

   Therefore, the area enclosed between the graphs of f(x) and g(x) is:

   A = ∫[0, 2] |f(x) - g(x)| dx + ∫[2, 3] |g(x) - f(x)| dx
   

   Evaluating the first integral, we have:

   ∫[0, 2] |x^2 - 5x + 6| dx = ∫[0, 2] (x^2 - 5x + 6) dx     (since the integrand is non-negative on [0, 2])
                            = ∫[0, 2] (x^2 - 5x + 6) dx
                            = [x^3/3 - (5x^2)/2 + 6x] |[0, 2]
                            = [2^3/3 - (5(2)^2)/2 + 6(2)] - [0^3/3 - (5(0)^2)/2 + 6(0)]
                            = 8/3 - 20/2 + 12 - 0/3 + 0 + 0
                            = 8/3 - 20/2 + 12
                            = (-32 + 60 - 72)/6
                            = (-44)/6
                            = -22/3
   

   Evaluating the second integral, we have:

   ∫[2, 3] |x^2 - 5x + 6| dx = ∫[2, 3] (5x - x^2 - 6) dx     (since the integrand is non-negative on [2, 3])
                            = ∫[2, 3] (5x - x^2 - 6) dx
                            = [5x^2/2 - (x^3)/3 - 6x] |[2, 3]
                            = [3^3/2 - (3(3)^2)/3 - 6(3)] - [2^3/2 - (2(2)^2)/3 - 6(2)]
                            = 27/2 - 27/3 - 18 - 8/2 + 8/3 + 12
                            = (54 - 54 - 108 - 16 + 16 + 36)/6
                            = (-128)/6
                            = -64/3
   

   Adding both integrals together, we get:

   A = -22/3 + (-64/3) = -86/3
   

   The area enclosed between the graphs of f(x) and g(x) on the interval [0, 3] is therefore -86/3 square units.