Question:

Consider the functions f(x) = 2x^3 - 9x^2 + 12x and g(x) = x^2 - 4x + 3.

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

b) Determine the area between the graphs of f and g over the interval [0, 3].

c) A line is drawn perpendicular to the x-axis from the point of intersection (x-coordinate) found in part a) and partitions the area between the graphs into two regions. Find the ratio of the areas of the two regions.

Answer:

a) To find the points of intersection between the graphs of f and g, we need to set f(x) equal to g(x) and solve for x.

2x^3 - 9x^2 + 12x = x^2 - 4x + 3

2x^3 - 9x^2 + 12x - x^2 + 4x - 3 = 0

2x^3 - 10x^2 + 16x - 3 = 0

We can solve this equation either by factoring or by using a numerical method such as Newton's method. Let's solve it by factoring:

(2x - 1)(x^2 - 9x + 3) = 0

Setting each factor equal to zero:

2x - 1 = 0 => x = 1/2

x^2 - 9x + 3 = 0 => no real solutions

Therefore, the only point of intersection between the graphs of f and g is (1/2, f(1/2)).

b) To find the area between the graphs of f and g over the interval [0, 3], we need to find the definite integral of the absolute difference between the two functions over the given interval:

Area = ∫[0,3] |f(x) - g(x)| dx

We divide the interval [0, 3] into two sub-intervals: [0, 1/2] and [1/2, 3].

For the sub-interval [0, 1/2], f(x) > g(x), so the absolute difference is f(x) - g(x):

∫[0,1/2] (f(x) - g(x)) dx = ∫[0,1/2] (2x^3 - 9x^2 + 12x - (x^2 - 4x + 3)) dx

= ∫[0,1/2] (2x^3 - 8x^2 + 16x - 3) dx

= [(1/2)x^4 - (8/3)x^3 + 8x^2 - 3x] [0,1/2]

= [(1/2)(1/16) - (8/3)(1/8) + 8(1/4) - 3(1/2)] - [0]

= 1/32 - 1/3 + 2 - 3/2

= -7/96

For the sub-interval [1/2, 3], g(x) > f(x), so the absolute difference is g(x) - f(x):

∫[1/2,3] (g(x) - f(x)) dx = ∫[1/2,3] ((x^2 - 4x + 3) - (2x^3 - 9x^2 + 12x)) dx

= ∫[1/2,3] (-2x^3 + 9x^2 - 16x + 3) dx

= [(-1/2)x^4 + (3/4)x^3 - 8x^2 + 3x] [1/2,3]

= [(-1/2)(81/16) + (3/4)(81/8) - 8(9) + 3(3)] - [(-1/2)(1/16) + (3/4)(1/8) - 8(1/4) + 3(1/2)]

= -27/32 + 81/32 - 72 + 9 + 1/32 + 3/32 - 2 + 3/2

= -47/32 + 81/32 - 74/32 + 6/2

= -40/32 + 6/2

= -1/4 + 3

= 11/4

Therefore, the area between the graphs of f and g over the interval [0, 3] is (-7/96) + (11/4) = 44/96 = 11/24.

c) The line that is perpendicular to the x-axis from the point of intersection (x-coordinate 1/2) splits the area between the graphs into two regions: a triangle and a region bounded by a curve. Let's denote the area of the triangular region as A1 and the area of the bounded region as A2.

To find A1, we need to calculate the area of the triangle formed by the x-axis, the line perpendicular to the x-axis, and the vertical segment from (1/2, f(1/2)) to the x-axis.

The height of the triangle is |f(1/2) - 0| = |f(1/2)| = |2(1/2)^3 - 9(1/2)^2 + 12(1/2)| = |1/4 - 9/4 + 6| = 3.

The width of the triangle is |1/2 - 0| = 1/2.

Therefore, A1 = (1/2)(3) = 3/2.

To find A2, we can subtract A1 from the total area we found in part b):

A2 = Area between the graphs - A1 = 11/24 - 3/2 = 11/24 - 36/24 = -25/24.

However, since areas cannot be negative, we take the absolute value and use the positive value: A2 = | -25/24 | = 25/24.

Therefore, the ratio of the areas of the two regions is A1 : A2 = 3/2 : 25/24.

To simplify the ratio, we multiply both the numerator and denominator of the second fraction by 2:

A1 : A2 = 3/2 : (25/12) = 36/24 : 25/12 = 3/2 : 25/12 = 3/2 : 25/12 = 18 : 25.

Hence, the ratio of the areas of the two regions is 18 : 25.