AP Calculus AB Exam Question:
Consider the function f defined by f(x) = x^2 - 3x + 2.
Find the x-coordinate(s) of the critical point(s) of f.
Find the value of x at which f achieves its minimum or maximum.
Find the value of f at the critical point(s) found in part 1.
Determine the area enclosed by the curve f(x), the x-axis, and the lines x = -1 and x = 4.
Note: You may use the Riemann sum or any other appropriate method to find the area.
Answer:
To find the x-coordinate(s) of the critical point(s), we need to find the derivative of f(x) and set it equal to zero.
Taking the derivative: f'(x) = 2x - 3.
Setting f'(x) = 0, we have: 2x - 3 = 0.
Solving for x, we find: x = 3/2.
Therefore, the critical point is at x = 3/2.
To find the value of x at which f achieves its minimum or maximum, we need to examine the concavity of the function. Since the second derivative of f(x) is positive for all values of x, the function is concave up. This means that the critical point is a local minimum.
Thus, x = 3/2 is the value at which f achieves its minimum.
To find the value of f at the critical point(s), we substitute x = 3/2 into the original function.
f(3/2) = (3/2)^2 - 3(3/2) + 2 = 9/4 - 9/2 + 2* = 1/4*.
Therefore, the value of f at the critical point x = 3/2 is 1/4.
To determine the area enclosed by the curve f(x), the x-axis, and the lines x = -1 and x = 4, we need to find the definite integral of f(x) over this interval.
The area can be calculated as follows:
∫[from -1 to 4] (x^2 - 3x + 2) dx = [1/3x^3 - 3/2x^2 + 2x] | -1 to 4
= (1/3(4)^3 - 3/2(4)^2 + 2(4)) - (1/3(-1)^3 - 3/2(-1)^2 + 2(-1))
= (64/3 - 48/2 + 8) - (-1/3 - 3/2 - 2)
= 66/3 + 15/2 + 2
= 22 + 7.5 + 2
= 31.5
The area enclosed by the curve f(x), the x-axis, and the lines x = -1 and x = 4 is 31.5 units squared.