Question

Use the first derivative test to locate the relative extrema of the function in the given domain, and determine the intervals of increase and decrease. f(x)=3x²−12x+3 with domain [0, 3].

A) Relative maximum at x=1, increasing on [0, 1] and decreasing on [1, 3]
B) Relative minimum at x=2, increasing on [0, 2] and decreasing on [2, 3]
C) Relative maximum at x=3, increasing on [0, 3]
D) None of the above

Answer 1

Using the first derivative test, the function f(x)=3x²-12x+3 has a relative minimum at x=2, increases on [0, 2], and decreases on [2, 3], which corresponds to option B.

Step-by-step explanation:

To find the relative extrema and the intervals where the function f(x) = 3x² - 12x + 3 is increasing or decreasing, we must first calculate its first derivative f′(x) and then apply the first derivative test. The first derivative of f(x) is f′(x) = 6x - 12. Setting f′(x) equal to zero, we solve for x to find the critical points: 6x - 12 = 0 which gives x = 2.

Next, we test the intervals around the critical point within the domain [0, 3]. For x < 2, say at x=1, f′(x) is positive, implying the function is increasing. For x > 2, say at x=3, f′(x) is negative, indicating the function is decreasing. Thus, there is a relative maximum at x=2 and the function increases on [0, 2] and decreases on [2, 3]. This matches option B.