1) Function f(x) whose derivative is f′(x)=(x−1)^2 (x+5).
a) To find the critical points of f, we need to find the values of x where f'(x) = 0 or f'(x) is undefined. In this case, f'(x) is defined for all real numbers, so we need to find where f'(x) = 0.
Setting f'(x) = 0:
(x−1)^2 (x+5) = 0
Solving for x, we get x = 1 and x = -5.
Therefore, the critical points of f are x = 1 and x = -5.
b) To determine where f is increasing or decreasing, we need to analyze the sign of f'(x) on different intervals. We can use the critical points as reference points.
For x < -5, f'(x) > 0, so f is increasing on this interval.
For -5 < x < 1, f'(x) < 0, so f is decreasing on this interval.
For x > 1, f'(x) > 0, so f is increasing on this interval.
Therefore, f is increasing on the intervals (-∞, -5) and (1, ∞), and it is decreasing on the interval (-5, 1).
c) To find the local maximum and minimum values, we need to examine the behavior of f around the critical points and look for points of inflection.
At x = 1, f changes from decreasing to increasing, indicating a local minimum.
At x = -5, f changes from increasing to decreasing, indicating a local maximum.
Therefore, f assumes a local minimum at x = 1 and a local maximum at x = -5.
2) Function f(x) whose derivative is f′(x)=(x−3)(x+1)(x−6).
a) To find the critical points of f, we need to find the values of x where f'(x) = 0 or f'(x) is undefined. In this case, f'(x) is defined for all real numbers, so we need to find where f'(x) = 0.
Setting f'(x) = 0:
(x−3)(x+1)(x−6) = 0
Solving for x, we get x = 3, x = -1, and x = 6.
Therefore, the critical points of f are x = 3, x = -1, and x = 6.
b) To determine where f is increasing or decreasing, we need to analyze the sign of f'(x) on different intervals. We can use the critical points as reference points.
For x < -1, f'(x) > 0, so f is increasing on this interval.
For -1 < x < 3, f'(x) < 0, so f is decreasing on this interval.
For x > 3, f'(x) > 0, so f is increasing on this interval.
Therefore, f is increasing on the intervals (-∞, -1) and (3, ∞), and it is decreasing on the interval (-1, 3).
c) To find the local maximum and minimum values, we need to examine the behavior of f around the critical points and look for points of inflection.
At x = 3, f changes from decreasing to increasing, indicating a local minimum.
At x = -1,
f changes from increasing to decreasing, indicating a local maximum.
Therefore, f assumes a local minimum at x = 3 and a local maximum at x = -1.
3) Function f(x) whose derivative is f'(x)=x(x+3).
a) To find the critical points of f, we need to find the values of x where f'(x) = 0 or f'(x) is undefined. In this case, f'(x) is defined for all real numbers, so we need to find where f'(x) = 0.
Setting f'(x) = 0:
x(x+3) = 0
Solving for x, we get x = 0 and x = -3.
Therefore, the critical points of f are x = 0 and x = -3.
b) To determine where f is increasing or decreasing, we need to analyze the sign of f'(x) on different intervals. We can use the critical points as reference points.
For x < -3, f'(x) < 0, so f is decreasing on this interval.
For -3 < x < 0, f'(x) > 0, so f is increasing on this interval.
For x > 0, f'(x) > 0, so f is increasing on this interval.
Therefore, f is increasing on the intervals (-∞, -3) and (0, ∞), and it is decreasing on the interval (-3, 0).
c) To find the local maximum and minimum values, we need to examine the behavior of f around the critical points and look for points of inflection.
At x = 0, f changes from decreasing to increasing, indicating a local minimum.
At x = -3, f changes from increasing to decreasing, indicating a local maximum.
Therefore, f assumes a local minimum at x = 0 and a local maximum at x = -3.