1b. Critical points at -3, -2, 0 (note 2 is not a critical point because it is not in the domain). Global max of 16 at -2. Global min of 0 at 0. The domain is a closed interval, so max and min exist.
1c. Critical points at -3, -2, -1, 0, 2, 3. Global max of 16 at -2. Global min of -16 at 2. The closed interval method also works if the domain is a union of finitely many closed intervals, because the global max/min are just the max/min of the maxes/mins on the intervals.
1d. On [0, 3), the range is [-16, 0]; the max of 0 occurs at 0, the min of -16 occurs at 2. On (0, 3], the range is [-16, 0); there is no max because we can always make f(x) larger by choosing x to make f(x) closer to 0, but the min of -16 occurs at 2.
2. Below I also included information about intervals of increase or decrease, global max/min, intervals of concavity, and inflection points. Note: According to Prof. Haiman, we should consider c to be an inflection point if f '(c) exists and f '' changes signs around c.
2a. Critical point at 0 (f ' undefined)(local min of 0). f increases on (0, infinity), decreases on (-infinity, 0). Global min of 0 at 0. No global max since f(x) goes to infinity as x goes to infinity or -infinity. f is concave down on (-infinity, 0) and (0, infinity). No inflection point.
2b. Critical points at 0 (f ' = 0)(local max of 2), at 1 (f ' undefined)(no local max or min), at 3 (f ' = 0)(local min of -3). f increases on (-infinity, 0) and (3, infinity). f decreases on (0, 3). No global max or min because f(x) goes to infinity as x goes to infinity, while f(x) goes to -infinity as x goes to -infinity. f is concave up on (1, infinity), concave down on (-infinity, 1). No inflection point since f ' is undefined at 1.
2c. Critical points at pi/2 + n*pi for n >= 0. Note f '(0) = -1 because f is continuous at 0 and has one-sided derivatives of -1 at 0, so 0 is not a critical point. f decreases on (-infinity, pi/2) and (3pi/2 + 2n*pi, 5pi/2 + 2n*pi) for n >= 0. f increases on (pi/2 + 2n*pi, 3pi/2 + 2n*pi) for n >= 0. f has local maxes of 1 at 3pi/2 + 2n*pi for n >= 0. f has global and local mins of -1 at pi/2 + 2n*pi for n >= 0. f has no global max because f(x) goes to infinity as x goes to -infinity. f is concave up on (2n*pi, pi + 2n*pi) for n >= 0, concave down on (pi + 2n*pi, 2pi + 2n*pi) for n >= 0. Because f ''(x) = 0 for x < 0, we do not consider f to be concave up or down in (-infinity, 0). f has inflection points (n*pi, 0) for n >= 1. We do not consider (0, 0) to be an inflection point because there is no concavity on one side of 0.
2d. Critical points at -pi/3 + 2n*pi for all n (f ' = 0)(local mins of -sqrt(3)/2 + pi/6 - n*pi), at pi/3 + 2n*pi for all n (f ' = 0)(local maxes of sqrt(3)/2 - pi/6 - n*pi). Here sqrt denotes square root. f increases on (-pi/3 + 2n*pi, pi/3 + 2n*pi) for all n. f decreases on (pi/3 + 2n*pi, 5pi/3 + 2n*pi) for all n. No global max or min because f(x) goes to infinity as x goes to -infinity, and f(x) goes to -infinity as x goes to infinity. Or we can just observe that the local maxes have no max among them, and the local mins have no min among them. (The converse is false! Even if f has a max among local maxes, f need not have a global max.) f is concave down on (2n*pi, pi + 2n*pi) for all n, concave up on (-pi + 2n*pi, 2n*pi) for all n. The inflection points are (n*pi, -n*pi/2) for all n.
2e. There is no local min or max at 0 because f(0) = 0 and the function oscillates between positive and negative as x approaches 0.
2f. Critical point at 0 (0 is on the boundary of the domain)(local min of 0). f increases on (0, pi/2). Global min of 0 at 0. No global max because f(x) goes to infinity as x goes to pi/2 on the left. f is concave up on (0, pi/2). There is no inflection point (although (0, 0) would have been one if the domain were extended to the left of 0).
2g. Critical points at 0 (0 is on the boundary of the domain)(local min of 0), at 1 (f ' = 0)(local max of 1). f increases on (0, 1), decreases on (1, infinity). No global min because f(x) goes to -infinity as x goes to infinity. Global max of 1 at 1. f is concave down on (0, infinity). There is no inflection point.
2h. Critical points at 0 (0 is on the boundary of the domain)(local max of 1) and at 1 (f ' undefined)(local min of 0). f decreases on (0, 1), increases on (1, infinity). Global min of 0 at 1. No global max since f(x) goes to infinity as x goes to infinity. f has no concavity anywhere because f ''(x) = 0 for x not equal to 1. There is no inflection point.
2i. Critical point at 0 (f ' undefined)(local max of 1). f decreases on (-infinity, 0), increases on (0, infinity). No global max since f(x) goes to infinity as x goes to infinity or -infinity. No global min since f(x) is always positive and we can always make f(x) smaller by making x closer but not equal to 0. f has no concavity anywhere because f ''(x) = 0 for x not equal to 0. There is no inflection point.
2j. Critical point at 1 (f ' undefined)(no local max or min). f increases on (-infinity, 1) and (1, infinity). No global max or min because f(x) goes to infinity as x goes to infinity, and f(x) goes to -infinity as x goes to -infinity. f is concave down on (1, infinity). f has no concavity on (-infinity, 1) due to f ''(x) = 0. There is no inflection point because on one side of (1, 0) there is no concavity.
2k. Critical point at 1 (f ' undefined)(local max of 2). f increases on (-infinity, 1) and (1, infinity). No global max or min because f(x) goes to infinity as x goes to infinity, and f(x) goes to -infinity as x goes to -infinity. f has no concavity anywhere because f ''(x) = 0 for x not equal to 1. There is no inflection point.
2l. Same as 2k, but no local max or min at 1.
3. By the Mean Value Theorem with endpoints 0 and x, there exists c between 0 and x such that f '(c) = (f(x) - f(0))/(x - 0) = f(x)/x. So f(x) = f '(c)x. Now c > 0, so s < f '(c) < t. Thus f(x) = f '(c)x > sx and f(x) = f '(c)x < tx.
4. Let t be the number of hours elapsed since the 1 pm escape. Let f(t) be the location of Jack on the road at time t, making one direction of the road positive and the other negative. Then f(0) = 0 since Jack is at the prison at time 0. Because Jack's maximum speed is 6 and the speed is just the absolute value of f '(t) , we know that -6 < f '(t) < 6 for all t. Hence we can apply problem 3 to show that -6t < f(t) < 6t. Now t = 0.25 at 1:15 pm, so -1.5 < f(0.25) < 1.5 and Jack is within 1.5 miles of the prison at 1:15 pm. Therefore the police should block off at 1.5 miles on either side of the prison.
5. Let f(t) be the location of the red car at time t. By the Mean Value Theorem on the interval [2 pm, 2:04 pm], there is some time c between 2 pm and 2:04 pm when f '(c) = (f(2:04) - f(2))/(2:04 - 2) = 5 miles/4 minutes = 75 miles/hour. In other words, the red car is travelling at 75 miles/hour and is exceeding the speed limit at time c, even though we do not know when c is. Give the car a ticket!
6. This problem can be done using the Mean Value Theorem, but it is not very easy. I can show you how if you are curious. However, you should be able to do this problem using L'Hospital's Rule: It is enough to show that rx^d - ln x has limit infinity as x goes to infinity.