Notes PDF
Introductory Problems
1. Find and classify the critical points of cos(x).
Answer
1. Even multiples of π are all local maxima. Odd multiples of π are all local minima.
Solution
1. The derivative is f'(x) = -sin(x). This is zero at every integer multiple of π, i.e. at every nπ. The second derivative is f”(x) = -cos(x). At an integer multiple of π, this is either +1 or -1: At even multiples of π, f”(x) = -1. These are all local maxima. At odd multiples of π, f”(x) = +1. These are all local minima. This answer makes sense: we know that cos(x) oscillates up and down with a period of 2π.
2. Find the global max/min of cos(x) on the interval [-π/3, 2π/3]
Answer
1. Global max is at x = 0, and global min is at x = 2π/3..
Solution
1. As we know (or have just seen in question 1), the critical points of cos(x) are at integer multiples of π. Only one of those, x = 0, is within our interval. The other two candidates for global max/min are at the endpoints, x = -π/3 and x = 2π/3. f(0) = 1 f(-π/3) = 1/2 f(2π/3) = – 1/2 Global max is at x = 0, and global min is at x = 2π/3.
3. Find global max/min of -3x4 + 4x3.
Answer
1. Global maximum at x = 1. No global min.
Solution
1. The derivative is f ‘ (x) = -12x3+12x2= 12x2(-x+1). This is zero at x = 0 and x = 1, so those are our critical point candidates. For x very far away from the origin (in both directions), the function goes down forever, so there is no global minimum. The global maximum is at one of the critical points. Comparing their values, we find that f(0)= 0 f(1)= 1 Global maximum is at the point x= 1.
4. Find global max/min of x3– 6×2 + 9x + 1 on the interval [0, 2].
Answer
1. Global min is at x = 0, global max is at x = 1.
Solution
1. The derivative is f ‘ (x) = 3 x2 – 12x + 9 = 3 (x2– 4x + 3)= 3 (x – 1) (x – 3). Zeroes are at x = 1 and at x = 3. The point x = 3 is not inside our interval, so our only critical point candidate is x = 0. We compare its value with the value at the endpoints: f(1) = 5 f(0) = 1 f(2) = 3 Global min is at x = 0, global max is at x = 1.
5. Find global max/min of √x.
Answer
1. Global min at x = 0. No global max.
Solution
1. First note that the function is only defined for x ≥ 0. The derivative is f ‘ (x) = 1/(2√x), which is never zero, and it is undefined for x = 0 (in addition to all x < 0). We do get one candidate from this, namely x = 0 where we cannot evaluate the derivative. This is also the endpoint of the interval where f(x) is defined. We do not have to check negative x, since the function is not defined there. For very large x, f(x) goes to infinity, so there is no global maximum. There is a global minimum, at it occurs at the one candidate we have, namely at x = 0.