Section 1

1. Minimum f(3) = 1, maximum f(1) = 5, since f(x) = 5 - (x - 1)2. No solution if y < 1 or if y > 5. One solution if 1 ≤ y < 4 or if y = 5. Two solutions if 4 ≤ y < 5.

2. Since x3 - 5 takes on all values between -4 and +3 as x varies from 1 to 2, for some x between 1 and 2 it takes on the value 0. Since x3 - 5 = 0 means x3 = 5, such an x must be.

3. The value of this function at x = 3 is negative, at x = 4 is positive, so it has a zero between 3 and 4.

4. Minimum f(5) = ⅕. There is no maximum because f(1/n) = n for n = 1, 2, 3, ··· are values of f. Note that f(0) is not defined.

5. In this case, every value of f is 3 (its graph is a horizontal line segment), so m = M = 3.

6. Minimum f(01) = 0. There is no maximum because 5 is not in the interval [0, 5).