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MATHEMATICS 120 PROBLEM SET 3 - SOLUTIONS 1. The function f (x) is defined by 0 if − 2 ≤ x ≤ −1, ax2 + bx if − 1 < x < 2, f (x) = c if x = 2, 2x if 2 < x ≤ 5. Find all values a, b, c such that f (x) is continuous on [−2, 5]. For f (x) to be continuous at −1, we need 0 = a − b, so that a = b. For f (x) to be continuous at 2, we need 4a + 2b = c = 4. Therefore c = 4, 4a + 2b = 6a = 4, a = 2/3, b = 2/3. 2. Use the intermediate value theorem to prove that the equation x3 − 7x + 2 = 0 has three solutions in the interval [−3, 3]. Let f (x) = x3 − 7x + 2, then f (−3) = −4 < 0, f (−2) = 8 > 0, f (2) = −4 < 0, f (3) = 8 > 0. Since f (x) is continuous everywhere, by the intermediate value theorem the equation f (x) = 0 has at least one solution in each of the intervals (−3, −2), (−2, 2), (2, 3). (Note: since f (x) is a cubic polynomial, there cannot be more than 3 solutions.) 3. Give an example of a function f (x), defined on an open interval, such that f (x) has a maximum value but does not have a minimum value. Also give an example of a function g(x), defined for all x, such that g(x) has both a maximum value and a minimum value. (You can give a formula or draw a graph – either is acceptable.) If f (x) = −x2 on (−1, 1), then f (x) has a maximum value f (0) = 0 but no minimum value. For g(x), we can take e.g. g(x) = sin x, with maximum value 1 and minimum value −1. (The maximum and minimum values may be attained more than once – in fact if g(x) = sin x, each of them is attained infinitely many times.) 4. Assume that the following is true for some function f (x), defined for all x: for every > 0 there is a δ > 0 such that |f (x) − L| < for some x such that 0 < |x − a| < δ (read this carefully!). Does this imply that limx→a f (x) = L? Explain why or why not. It does not. In the definition of the limit at a we have “for all x such that 0 < |x − a| < δ”, not “for some x”. For example, let f (x) = sin(1/x) if x 6= 0 and f (0) = 0. Then f (x) does not have a limit at 0. But if a = 0 and L = 0, then the statement in the problem is true: for every > 0 there is a δ > 0 such that |f (x) − L| < for some x such that 0 < |x − a| < δ, for example for x = 1/nπ with n large enough. 1