Mathematics 242 – Principles of Analysis Selected Solutions – Problem Set 4 November 3, 2000

Section 3.5/7. The function f(x) = \frac{x^{3}}{x^{2}−1} has infinite discontinuities (vertical asymptotes) at x = ±1. We have

lim_{x→−1^{−}} f(x) = −∞ lim_{x→−1^{+}} f(x) = +∞
lim_{x→1^{−}} f(x) = −∞ lim_{x→1^{+}} f(x) = +∞

Furthermore f(0) = 0 and f(2) = 8/3.

We get an x_{1} ∈ (−1, 0) with f(x_{1}) = 10. Reason: Since lim_{x→−1^{+}} f(x) = +∞, there exists 0 > b > −1 such that f(b) = k > 10. Then the IVT applies to f(x) on the interval [b, 0] since that interval does not contain any of the discontinuities of f. Since 0 = f(0) < 10 < f(b) = k, there exists x_{1} ∈ [b, 0] such that f(x_{1}) = 10.

Similarly, since lim_{x→1^{+}} f(x) = +∞ and f(2) = 8/3 < 10, we get a second point x_{2} ∈ (1, 2) such that f(x_{2}) = 10.

Finally, we also have lim_{x→+∞} f(x) = +∞. So there exists some c > 2 such that f(c) > 10. The IVT also applies on the interval [2, c] (because it misses both discontinuities of f, and hence there exists x_{3} > 2 such that f(x_{3}) = 10.

14b. (This proof follows the same ideas as the proofs of 13a and 14a.) We assume that f is continuous on [0, 1], with f(0) < 0 and f(1) > 1. Also assume that g is continuous on [0, 1] with g(0) ≥ 0 and g(1) ≤ 1. Consider the function h(x) = f(x) − g(x). Since f, g are continuous, so is h. We also have by hypothesis:

\begin{matrix}h(0) = f(0) − g(0) < 0 \\ h(1) = f(1) − g(1) > 0 \end{matrix}

If m is the minimum of h on [0, 1] and M is the maximum of h on [0, 1], then m < 0 and M > 0. Hence by the IVT, there exists c ∈ [0, 1] such that h(c) = f(c) − g(c) = 0, or f(c) = g(c).

17.

a) This statement is true. Proof: Aiming for a contradiction, suppose that f is continuous on [a, b], that f(x) ≠ 0 for all x ∈ [a, b], but f(x) takes both positive and negative values on the interval [a, b]. Then the minimum m of f on the interval [a, b] must be negative and the maximum M must be positive, so we get a contradiction to the IVT: m < 0 < M, but there is no x ∈ [a, b] such that f(x) = 0.

b) This is also true. Proof: By the EVT, if m is the minimum value of f on [a, b], then m = f(c) for some c ∈ [a, b]. But then f(c) > 0 and f(x) ≥ f(c) = m > 0 for all x ∈ [a, b].

a) This statement is true. By the IVT and the EVT, if m is the minimum of f on [a, b] and M is the maximum, then f([a, b]) = [m, M] (every y with m ≤ y ≤ M is f(x) for some x ∈ [a, b]).

b) This statement is false. Counterexample: f(x) = x^{2}, (a, b) = (−1, 1). The image f((−1, 1)) equals [0, 1) which is not an open interval. (The open interval (−1, 1) gets “folded over” at its midpoint under the mapping f, so the image is not an open interval.)

19. This statement is false. We will construct a function that is continuous on the halfopen interval (0, 1], but that has neither a maximum nor a minimum on that interval. The idea is to look for a function that has oscillations of greater and greater amplitude as x → 0^{+}, but with shorter and shorter distance between each successive max and min as x → 0^{+}, so there are infinitely many local maximum and minimum points in all intervals (0, b]. An example of this type is given by the function

f(x) = \frac{1}{x} sin (\frac{1}{x})

(see graph on page 74, and multiply that by \frac{1}{x} so that the amplitudes of the oscillations increase without bound as x → 0^{+}.)