Mathematics 242 – Principles of Analysis Solutions for Midterm Exam 3 May 2, 2014

I. (15) Let

f(x) = \{ \begin{matrix}x^{4/5} sin(1/x) & if x ≠ 0 \\ 0 & if x = 0. \end{matrix}

Is f continuous at x = 0? Is f differentiable at x = 0? Give complete reasons for your assertions.

Solution: Since | sin(1/x)| ≤ 1 for all x ≠ 0, we have

−x^{4/5} ≤ f(x) ≤ x^{4/5}

for all x ≠ 0. Also lim_{x→0} ±x^{4/5} = 0. By the limit squeeze theorem, lim_{x→0} f(x) = 0 = f(0). Therefore, f is continuous at x = 0. However, f is not differentiable at x = 0 because

lim_{x→0} \frac{f(x) − f(0)}{x − 0} = lim_{x→0} \frac{sin(1/x)}{x^{1/5}}

does not exist. For instance, at x_{n} = \frac{2}{(4n+1)π} (a sequence converging to 0), we have

\frac{sin(1/x_{n})}{x_{n}^{1/5}} = sin((4n + 1)π/2) (\frac{(4n + 1)π}{2} )^{1/5} = (2nπ + \frac{π}{2} )^{1/5} → +∞

as n → ∞.

II. Both parts of this question refer to the function f : R → R defined by f(x) = 1 − x^{2}.

A) (20) Consider the regular partitions P_{n} of the interval [1, 3] and show directly, using the upper and lower sums, that f is integrable on [1, 3].

Solution: Note that f is decreasing on [1, 3] since f′(x) = −2x < 0 for all x with 1 ≤ x ≤ 3. This means the desired statement can either be shown by following the proof of our theorem that monotone functions are integrable, or directly. Here is the direct way:

The partition is

P_{n} = {1, 1 + 2/n, 1 + 4/n, …, 3},

with x_{i} = 1 + 2i/n for i = 0, 1, …, n. Hence, since f is smallest at the right endpoint in each subinterval,

L_{P_{n}}(f) = ∑_{i=1}^{n} (1 − (1 + 2i/n)^{2}) \frac{2}{n}

= −\frac{8}{n^{3}} ∑_{i=1}^{n} i^{2} − \frac{8}{n^{2}} ∑_{i=1}^{n} i

= −\frac{8}{n^{3}} · \frac{n(n + 1)(2n + 1)}{6} − \frac{8}{n^{2}} · \frac{n(n + 1)}{2}

= −\frac{20}{3} − \frac{8}{n} − \frac{4}{3n^{2}} .

Similarly, f is largest at the left endpoint in each subinterval, so

\begin{aligned}U_{P_{n}}(f) & = ∑_{i=1}^{n} (1 − (1 + 2(i − 1)/n)^{2}) \frac{2}{n} \\ & = −\frac{20}{3} + \frac{8}{n} − \frac{4}{3n^{2}} . \end{aligned}

Therefore, for any given ε > 0, if n > 16/ε,

U_{P_{n}}(f) − L_{P_{n}}(f) = \frac{16}{n} < ε.

This shows that f is integrable.

B) (15) Explain why the hypothesis of the Mean Value Theorem is satisfied for f on the interval [1, 3] and find the number c mentioned in the conclusion. Solution: f is a polynomial function, so it is differentiable, hence continuous everywhere. On the interval [1, 3], f(3) − f(1) = −8 − 0 = −8. The MVT says that there is some c ∈ (1, 3) where −8 = f′(c) · (3 − 1). Since −8 = 2f′(c) = −4c, this is true for c = 2.

III. (20) Show that if f is continuous on [a, b], then f is integrable on [a, b].

Solution: By a previous result we know that f(x) continuous on [a, b] implies that f(x) is uniformly continuous on [a, b]. Therefore, given ε > 0, there exists a δ > 0 such that |f(x) − f(x′)| < ε/(b − a) whenever |x − x′| < δ (with x, x′ ∈ [a, b], of course). Now, let P be any partition of [a, b] with Δx_{i} < this δ for all i. By the EVT, on the interval [x_{i−1}, x_{i}] from the partition P, f attains a maximum M_{i} = f(c_{i}) and a minimum m_{i} = f(d_{i}) at some c_{i}, d_{i} ∈ [x_{i−1}, x_{i}]. But then M_{i} − m_{i} < ε/(b − a) since |c_{i} − d_{i}| < δ. Hence

U(f, P) − L(f, P) < ∑_{i=1}^{n} \frac{ε}{(b − a)}Δx_{i} = \frac{ε}{b − a} ∑_{i=1}^{n} Δx_{i} = ε.

(This follows since ∑_{i} Δx_{i} = b − a.) Therefore f is integrable on [a, b].

IV. True-False. Say whether each of the following statements is true or false. For true statements, give short proofs; for false ones give reasons or counterexamples. Do any three parts. If you submit solutions for all four, then I will consider the other one for Extra Credit.

A) (10) Let f(x) = e^{2x} − e^{x}. There exists some c ∈ (0, ln(2)) such that f(c) = 1.

Solution: The statement is TRUE. We apply the IVT. First, f is continuous everywhere since the exponentials e^{x} and e^{2x} are differentiable everywhere. On the interval [0, ln(2)], we have f(0) = 1 − 1 = 0 and f(ln(2)) = 4 − 2 = 2. Since 1 is in the range between the endpoint values, the (“weak form” of) the IVT implies that there exists c ∈ (0, ln(2)) such that f(c) = 1.

B) (10) The function f(x) = arctan(x) is uniformly continuous on the interval (−1, 1).

Solution: This is TRUE. Method 1: f(x) = arctan(x) is continuous at all x ∈ R, hence on the closed interval [−1, 1]. Our general theorem implies that f(x) is uniformly continuous on [−1, 1], hence also on the subset (−1, 1).

Method 2: We can also apply the MVT to f on the interval [x, x′] where −1 < x < x′ < 1 are arbitary. Then there exists a c ∈ (x, x′) such that f(x) − f(x′) = f′(c)(x − x′). But |f′(x)| = \frac{1}{1+x^{2}} is bounded above by 1 on R, hence on this interval. Therefore |f(x) − f(x′)| ≤ |x − x′|, so f is Lipschitz continuous (with Lipschitz constant k = 1), hence uniformly continuous.

C) (10) There are continuous functions f(x) on [a, b] for which there exist no differentiable function F (x) on [a, b] with F′(x) = f(x).

Solution: This is FALSE. The Fundamental Theorem of Calculus (part 1) implies that F (x) = ∫_{a}^{x} f(t) dt is always an antiderivative of f(x) on [a, b].

D) (10) Let f be differentiable on an open interval I with [a, b] ⊂ I. If f′(a) > 0 and f′(b) < 0, then there must exist some c ∈ (a, b) where f′(c) = 0.

Solution: This is TRUE. Since f is differentiable everywhere on [a, b], it is also continuous on that interval. By the Extreme Value Theorem, f reaches a maximum value M = f(c) for c somewhere in that interval. On the other hand, the inequalities f′(a) > 0 and f′(b) > 0 imply that f must take values larger than both f(a) and f(b) in the interval. For instance, to get

f′(a) = lim_{x→a} \frac{f(x) − f(a)}{x − a} > 0,

it must be true that f(x) > f(a) for x in some interval (a, a + δ) for δ > 0. Similarly f′(b) < 0 implies f(x) > f(b) for x in some (b − δ, b). Hence the location where the maximum is attained is c ∈ (a, b). It follows that f(c) must be a local maximum (not an endpoint maximum) and hence f′(c) = 0 since we are assuming f is always differentiable on I ⊃ [a, b].