Mathematics 242 – Principles of Analysis Solutions for Problem Set 7 – due: Friday, April 4

‘A’ Section

1. Let f(x) = \frac{15x}{x^{4}+3x^{2}+1}. Use the Intermediate Value and/or Extreme Value Theorems to show the following: A) For all k ∈ [−3, 3], there exist c ∈ [−1, 1] such that f(c) = k.

Solution: First, f(x) is a rational function and x^{4} + 3x^{2} + 1 ≥ 1 for all x ∈ R, so it follows that f(x) is continuous at all c ∈ R. We have f(−1) = −3 and f(1) = 3. Therefore, by the IVT, for all k ∈ [−3, 3], there exist c ∈ [−1, 1] such that f(c) = k.

B) For all k with 0 < k < 3, there exist some c ∈ (1, ∞) such that f(c) = k.

Solution: We see by the “Big Theorem” on function limits that lim_{x→∞} f(x) = 0. Hence if we have any k with 0 < k < 3, then there exists a B > 1 such that f(B) < k < 3 = f(1). Applying the IVT on the interval [1, B], we see there is a c ∈ (1, B) ⊂ (1, ∞) such that f(c) = k.

C) There is a c ∈ (0, 1) where f(c) = 3.

Solution: Consider the function

g(x) = f(x) − 3 = \frac{15x − 3x^{4} − 9x^{2} − 3}{x^{4} + 3x^{2} + 1}

We see that g(0) = −3 < 0 and g(1/2) = \frac{33}{29} > 0. g(x) is also continuous on [0, 1/2]. Therefore, by the IVT, there exists c ∈ (0, 1/2) ⊂ (0, 1) where g(x) = 0.

D) There is a d ∈ (0, 1) where f′(d) = 0.

Solution: Consider the number c from part c. We have f(c) = f(1) = 3, but f is certainly not constant between c and 1. Therefore by the Extreme Value Theorem, there must exist a maximum value f(d) > 3 or a minimum value f(d) < 3 for f on the interval [c, 1] attained at some d ∈ (c, 1). By plotting the function f, we can see that in fact the first statement is the one that is true – there is a maximum. We claim that f′(d) = 0. First note that

lim_{x→d^{−}} \frac{f(x) − f(d)}{x − d} ≥ 0

since f has a maximum at d. Similarly,

lim_{x→d^{+}} \frac{f(x) − f(d)}{x − d} ≤ 0

Since f is a rational function whose denominator never equals zero, f is differentiable at d, so these one-sided limits must both exist and be equal. Therefore f′(d) = 0.

2. Show that there is a solution of the equation tan(x) = x in the interval (\frac{(2k−1)π}{2} , \frac{(2k+1)π}{2} ) for every k ∈ Z.

Solution: We know that tan(x) has infinite discontinuities at \frac{(2k−1)π}{2} and \frac{(2k+1)π}{2} and hence

lim_{x→ \frac{(2k−1)π}{2} ^{+}} tan(x) = −∞ lim_{x→ \frac{(2k+1)π}{2} ^{−}} tan(x) = +∞

It follows that we also have

lim_{x→ \frac{(2k−1)π}{2} ^{+}} tan(x) − x = −∞ lim_{x→ \frac{(2k+1)π}{2} ^{−}} tan(x) − x = +∞

Therefore, tan(x) − x must change sign on some interval [a, b] ⊂ (\frac{(2k−1)π}{2} , \frac{(2k+1)π}{2} ) and the Intermediate Value Theorem implies tan(x) − x = 0 somewhere in that interval.

3. Using the definition of the derivative, find the value of f′(c), or say why f is not differentiable at x = c:

A) f(x) = x^{3} + 2x − 4 at c = 2.

Solution: We have

f′(2) = lim_{x→2} \frac{x^{3} + 2x − 12}{x − 2}

= lim_{x→2} \frac{(x − 2)(x^{2} + 2x + 6)}{x − 2}

= lim_{x→2} x^{2} + 2x + 6

= 14.

B) f(x) = sin(|x|) at c = 0. Hint: Look back at Problem Set 6, B 2. Solution: f′(0) does not exist for this function because

lim_{x→0^{+}} \frac{sin(|x|)}{x} = +1

by the indicated problem on Problem Set 6, while

lim_{x→0^{−}} \frac{sin(|x|)}{x} = lim_{x→0^{−}} \frac{− sin(x)}{x} = −1.

Since the one-sided limits are not equal, the derivative at 0 does not exist. C) The function defined by

f(x) = \{ \begin{matrix}x^{2} & if x > 1 \\ 2x − 1 & if x ≤ 1 \end{matrix}

at c = 1.

Solution: We have

lim_{x→1^{+}} \frac{f(x) − f(1)}{x − 1} = lim_{x→1^{+}} \frac{x^{2} − 1}{x − 1} = lim_{x→1^{+}} x + 1 = 2.

On the other hand,

lim_{x→1^{−}} \frac{f(x) − f(1)}{x − 1} = lim_{x→1^{−}} \frac{2x − 2}{x − 1} = lim_{x→1^{−}} 2 = 2.

Since the one-sided limits exist and are equal, f′(1) exists and equals 2.

D) The function defined by

f(x) = \{ \begin{matrix}x^{2} & if x ∈ Q \\ 0 & if x ∈ Q^{c} \end{matrix}

at c = 0.

Solution: We have for x ≠ 0,

\frac{f(x) − f(0)}{x − 0} = \{ \begin{matrix}x & if x ∈ Q \\ 0 & if x ∈ Q^{c} \end{matrix}

Given any ε > 0, if we take δ = ε, then for all x in the deleted neighborhood defined by 0 < |x| < ε,

|\frac{f(x) − f(0)}{x − 0} − 0| < ε.

It follows that

lim_{x→0} \frac{f(x) − f(0)}{x − 0} = 0 = f′(0).

(It is not too hard to show that f′(c) exists only for this one c = 0. This function is not differentiable anywhere else.)

‘B’ Section

1. Let f be continuous on [0, 1] with f(0) < 0 and f(1) > 1. Suppose that g is another continuous function on [0, 1] such that g(0) ≥ 0 and g(1) ≤ 1. Show that there exists some c ∈ (0, 1) such that f(x) = g(x).

Solution: Let h(x) = f(x)−g(x). Since f, g are continuous on [0, 1], the same is true for h. By the given information, h(0) = f(0) − g(0) < 0 and h(1) = f(1) − g(1) > 0. Therefore, the IVT implies that h(c) = 0 for some c ∈ (0, 1). But then 0 = h(c) = f(c) − g(c), so f(c) = g(c).

2. Let f be continuous on [a, b] with f(a) < k < f(b). Here is a variation on our proof of the Intermediate Value Theorem.

A) Let

T = {x ∈ [a, b] | f(x) > k}.

Show that T has a greatest lower bound and that f(glb(T)) = k.

Solution: T is contained in the interval [a, b], so it is a bounded subset of R. Then c = glb(T) exists by the LUB Axiom. Note that a < c since f(a) < k. Hence the interval [a, c) is contained in the complement of T. If we let {x_{n}} be any sequence contained in [a, c) converging to c, then since f is continuous, lim_{n→∞} f(x_{n}) = f(c). But f(x_{n}) ≤ k for all n, so

(1)  f(c) = lim_{n→∞} f(x_{n}) ≤ k

also (by Corollary 2.2.8 in the text). On the other hand, given any ε > 0, c + ε is not a lower bound for T, so there exists some x ∈ T such that c ≤ x < c + ε. Apply this for ε = \frac{1}{n} for each natural number. Then we get a sequence x′_{n} such that x′_{n} ∈ T for all n and c ≤ x′_{n} < c + \frac{1}{n}. It follows easily that x′_{n} → c as n → ∞. Therefore since f is continuous at c, lim_{n→∞} f(x′_{n}) = f(c). But x′_{n} ∈ T for all n, so f(x′_{n}) > k. Hence

(2)  f(c) = lim_{n→∞} f(x′_{n}) ≥ k.

The two inequalities (1) and (2) show that f(c) = k.

B) Will this glb(T) always be the same as the c we found in our proof of the IVT with f(c) = k? If so, prove they are the same; if not, give a counterexample.

Solution: In the proof we did in class we considered

S = {x ∈ [a, b] | f(x) ≤ k}

and we showed that if c′ = lub(S), then f(c′) = k. The c found in part A and the c′ here do not have to be the same. For instance, let f(x) = x^{3} − 2x + 1 on [−2, 2]. We have f(−2) = −3 and f(2) = 5. So the IVT will apply for any k with −3 < k < 5. Consider k = 0. The equation x^{3} − 2x + 1 = 0 actually has three different roots in the interval [−2, 2]: One between −2 and −1 (call this one α), a second between 1/2 and 1 (call this one β), and a third at x = 1. The set T as in part A is the union T = (α, β) ∪ (1, 2), so c = glb(T) = α. On the other hand, the set S as in the proof we did in class is S = [−2, α] ∪ [β, 1], so c′ = lub(S) = 1.

3. This property deals with another property of real-valued functions of a real variable sometimes called Lipschitz continuity.

A) Let f be a function on an interval I with the property that there exists a strictly positive constant k such that |f(x) − f(x′)| ≤ k|x − x′| for all x, x′ ∈ I (this is the definition of Lipschitz continuity). Show that f is uniformly continuous on I.

Solution: Given ε > 0, let δ = ε/k. Then for any x, x′ ∈ I such that |x−x′| < δ = ε/k, it follows that

|f(x) − f(x′)| ≤ k|x − x′| < k · ε/k = ε.

This shows that the definition of uniform continuity is satisfied for f on I.

B) The converse of the statement in part A is not true: Show that f(x) = x^{1/3} is uniformly continuous on [−1, 1], but there is no constant k such that |f(x) − f(x′)| ≤ k|x − x′| for all x, x′ ∈ [−1, 1]. Hint: Think slopes of secant lines to the graph y = x^{1/3}.

Solution: First, f(x) is continuous on [−1, 1], hence it is uniformly continuous by the result of Theorem 3.6.8 (proved in class before Easter break). Let x′ = 0 and take arbitrary x > 0 we have

\frac{f(x) − f(0)}{x − 0} = \frac{x^{1/3}}{x} = \frac{1}{x^{\frac{2}{3}}} .

But lim_{x→0^{+}} \frac{1}{x^{\frac{2}{3}}} = +∞. In other words, the value of the difference quotient will get

unboundedly large as x → 0^{+}. Hence there is no single k such that

|\frac{f(x) − f(0)}{x − 0}| ≤ k

for all x in [−1, 1]. But that shows that there is no k such that |f(x)−f(0)| ≤ k|x−0| for all x in [−1, 1].