Mathematics 242 – Principles of Analysis Solutions for Problem Set 7 – due: Friday, April 12

’A’ Section

1. Let f(x) = \frac{15x}{x^{4}+x^{2}+1}. Using the Intermediate Value Theorem, A) Show: For all k ∈ [−5, 5], there exist c ∈ [−1, 1] such that f(c) = k.

Solution: First, f(x) is a rational function and x^{4} + x^{2} + 1 > 1 for all x ∈ R, so it follows that f(x) is continuous at all c ∈ R. We have f(−1) = −5 and f(1) = 5. Therefore, by the IVT, for all k ∈ [−5, 5], there exist c ∈ [−1, 1] such that f(c) = k.

B) Show: For all k with 0 < k < 5, 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 < 5, then there exists a B > 1 such that f(B) < k < 5 = 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) Show that if 0 < m < 15, then f(x) = mx has two real solutions other than x = 0. Solution: Note that

f(x) − mx = \frac{15x}{x^{4} + x^{2} + 1} − mx = \frac{−mx^{5} − mx^{3} − (m − 15)x}{x^{4} + x^{2} + 1}.

The numerator factors as

−mx (x^{4} + x^{2} + \frac{m − 15}{m}) .

Method 1: The polynomial g(x) = x^{4}+x^{2}+ \frac{m−15}{m} is continuous at all real x. Moreover, g(0) = \frac{m−15}{m} < 0 (since m < 15 while g(b) > 0 if b is sufficiently large because the x^{4} term occurs with a positive coefficient. Therefore, by the IVT, there is a c with g(c) = 0 on the interval (0, b). Since g(x) is an even function, we also have g(−c) = 0.

Method 2: Applying the quadratic formula to the quadratic in x^{2}, the second factor is zero when

x^{2} = \frac{−1 ± \sqrt{1 − 4 (\frac{m−15}{m})}}{2} = \frac{−\sqrt{m} ± \sqrt{60 − 3m}}{2\sqrt{m}}

In order for this to be defined, we must have m > 0. For x^{2} to be positive we must take the + sign and we must have 60 − 3m > m, so m < 15. For 0 < m < 15, there are two solutions found by taking the positive and negative square roots of the above:

x = ±\sqrt{\frac{−\sqrt{m} + \sqrt{60 − 3m}}{2\sqrt{m}}}.

Neither of these is equal to zero.

2. Show that there are at least three real solutions of the equation sin(x) + 2 cos(x) = x/2. Hint: Look at the values of g(x) = sin(x) + 2 cos(x) − x/2 at “nice” multiples of \frac{π}{2} .

Solution: From calculus we know that sin(x) and cos(x) are differentiable for all x and the same is true for x/2. Therefore, the function g(x) is differentiable everywhere and consequently continuous everywhere. We have

\begin{aligned}g (\frac{−3π}{2} ) & = 1 + \frac{3π}{4} > 0 \\ g(−π) & = −2 + \frac{π}{2} < 0 \\ g (\frac{−π}{2} ) & = −1 + \frac{π}{4} < 0 \\ g(0) & = 2 > 0 \\ g (\frac{π}{2} ) & = \frac{1}{2} > 0 \\ g(π) & = −2 − \frac{π}{2} < 0 \end{aligned}

Therefore, the IVT implies that g has one zero in the interval (−3π/2, −π), another in the interval (\frac{−π}{2} , 0), and a third in the interval (π/2, π).

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 + 1 at c = 2. Solution: We have

f′(2) = lim_{x→2} \frac{(x^{3} + 2x + 1) − 13}{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.)

4. Suppose f, g are differentiable functions with f(g(x)) = \frac{x}{x^{2}+1} and that

\begin{matrix}g(1) = 1 & g(2) = 4 \\ g′(1) = 2 & g′(2) = −1 \end{matrix}

Determine the equation of the tangent line to the given graph at the given point. A) y = f(x) at (1, f(1)).

Solution: First, f(1) = f(g(1)) = \frac{1}{2}. By the Chain Rule (f ∘ g)′(1) = f′(g(1))g′(1) = 2f′(1). On the other hand, by the Quotient Rule,

(f ∘ g)′(x) = \frac{(x^{2} + 1)(1) − (x)(2x)}{(x^{2} + 1)^{2}} = \frac{1 − x^{2}}{(x^{2} + 1)^{2}}

So (f ∘ g)′(1) = 0 and hence f′(1) = 0. The tangent line to y = f(x) at (1, f(1)) is the horizontal line y = \frac{1}{2}.

B) y = (f ∘ g)(x) at (2, (f ∘ g)(2)).

Solution: By the computations in part A, f(g(2)) = \frac{2}{5} and (f ∘ g)′(2) = \frac{−3}{25}. So the tangent line is

y = \frac{2}{5} − \frac{3}{25}(x − 2).

’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].

4. Let f and g be differentiable on (a, c) and let b ∈ (a, c). Assume f(b) = g(b). Define a new function by

p(x) = \{ \begin{matrix}f(x) & if x ∈ (a, b) \\ g(x) & if x ∈ [b, c) \end{matrix}

Show that p is differentiable on (a, c) if and only if f′(b) = g′(b).

Solution: Assume first that f′(b) = g′(b). Then

\begin{aligned}lim_{x→b^{−}} \frac{p(x) − p(b)}{x − b} & = lim_{x→b^{−}} \frac{f(x) − f(b)}{x − b} \\ & = f′(b) \end{aligned}

By our assumption, this is also

= g′(b)

= lim_{x→b^{+}} \frac{g(x) − g(b)}{x − b}

= lim_{x→b^{+}} \frac{p(x) − p(b)}{x − b}.

This shows that p is differentiable at b since the one sided limits of the difference quotients for p exist and are equal. Differentiability of p at all x ≠ b in (a, c) follows from the way p(x) is defined. At those x, the values of p are either the same as the values of f or the values of g on some interval containing x. Hence, for instance, if a < x_{0} < b, then since p(x) = f(x) for all x in an interval containing x_{0},

p′(x_{0}) = lim_{x→x_{0}} \frac{p(x) − p(x_{0})}{x − x_{0}} = lim_{x→x_{0}} \frac{f(x) − f(x_{0})}{x − x_{0}} = f′(x_{0}),

since f is differentiable at x_{0}. Similarly, if b < x_{0} < c, then p′(x_{0}) = g′(x_{0}).

Conversely, suppose that p is differentiable at all x in (a, c). This implies in particular that p is differentiable at x = b, so

f′(b) = lim_{x→b^{−}} \frac{f(x) − f(b)}{x − b}

= lim_{x→b^{−}} \frac{p(x) − p(b)}{x − b}

= lim_{x→b^{+}} \frac{p(x) − p(b)}{x − b}

= lim_{x→b^{+}} \frac{g(x) − g(b)}{x − b}

= g′(b).