Sample Questions: Solutions Midterm II Math 127B. Winter, 2005

1. For each of the following statements, say if it is true or false. (No explanation is required.)

(a) If f is differentiable and f′ > 0, then f is strictly increasing.

(b) If f is strictly increasing and differentiable, then f′ > 0.

(c) If f is the sum of a convergent Taylor series in an open interval containing the origin, then f is infinitely differentiable.

(d) If f is infinitely differentiable in an open interval containing the origin, then the Taylor series of f converges.

(e) There exists 0 < x < 1 such that e^{x} sin 1 = cos x (e − 1).

Solution.

• (a) True. (Follows from the mean value theorem.)

• (b) False. (For example, f(x) = x^{3}.)

• (c) True. (Power series can be differentiated term-by-term inside their interval of convergence.)

• (d) False. (For example, we saw in class that

f(x) = ∑_{n=1}^{∞} e^{−n} cos(n^{2}x)

is infinitely differentiable on ℝ, but its Taylor series (at 0) diverges.)

• (e) True. (Apply the generalized mean value theorem to e^{x} and sin x on [0, 1].)

2. Define the derivative. Consider

f(x) = \{ \begin{matrix}|x|^{a} & for x irrational, \\ 0 & for x rational. \end{matrix}

For what values of a > 0 is f differentiable at 0? Is f differentiable at x ≠ 0?

Solution.

• A function f : (a, b) → ℝ is differentiable at x_{0} ∈ (a, b) with derivative f′(x_{0}) if the following limit exists

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

• This function is differentiable at 0 if and only if a > 1. This follows because

\frac{f(x) − f(0)}{x − 0} = 0 if x is rational and nonzero,

\frac{f(x) − f(0)}{x − 0} = \frac{|x|^{a}}{x} if x is irrational,

and

lim_{x→0} \frac{|x|^{a}}{x} = 0

if and only if a > 1.

• The function f is discontinuous at every x ≠ 0, therefore it is not differentiable at any x ≠ 0.

3. State Taylor’s theorem. Prove that

log(1 + x) < x

for all x > 0.

Solution.

• If f : (a, b) → ℝ is an n-times differentiable function on an open interval (a, b) containing the origin, then for any x ∈ (a, b) there exists a y between 0 and x such that

f(x) − ∑_{k=0}^{n−1} \frac{f^{(k)}(0)}{k!}x^{k} = \frac{f^{(n)}(y)}{n!}x^{n}.

• First proof. The function f(x) = log(1 + x) is infinitely differentiable in (−1, ∞), and

f′(x) = \frac{1}{1 + x}, f′′(x) = −\frac{1}{(1 + x)^{2}} .

Since f(0) = 0 and f′(0) = 1, the Taylor polynomial of f of degree 1 is x. If x > 0, then using Taylor’s theorem with n = 2 we find that there exists 0 < y < x such that

f(x) − x = −\frac{1}{2!(1 + y)^{2}} x^{2} < 0,

so

log(1 + x) < x.

• Second proof. For any x > 0, the function g(x) = x − log(1 + x) is continuous on the closed interval [0, x] and differentiable in the open interval (0, x). By the mean value theorem, there exists 0 < y < x such that

g(x) − g(0) = g′(y)(x − 0).

Since g(0) = 0 and

g′(y) = 1 − \frac{1}{1 + y} > 0 for y > 0,

we conclude that

g(x) > 0 for x > 0

which proves the result.

4. Carefully state a version of L’Hospital’s rule that applies to the following limit. Use it to prove that the limit exists, and find its value:

lim_{x→0} \frac{1 − cos x}{x^{2}}.

Solution.

• Suppose that f, g are functions that are differentiable in an open interval containing x_{0}, with g′(x) nonzero in the interval, and

lim_{x→x_{0}} f(x) = 0,  lim_{x→x_{0}} g(x) = 0.

If the limit

lim_{x→x_{0}} \frac{f′(x)}{g′(x)} = L

exists, then so does the limit

lim_{x→x_{0}} \frac{f(x)}{g(x)} = L.

• The functions f(x) = sin x and g(x) = 2x satisfy the hypotheses of L’Hospital’s theorem on any interval containing the origin, and

lim_{x→0} \frac{f′(x)}{g′(x)} = lim_{x→0} \frac{cos x}{2} = \frac{1}{2}

since cos x is continuous at x = 0. Hence

lim_{x→0} \frac{sin x}{2x} = \frac{1}{2}

Applying L’Hospital’s theorem again to the functions F (x) = 1 − cos x and G(x) = x^{2}, which also satisfy its hypotheses, with

lim_{x→0} \frac{F′(x)}{G′(x)} = \frac{1}{2}

we conclude that

lim_{x→0} \frac{F (x)}{G(x)} = \frac{1}{2},

which proves the result.

5. Define the hyperbolic sine

sinh x = \frac{e^{x} − e^{−x}}{2} .

Prove that sinh x is strictly increasing on ℝ and hence has an inverse. Prove that the inverse is differentiable and compute its derivative.

Solution.

• The functions

cosh x = \frac{e^{x} + e^{−x}}{2} , sinh x = \frac{e^{x} − e^{−x}}{2} ,

satisfy the following easily verified identities:

(cosh x)′ = sinh x, (sinh x)′ = cosh x, cosh^{2} x − sinh^{2} x = 1.

• Since (sinh x)′ = cosh x > 0 on ℝ, the function sinh x is strictly increasing, and therefore invertible on its range. Moreover, since e^{x} → ∞ and e^{−x} → 0 as x → ∞, we have sinh x → ∞ as x → ∞; similarly sinh x → −∞ as x → −∞. Thus, the range of sinh : ℝ → ℝ is all of ℝ, and sinh^{−1} : ℝ → ℝ

• Since sinh x is differentiable on ℝ and (sinh x)′ ≠ 0, the inverse function is differentiable on ℝ, and

(sinh^{−1})′(sinh x) = \frac{1}{(sinh x)′} = \frac{1}{cosh x}.

Setting y = sinh x, and writing

cosh x = \sqrt{1 + sinh^{2} x} = \sqrt{1 + y^{2}},

we conclude that

(sinh^{−1})′(y) = \frac{1}{\sqrt{1 + y^{2}}} .

6. A function f has a jump discontinuity at x_{0} if both the left and right limits

lim_{x→x_{0}^{+}} f(x),  lim_{x→x_{0}^{−}} f(x)

exist but have different values. Suppose that f : (a, b) → ℝ is differentiable in (a, b). Prove that f′ does not have a jump discontinuity in (a, b).

Solution.

• Suppose, for contradiction, that the limits

L_{±} = lim_{x→x_{0}^{±}} f′(x)

exist and are distinct. Let

ϵ = \frac{L_{+} − L_{−}}{3} .

Then there exists δ > 0 such that

|f′(x) − L_{+}| < ϵ for x_{0} < x < x_{0} + δ

and

|f′(x) − L_{−}| < ϵ for x_{0} − δ < x < x_{0}.

For definiteness, suppose that L_{+} > L_{−} (otherwise, we can consider −f instead of f). It follows that f′(x) cannot take values in the interval [L_{−} + ϵ, L_{+} − ϵ] of width ϵ > 0 when 0 < |x − x_{0}| < δ. Since f′(x) > L_{+} − ϵ for x > x_{0} and f′(x) < L_{−} + ϵ for x < x_{0}, this contradicts the intermediate value property of the derivative, whatever the value of f′(x_{0}).