Calculus I, Fall 2015

Midterm Exam 2

Dec 08, 2015, 10:10AM

1. (10 pts) True or False? If true, prove it. If false, give a counter example.

If |f(x) − (3x + 2)| ≤ |x|^{1.5} for all x ∈ R, then f is differentiable at x = 0.

Ans: True. (2 points) Take x = 0, then |f(0) − 2| ≤ 0 ⇒ f(0) = 2. (2 points) Consider |\frac{f(x)−f(0)}{x} − 3| = |\frac{f(x)−(3x+2)}{x}| ≤ |x|^{0.5} → 0 as x → 0. (6 points)

2. (a) (6 pts) Graph f(x) = \frac{x}{\sqrt{x^{2} + 1}}. Give all details including possible asymptotes.

(b) (6 pts) The function y = f(x) is odd (f(−x) = −f(x)) and the root x^{∗} to the equation f(x) = 0 is x^{∗} = 0. Give formula of Newton’s method for finding this root.

(c) (6 pts) The Newton’s method does not always converge. There is an a > 0 such that Newton’s method converges if and only if −a < x_{0} < a. Take this fact for granted and find a (show how to find a, but need NOT prove that Newton’s method converge if and only if −a < x_{0} < a).

Ans:

(a) f′(x) = \frac{1}{(x^{2}+1)^{3/2}} . (1 point) f′′(x) = \frac{−3x}{(x^{2}+1)^{5/2}} . (1 point) Graphing-Inflection point: (0, 0) (1 point) Increasing on: (−∞, ∞) (1 point) Concave up on: (−∞, 0); concave down on: (0, ∞) (1 point) f(x) → ±1 as x → ±∞ ⇒ y = ±1 are horizontal asymptotes. (1 point)

(b) x_{n+1} = x_{n} − \frac{f(x_{n})}{f′(x_{n})} (2 points) = x_{n} − \frac{\frac{xn}{\sqrt{x^{2}_{n}+1}}}{\frac{1}{(x^{2}_{n}+1)^{3/2}}} (2 points) = −x_{n}^{3} (2 points).

(c) |x_{n} − x_{∗}| = |x_{n}| = |x_{n−1}|^{3} = … = |x_{0}|^{3^{n}} → 0 (3 points) if and only if |x_{0}| < 1. Thus a = 1 (3 points).

3. (12 pts) Let f be a differentiable function defined on {x ≥ 0} satisfying

(a): f(0) = −1,

(b): f′(x) ≥ 1/2 for all x ≥ 0.

Show that f(x) = 0 has \underline{one and only one} solution on \{x ≥ 0\}.

Ans:

Step 1: f(x) = 0 has \underline{one} solution on \{x ≥ 0\}:

Since

f(2) = f(0) + ∫_{0}^{2} f′(x)dx ≥ −1 + (2 − 0) min_{0≤x≤2} f′(x) ≥ −1 + 2 × \frac{1}{2} = 0 (2points)

and f(0) = −1 < 0. (1 point)

Moreover, f′(x) exist for all x ≥ 0, which implies f(x) is continuous on x ≥ 0 (1 point). By Intermediate Value Theorem, there exists c between 0 and 2 such that f(c) = 0 (2 points).

Step 2:

f(x) = 0 has \underline{only one} solution on \{x ≥ 0\}:

If there exist c_{1} ≥ 0, c_{1} ≠ c such that f(c_{1}) = 0 (2 points), by Rolle’s Theorem, there exist c_{2} between c and c_{1} such that

f′(c_{2}) = 0 (2points)

which is contradict to f′(x) ≥ \frac{1}{2} for all x ≥ 0. Hence f(x) = 0 has only one solution on x ≥ 0 (2 points).

4. (18 pts) Find the limits of the following expressions:

(a) lim_{x→0^{+}} x^{x} (b) lim_{x→0^{+}} \frac{e^{\frac{−1}{x}}}{x} (c) lim_{x→0} \frac{x^{2} cos \frac{1}{x}}{sin x}

Ans:

(a) lim_{x→0^{+}} x^{x} = lim_{x→0^{+}} e^{x ln x} (2points) = e^{lim_{x→0^{+}} x ln x} (2points) = 1 (2points).

(b) lim_{x→0^{+}} \frac{e^{\frac{−1}{x}}}{x} = lim_{x→0^{+}} \frac{\frac{1}{x}}{e^{\frac{1}{x}}} (2points) = lim_{x→0^{+}} \frac{1}{e^{\frac{1}{x}}} (2points) = 0 (2points).

(c) lim_{x→0} \frac{x^{2} cos \frac{1}{x}}{sin x} = lim_{x→0} \frac{x}{sin x} lim_{x→0} x cos \frac{1}{x} (2points) = 1·0 (2points) = 0 (2points).

5. (8 pts) Solve for y(x) on x < 0 from

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

Ans: ∫_{−1}^{x} y′′(t) dt = ∫_{−1}^{x} t^{−2} dt = −\frac{1}{x} − 1 (2 points) ⇒ y′(x) = 1 − \frac{1}{x} (2 points). ∫_{−1}^{x} y′(t) dt = ∫_{−1}^{x}(1 − \frac{1}{t}) dt = − ln |x| + x + 1 (2 points) ⇒ y(x) = − ln |x| + x + 2 (2 points).

6. (8 pts) Evaluate lim_{n→∞} ∑_{k=n}^{2n} \frac{n}{k^{2}}

Ans:

lim_{n→∞} ∑_{k=n}^{2n} \frac{n}{k^{2}} = lim_{n→∞}(\frac{n}{n^{2}} + ∑_{k=n+1}^{2n} \frac{1}{n} \frac{1}{(\frac{k}{n})^{2}} )(3points) = ∫_{1}^{2} \frac{1}{x^{2}} dx(3points) = \frac{1}{2}(2points)

2 7. (14 pts) State both parts of Fundamental Theorem of Calculus, prove that ’part 1 implies

part 2’. If you can’t prove this, you could prove ’part 1’ instead. Ans: The Fundamental Theorem of Calculus, Part 1: If f is continuous on [a, b](2 points), then F (x) = ∫_{a}^{x} f(t)dt is continuous on [a, b] and differentiable on (a, b) and its derivative is f(x)(2 points):

F′(x) = \frac{d}{dx} ∫_{a}^{x} f(t)dt = f(x)  (1)

The Fundamental Theorem of Calculus, Part 2: If f is continuous at every point in [a, b](2 points) and F is any antiderivative of f on [a, b], then

∫_{a}^{b} f(t)dt = F (b) − F (a)(2points)  (2)

Proof of part 1 implies part 2: Let G(x) be any antiderivative of f(x), which is

\frac{d}{dx}G(x) = f(x)

Since \frac{d}{dx} ∫_{a}^{x} f(t)dt = f(x) = \frac{d}{dx}G(x), we have

∫_{a}^{x} f(t)dt + c = G(x)

for some constant c ∈ R.(3 points) Then G(b) − G(a) = ∫_{a}^{b} f(t)dt + c − (∫_{a}^{a} f(t)dt + c) = ∫_{a}^{b} f(t)dt(3 points).

8. (12 pts) Evaluate

(a) ∫_{1}^{2} \frac{1}{x(1 + ln^{2} x)}dx (b) ∫_{0}^{4} x\sqrt{2x + 1} dx

Ans: (a) Let u = ln x

∫_{1}^{2} \frac{1}{x(1 + ln^{2} x)}dx = ∫_{0}^{ln 2} \frac{1}{1 + u^{2}} du(3points) = tan^{−1}(ln 2)(3points)

(b) Let z = 2x + 1

∫_{0}^{4} x\sqrt{2x + 1} dx = ∫_{1}^{9} \frac{1}{4}(z − 1)\sqrt{z}dz(3points) = \frac{121}{5} − \frac{13}{3} = \frac{298}{15}(3points)

9. True or False? If true, prove it. If false, give a counter example. (a) (8 pts) If y = f(x) is differentiable at x = c then it is continuous at x = c.

(b) (8 pts) If y = f(x) is continuous at x = c then it is differentiable at x = c. Ans:

(a) True. Suppose f′(c) exitst.

lim_{x→c}[f(x)−f(c)] = lim_{x→c} \frac{f(x) − f(c)}{x − c}·(x−c) = lim_{x→c} \frac{f(x) − f(c)}{x − c}·lim_{x→c}(x−c) = f′(c)·0 = 0.

(8 pts)

(b) False. Let f(x) = |x| and c = 0. Then f is continuous at c but not differentiable at

c. (∵ lim_{x→0+} \frac{|x|}{x} = 1 ≠ −1 = lim_{x→0−} \frac{|x|}{x})

(8 pts)

10. (8 pts) Start with domain and range for csc and csc^{−1}, derive the formula for the derivative of csc^{−1}. Ans:

csc y : [−\frac{π}{2} , 0) ∪ (0, \frac{π}{2} ] → (−∞, −1] ∪ [1, ∞),

csc^{−1} x : (−∞, −1] ∪ [1, ∞) → [−\frac{π}{2} , 0) ∪ (0, \frac{π}{2} ]. (2 pts)

y = csc^{−1} x ⇒ csc y = x ⇒ − csc y cot y \frac{dy}{dx} = 1 ⇒ \frac{dy}{dx} = −\frac{1}{csc y cot y} (3 pts)

⇒ \frac{dy}{dx} = \{ \begin{matrix}−\frac{1}{x\sqrt{x^{2}−1}}, & x > 1 \\ \frac{1}{x\sqrt{x^{2}−1}}, & x < −1 \end{matrix} = −\frac{1}{|x|\sqrt{x^{2} − 1}}, |x| > 1. (3 pts)