Calculus I, Fall 2015

Midterm Exam 1

Oct 27, 2015, 10:10AM

1. (10 pts) Evaluate lim_{θ→ \frac{π}{3}} \frac{cos θ − \frac{1}{2}}{θ − \frac{π}{3}} .

Ans:

lim_{θ→ \frac{π}{3}} \frac{cos θ − \frac{1}{2}}{θ − \frac{π}{3}} = (cos θ)′|_{θ= \frac{π}{3}} (5 pts) = − sin θ|_{θ= \frac{π}{3}} = −\frac{\sqrt{3}}{2}.(5 pts)

2. (10 pts) Let f_{i}(x) = a_{i}x^{2} + b_{i}x + c_{i}, i = 1, 2, 3. Evaluate

\frac{d}{dx} | \begin{matrix}f_{1}(x) & f′_{1}(x) & f′′_{1}(x) \\ f_{2}(x) & f′_{2}(x) & f′′_{2}(x) \\ f_{3}(x) & f′_{3}(x) & f′′_{3}(x) \end{matrix} |

Ans:

= | \begin{matrix}f′_{1}(x) & f′_{1}(x) & f′′_{1}(x) \\ f′_{2}(x) & f′_{2}(x) & f′′_{2}(x) \\ f′_{3}(x) & f′_{3}(x) & f′′_{3}(x) \end{matrix} |+| \begin{matrix}f_{1}(x) & f′′_{1}(x) & f′′_{1}(x) \\ f_{2}(x) & f′′_{2}(x) & f′′_{2}(x) \\ f_{3}(x) & f′′_{3}(x) & f′′_{3}(x) \end{matrix} |+| \begin{matrix}f_{1}(x) & f′_{1}(x) & f′′′_{1}(x) \\ f_{2}(x) & f′_{2}(x) & f′′′_{2}(x) \\ f_{3}(x) & f′_{3}(x) & f′′′_{3}(x) \end{matrix} | (6 pts)

= | \begin{matrix}f_{1}(x) & f′_{1}(x) & f′′′_{1}(x) \\ f_{2}(x) & f′_{2}(x) & f′′′_{2}(x) \\ f_{3}(x) & f′_{3}(x) & f′′′_{3}(x) \end{matrix} | (2 pts)

= | \begin{matrix}f_{1}(x) & f′_{1}(x) & 0 \\ f_{2}(x) & f′_{2}(x) & 0 \\ f_{3}(x) & f′_{3}(x) & 0 \end{matrix} | = 0. (2 pts)

3. (10 pts) Find \frac{dy}{dx} where y = x^{(x^{x})}, x > 0.

Ans: = x^{(x^{x})}x^{x}(ln x(1 + ln x) + \frac{1}{x}).(3 pts+3 pts+4 pts)

4. (10 pts) Find y′(1) and y′′(1) where y(x) is implicitly given by tan(x+y)+sin^{−1}(x^{2}+y) = 0 near (x, y) = (1, −1). Ans: \frac{d}{dx} once:

[sec^{2}(x + y)](1 + y′) + \frac{2x + y′}{\sqrt{1 − (x^{2} + y)^{2}}} = 0, (3 pts)

evaluate at x = 1, y = −1, one gets y′(1, −1) = −\frac{3}{2}.(2 pts) \frac{d}{dx} twice: [2 sec^{2}(x + y) tan(x + y)](1 + y′)^{2} + [sec^{2}(x + y)]y′′ + \frac{(2+y′′)(1−(x^{2}+y)^{2})+(x^{2}+y)(2x+y′)^{2}}{(1−(x^{2}+y)^{2})^{\frac{3}{2}}} = 0(3 pts) evaluate at x = 1, y = −1, one gets y′′(1, −1) = −1.(2 pts)

5. (10 pts) Let f^{−1} be the inverse function of f. Evaluate \frac{d^{2}}{dy^{2}} f^{−1}(y) in terms of f′ and f′′. Show all details. Ans:

\frac{d}{dy} f^{−1}(y) = \frac{1}{f′(f^{−1}(y))}. (2 pts)

\frac{d^{2}}{dy^{2}} f^{−1}(y) = −\frac{f′′(f^{−1}(y)) · \frac{d}{dy}f^{−1}(y)}{[f′(f^{−1}(y))]^{2}} (6 pts)

= −\frac{f′′(f^{−1}(y))}{[f′(f^{−1}(y))]^{3}} . (2 pts)

6. (10 pts) Let f(x) : [0, 1] ↦ [0, 1] be a continuous function. Prove that f(x) = x has at least one solution. Ans: Let g(x) = f(x) − x, then g(x) is continuous on [0, 1].(3 pts) Since 0 ≤ f(x) ≤ 1, we have

g(0) = f(0) − 0 ≥ 0(2 pts)
g(1) = f(1) − 1 ≤ 0(2 pts)

By Intermediate Value Theorem, ∃c ∈ [0, 1] s.t. g(c) = 0, i.e. f(c) = c.(3 pts) Hence f(x) = x has at least one solution on [0, 1].

7. (10 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} (4 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. (4 pts)

8. (10 pts) True or False? If true, prove it. If false, give a counter example. (a) If y = f(x) is differentiable at x = c then it is continuous at x = c.

(b) 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.

(5 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})

(5 pts)

9. (10 pts) Give formal definition of lim_{x→∞}f(x) = L. Then verify that lim_{x→∞} \frac{1}{x} = 0 using the ε − δ argument.

Ans: lim_{x→∞} f(x) = L if given any ϵ > 0, ∃N > 0 such that for all x, x > N ⇒ |f(x)−L| < ϵ.(5 pts)

Given any ϵ > 0, pick N = \frac{1}{ϵ} > 0(3 pts), then for all x, x > N, |\frac{1}{x} − 0| = \frac{1}{x} < \frac{1}{N} = ϵ(2 pts)

10. (16 pts) Give formal definition of lim_{x→c} f(x) ≠ L. Then verify that lim_{x→0} sin x ≠ 1 using ε − δ argument.

Ans: lim_{x→c} f(x) ≠ L if ∃ ϵ > 0 such that for any δ > 0, there exists x such that 0 < |x − c| < δ and |f(x) − L| ≥ ϵ. (4 pts) Take ϵ = \frac{1}{2}. Given any δ > 0. Take x = min\{\frac{δ}{2}, \frac{π}{6}\}. (6 pts) Then 0 < |x − 0| = x < δ and | sin x − 1| = 1 − sin x ≥ 1 − \frac{1}{2} = ϵ. (∵ 0 < x ≤ \frac{π}{6} ⇒ 0 < sin x ≤ \frac{1}{2}) (6 pts)