Calculus I, Fall 2021

Brief solutions to Midterm 02 (average = 57.88 pts, total 120 pts)

Dec 09, 2021. 10 pts each problem.

1. (5 pts + 5 pts) (average = 6.42 pts) Write down domains and ranges of all six inverse trigonometric functions. Derive the formula for (csc^{−1})′ (pay attention to the ′+′ or ′−′ sign in your formula and explain). Ans: See quiz 4 solution (please do read).

2. (5 pts + 5 pts) (average = 6.93 pts) Use the linear approximation to find an approximate value of \sqrt{1.001} and give an estimate of the error for this approximation. Need not derive the formula you use. Ans: Linear approximation = 1 + \frac{1}{2}0.001 = 1.0005. Error ≤ |\frac{1}{2} \frac{1}{2}(\frac{−1}{2})0.001^{2}| = 1.25 · 10^{−7}.

3. (2 pts + 8 pts) , average = 1.72 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. Since lim_{x→0} \frac{f(x) − (3x + 2)}{|x − 0|} = 0. Therefore 3x + 2 is the linear approximation of f near x = 0 and f′(0) = 3.

4. (5 pts + 5 pts) , average = 1.53 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{exactly one} solution on \{x ≥ 0\}. Ans: First, we show that f(3) > 0. Otherwise, if f(3) ≤ 0, then by Mean Value Theorem, there exists a c ∈ (0, 3) such that f′(c) = \frac{f(3) − f(0)}{3 − 0} ≤ \frac{0 − (−1)}{3 − 0} = \frac{1}{3}, a contradiction. Secondly, since f(0) < 0, f(3) > 0, by Intermediate Value Theorem, there exists a c_{1} ∈ (0, 3) such that f(c_{1}) = 0. Finally, we show that f(x) = 0 has only one solution. Suppose not, we will have c_{1} ≥ 0, c_{2} ≥ 0, c_{1} ≠ c_{2} such that f(c_{1}) = f(c_{2}) = 0. From Rolle’s Theorem, there exists an α ∈ (c_{1}, c_{2}) such that f′(α) = 0, a contradiction.

5. (3 pts + 2 pts + 5 pts) , average = 6.80 pts) Find all critical points of f(x) = x^{\frac{1}{3}} (x − 4) on [−1, 5]. For each one of them, use first derivative test to determine whether it corresponds to a local minimum, a local maximum or neither. Then find the absolute max and min on [−1, 5]. Ans: f′(x) = \frac{4}{3}x^{\frac{−2}{3}} (x − 1). Critical points: x = 0, x = 1. f′(x) < 0 on [−1, 1), f′(x) > 0 on (1, 5). So f(1) = only local min, therefore absolute min. Local max: f(−1) = 5, f(5) = 5^{\frac{1}{3}} . Compare values of f(−1) and f(5), we conclude that f(−1) = absolute max. x = 0 is neither local min nor local max.

6. (4 pts + 6 pts) , average = 4.02 pts) Let

f(x) = \{ \begin{matrix}e^{( \frac{−1}{x^{2}} )}, & x ≠ 0, \\ 0, & x = 0. \end{matrix}

Is f differentiable at x = 0? Is f twice differentiable at x = 0? Start with definitions of f′(0) and f′′(0) and explain. Ans: Yes. Yes.

f′(0) = lim_{x→0} \frac{f(x) − f(0)}{x − 0} = lim_{x→0} \frac{\frac{1}{x}}{e^{x^{−2}}} (apply L’Hôpital’s Rule for \frac{±∞}{∞} once) = lim_{x→0} \frac{−x^{−2}}{−2x^{−3}e^{x^{−2}}} =

lim_{x→0} \frac{x}{2e^{x^{−2}}} = 0.

f′(x) = \frac{2}{x^{3}} e^{( \frac{−1}{x^{2}} )} for x ≠ 0.

f′′(0) = lim_{x→0} \frac{f′(x) − f′(0)}{x − 0} = lim_{x→0} \frac{\frac{2}{x^{4}}}{e^{\frac{1}{x^{2}}}} = lim_{x→0} \frac{2x^{−4}}{e^{x^{−2}}} (apply L’Hôpital’s Rule for \frac{±∞}{∞} twice) =

lim_{x→0} \frac{−8x^{−5}}{−2x^{−3}e^{x^{−2}}} = lim_{x→0} \frac{4x^{−2}}{e^{x^{−2}}} = lim_{x→0} \frac{−8x^{−3}}{−2x^{−3}e^{x^{−2}}} = lim_{x→0} \frac{4}{e^{x^{−2}}} = 0

7. (4 pts + 6 pts) , average = 4.09 pts) Evaluate (a) lim_{x→0^{+}} x^{x} (b) lim_{x→0} (\frac{1}{x^{2}} − \frac{1}{sin^{2} x}) Ans: (a): See quiz 6 solution. (b): Let f(x) = x^{2}, g(x) = sin^{2}(x) and denote k-th derivative of f, g by f_{k}(x) = f^{(k)}(x), g_{k}(x) = g^{(k)}(x). We have

f_{0}(x) = x^{2}, f_{1}(x) = 2x, f_{2}(x) = 2, f_{3}(x) = 0, f_{4}(x) = 0,

g_{0}(x) = sin^{2} x, g_{1}(x) = sin 2x, g_{2}(x) = 2 cos 2x, g_{3}(x) = −4 sin 2x, g_{4}(x) = −8 cos 2x

lim_{x→0} (\frac{1}{x^{2}} − \frac{1}{sin^{2} x}) = lim_{x→0} \frac{g_{0}(x) − f_{0}(x)}{f_{0}(x)g_{0}(x)} (” = (\frac{0}{0})”) = lim_{x→0} \frac{g_{1}(x) − f_{1}(x)}{f_{0}(x)g_{1}(x) + f_{1}(x)g_{0}(x)} (” = (\frac{0}{0})”)

= lim_{x→0} \frac{g_{2}(x) − f_{2}(x)}{f_{0}(x)g_{2}(x) + 2f_{1}(x)g_{1}(x) + f_{2}(x)g_{0}(x)} (” = (\frac{0}{0})”)

= lim_{x→0} \frac{g_{3}(x) − f_{3}(x)}{f_{0}(x)g_{3}(x) + 3f_{1}(x)g_{2}(x) + 3f_{2}(x)g_{1}(x) + f_{3}(x)g_{0}(x)} (” = (\frac{0}{0})”)

= lim_{x→0} \frac{g_{4}(x) − f_{4}(x)}{f_{0}(x)g_{4}(x) + 4f_{1}(x)g_{3}(x) + 6f_{2}(x)g_{2}(x) + 4f_{3}(x)g_{1}(x) + f_{4}(x)g_{0}(x)} = \frac{−8 cos 0}{6 · 2 · 2 cos 0} = \frac{−1}{3}

8. (5 pts + 5 pts) , average = 6.88 pts) Solve for y(x) on x < 0 from y′′(x) = x^{−2}, y(−1) = 1, y′(−1) = 2. Ans:

y′(x) = y′(−1) + ∫_{−1}^{x} t^{−2}dt = 2 + (−t^{−1})|_{−1}^{x} = −\frac{1}{x} + 1

y(x) = y(−1) + ∫_{−1}^{x} ( − \frac{1}{t} + 1)dt = 1 + ( − ln |t| + t)|_{−1}^{x} = − ln(−x) + x + 2

9. (2 pts + 2 pts + 6 pts) , average = 4.50 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: See the textbook or Lecture 19 note.

10. (5 pts + 5 pts) , average = 6.40 pts) Evaluate (a) ∫_{1}^{2} \frac{1}{x(1 + ln^{2} x)}dx (b) ∫_{0}^{4} x\sqrt{2x + 1} dx

Ans: (a): = ∫_{1}^{2} \frac{1}{(1 + ln^{2} x)}d ln x = ∫_{1}^{2} tan^{−1}(ln x) = tan^{−1}(ln 2)

(b): See Section 5.5, Example 6 for the indefinite integral part, then substitute to evaluate the definite integral.

11. (2 pts + 8 pts) , average = 3.72 pts) True or False? If true, prove it. If false, give a counter example. If y = f(x) is differentiable at x = c then it is continuous at x = c. Ans: See midterm 1 solution (same problem, please do read).

12. (average = 4.86 pts) Evaluate \frac{d}{dx}(tan x)^{sin x}, 0 < x < \frac{π}{2}. Ans: See midterm 1 solution (similar (almost the same) problem, please do read).