108學年度｜微積分模組班｜Class 01｜Note 4.

-

Note 4: Review

  • Date: 108/10/28 (Week 08)
  • Place: 新生教學館 301

Part 1: Graphing Problem

  • Reference.
    1. 微積分用書 Calculus: Early Transcendetals (8E), James Stewart, Section 2.6, Limits at infinity; Horizontal Asymptotes (P.126 - P.140)
    2. 微積分用書 Calculus: Early Transcendetals (8E), James Stewart, Section 4.5, Summary of Curve Sketching (P.315 - P.323)
Problem 1.1. (95.上微積分甲統一教學一組.第5題)
    Study the function $f(x) = \dfrac{(x-2)^2}{x+1}$, and answer the following questions.
  1. The domain of $y = f(x)$.
  2. $f'(x)$.
  3. $y = f(x)$ has critical point(s) at $x = $?
  4. $y = f(x)$ is increasing / decreasing on intervals?
  5. $f''(x)$.
  6. $y = f(x)$ is concave upward / down on intervals?
  7. Find the $(x,y)$ coordinates of the following points if exist.
    1. local maximum point(s)
    2. local minimum point(s)
    3. inflection point(s)
  8. Find the asymptotes of the graph $y = f(x)$ if exist.
    1. Vertical asymptotes(s)
    2. Horizontal asymptotes(s)
    3. Slanted asymptotes(s)
  9. Sketch the graph of $y = f(x)$ below.
Solution.
  1. Obviously, the domain is $\mathbb{R} \backslash \{-1\}$.
  2. $f'(x) = \dfrac{2(x-2)(x+1)-(x-2)^2 \cdot 1}{(x+1)x^2} = \dfrac{(x-2)(x+4)}{(x+1)^2}$.
  3. $f(x)$ has critical point at $x = 2$ and $x = -4$ by setting $f'(x) = 0$.
  4. $\begin{cases} f'(x) > 0 \Longrightarrow f \text{ increasing} &, x \in (-\infty, -4) \cup (2, \infty) \\ f'(x) < 0 \Longrightarrow f \text{ decreasing} &, x \in (-4, -1) \cup (-1, 2) \end{cases}$
  5. $f''(x) = \dfrac{(2x+2)(x+1)^2-(x^2+2x-8)\cdot 2(x+1)}{(x+1)^4} = \dfrac{18}{(x+1)^3}$.
  6. $\begin{cases} f''(x) > 0 \Longrightarrow f \text{ concave upward} &, x \in (-1, \infty) \\ f''(x) < 0 \Longrightarrow f \text{ concave downward} &, x \in (-\infty, -1) \end{cases}$
  7. For critical points:
    1. Note that $f''(2) = \dfrac{2}{3} > 0$, so $f(2) = 0$ is a local minimum.
    2. Note that $f''(-4) = -\dfrac{2}{3} < 0$, so $f(-4) = -12$ is a local maximum.
    3. There is no inflection points.
  8. For asymptotes(s):
    1. Note that $\displaystyle \lim_{x \to (-1)^-}f(x) = -\infty$ and $\displaystyle \lim_{x \to (-1)^+}f(x) = +\infty$, there is a vertical asymptote $x = -1$. (Also the only one)
    2. Note that $\displaystyle \lim_{x \to \infty}f(x) = \infty$ and $\displaystyle \lim_{x \to -\infty}f(x) = -\infty$, there are no horizontal asymptote.
    3. Note that $\dfrac{(x-2)^2}{x+1} = (x-5) + \dfrac{9}{x+1}$, so we have:$$\lim_{x \to \infty}|f(x)-(x-5)| = \lim_{x \to -\infty} |f(x)-(x-5)| = 0$$, so $y = x-5$ is a slant asymptote.
  9. The graph:
$\square$
Remark 1.2.
  • A general method to determine if $y=f(x)$ has a slant asymptote:
  1. Compute $\displaystyle \lim_{x \to \infty }\dfrac{f(x)}{x}$, if it converges to some limit $m \neq 0$, then $y = mx + b$ is a \textbf{candidate} of slant asymptote for $f(x)$. (Similar if we are finding $x \to -\infty$)
  2. Solve $b$ by the equation $\displaystyle \lim_{x \to \infty }\left[f(x)-(mx+b)\right] = 0$. (Similar if we are finding $x \to -\infty$)
Exercise 1.3. (107.上微積分微甲統一教學.第7題)
  • Consider the function $f \colon (1,\infty) \to \mathbb{R}$, $f(x) = \ln x − \ln(\ln x)$.
  1. Compute $f'(x)$. Find the intervals in the domain of $f$ on which $f$ is increasing and those on which $f$ is decreasing. What are the extreme values of $f$?
  2. Compute $f''(x)$. Find the intervals on which the graph of $f$ is concave upward and those on which the graph of $f$ is concave downward. Is there any inflection point of the curve $y = f(x)$?
  3. Find the vertical and horizontal asymptotes of the curve $y = f(x)$ if any. Find $\displaystyle \lim_{x \to \infty}\dfrac{f(x)}{x}$. Does the curve $y = f(x)$ have any slant asymptote?
  4. Draw the graph of $f(x)$.
Exercise 1.4.
  • Investigate the following function, answer the list below for each function:
    • The domain of $y = f(x)$.
    • (First derivative)
      • What is $f'(x)$?
      • $y = f(x)$ has critical point(s) at $x = $?
      • $y = f(x)$ is increasing / decreasing on intervals?
    • (Second derivatives)
      • What is $f''(x)$?
      • $y = f(x)$ is concave upward / down on intervals?
    • Find the $(x,y)$ coordinates of the following points if exist.
      • local maximum point(s).
      • local minimum point(s).
      • inflection point(s).
    • Find the asymptotes of the graph $y = f(x)$ if exist.
      • Vertical asymptotes(s).
      • Horizontal asymptotes(s).
      • Slanted asymptotes(s).
    • Sketch the graph of $y = f(x)$.
  1. ([J] Sec 4.5, Exer.11) $y = \dfrac{x-x^2}{2-3x+x^2}$.
  2. ([J] Sec 4.5, Exer.13) $y = \dfrac{x}{x^2-4}$.
  3. ([J] Sec 4.5, Exer.18) $y = 1+\dfrac{1}{x}+\dfrac{1}{x^2}$.
  4. ([J] Sec 4.5, Exer.46) $y = x-\ln x$.
  5. ([J] Sec 4.5, Exer.51) $y = xe^{-\frac{1}{x}}$.

Part 2: Limit Problem

  • Reference.
    1. 微積分用書 Calculus: Early Transcendetals (8E), James Stewart, Section 2.2, The Limit of a Function (P.83 - P.94)
    2. 微積分用書 Calculus: Early Transcendetals (8E), James Stewart, Section 2.3, Calculating Limits Using the Limit Laws (P.95 - P.104)
    3. 微積分用書 Calculus: Early Transcendetals (8E), James Stewart, Section 4.4, Indeterminate Forms and l'Hospital's Rule (P.304 - P.314)
Exercise 2.1. (105.上微積分甲統一教學一組.第2題)
    Find the following limits if they exist.
  1. $\displaystyle \lim_{x \to 0^+} (1-\cos x)^\frac{1}{\ln(x)}$.
  2. $\displaystyle \lim_{x \to 1}\left(\dfrac{1}{\ln x} - \dfrac{1}{x-1}\right)$.
Solution.
  1. By direct computation: \begin{align*} \lim_{x \to 0^+} (1-\cos x)^\frac{1}{\ln(x)} &= \lim_{x \to 0^+} \exp \left(\dfrac{\ln(1-\cos x)}{\ln x}\right) \\ & \overset{(*)}{=} \exp \left( \lim_{x \to 0^+}\dfrac{\ln(1-\cos x)}{\ln x}\right) \\ & \overset{L'H}{=} \exp \left( \lim_{x \to 0^+}\dfrac{\dfrac{\sin x}{1-\cos x}}{\dfrac{1}{x}}\right) \\ & = \exp \left( \lim_{x \to 0^+}\dfrac{x\sin x}{1-\cos x} \right) \\ & \overset{L'H}{=} \exp \left( \lim_{x \to 0^+} \dfrac{\sin x + x\cos x}{\sin x}\right) \\ & \overset{L'H}{=} \exp \left( \lim_{x \to 0^+} \dfrac{\cos x + \cos x - x\sin x}{\cos x}\right) \\ & = \exp (2) = e^2 \end{align*} $(*)$ holds since $\exp(x)$ is continuous. I want to show here that don't be scared of L'Hospital's Rule, you can always finish the calculation of limit some day. Of course there are various "neat" way in comparison to  L'Hospital's Rule solving limit problem, but during test, it's more efficient using  L'Hospital's Rule if the calculation isn't too complicated.
  2. By direct computation: \begin{align*} \lim_{x \to 1}\left(\dfrac{1}{\ln x} - \dfrac{1}{x-1}\right) &= \lim_{x \to 1}\dfrac{(x-1)-\ln x}{(x-1) \ln x} \\ & \overset{L'H}{=} \lim_{x \to 1}\dfrac{1-\dfrac{1}{x}}{\ln x + (x-1) \cdot \dfrac{1}{x}} \\ & = \lim_{x \to 1}\dfrac{x-1}{x\ln x + (x-1)} \\ & \overset{L'H}{=} \lim_{x \to 1} \dfrac{1}{\ln x + x \cdot \dfrac{1}{x}+1} \\ & = \dfrac{1}{2} \end{align*}
    3. $\square$
Exercise 2.2. (107.上微積分甲統一教學.第1題)
    Find the following limits.
  1. $\displaystyle \lim_{x \to 0} \sqrt[3]{x} \sin \left( \sin \left( \dfrac{1}{|x|} \right) \right)$.
  2. $\displaystyle \lim_{x \to \infty} \sqrt[3]{x} \sin \left( \sin \left( \dfrac{1}{|x|} \right) \right)$.
  3. $\displaystyle \lim_{x \to -\infty} \left(\sqrt{x^2+2x-1}-x\right)$.
Exercise 2.3. (106.上微積分甲統一教學.第2題)
  1. Compute the limit if it exists or explain why it doesn't exist. $$\lim_{x \to \infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} \right)$$
  2. Determine for what values of $a$, $0<a<1$, does $\displaystyle \lim_{x \to \infty} f(x)$ exist, where: $$f(x) = \sqrt{x+x^a}-\sqrt{x-x^a}$$
Exercise 2.4.
    Find the limit. Use l'Hospital's Rule where appropriate.
  1. ([J] Sec 4.4, Exer.28) $\displaystyle \lim_{x \to 0}\dfrac{\sinh x - x}{x^3}$.
  2. ([J] Sec 4.4, Exer.40) $\displaystyle \lim_{x \to 0}\dfrac{e^x-e^{-x}-2x}{x-\sin x}$.
  3. ([J] Sec 4.4, Exer.53) $\displaystyle \lim_{x \to 0^+}\left(\dfrac{1}{x}-\dfrac{1}{e^x-1}\right)$.
  4. ([J] Sec 4.5, Exer.64) $\displaystyle \lim_{x \to \infty} x^{e^{-x}}$.
  5. ([J] Sec 4.5, Exer.68) $\displaystyle \lim_{x \to \infty} \left(\dfrac{2x-3}{2x+5}\right)^{2x+1}$.

Part 3: Glueing Problem

  • Reference.
    1. 微積分用書 Calculus: Early Transcendetals (8E), James Stewart, Section 2.5, Continuity (P.114 - P.126)
    2. 微積分用書 Calculus: Early Transcendetals (8E), James Stewart, Section 2.8, The Derivative as a Function (P.152 - P.165)
Problem 3.1. (107.上微積分甲統一教學.第3題)
  • Let $f(x) = \begin{cases} (1+x)^{\frac{1}{x}} &, x \neq 0, x>-1\\ a &, x = 0\end{cases}$
  1. Find the value of a such that $f(x)$ is continuous at $x = 0$.
  2. Find $\displaystyle \lim_{x \to \infty} f(x)$.
  3. Compute $f'(x)$, for $x \neq 0$.
  4. Is $f(x)$ differentiable at $x = 0$? If $f(x)$ is differentiable at $x = 0$, then find $f'(0)$.
Solution.
  1. Note that we have: \begin{align*} \lim_{x \to 0}(1+x)^\frac{1}{x} &= \lim_{x \to 0}\exp \left(\dfrac{\ln(1+x)}{x}\right) \\ & \overset{(*)}{=} \exp \left(\lim_{x \to 0}\dfrac{\ln(1+x)}{x}\right) \\ & \overset{L'H}{=} \exp \left(\lim_{x \to 0}\dfrac{\dfrac{1}{1+x}}{1}\right) \\ & = \exp(1) = e \end{align*} $(*)$ holds since $\exp(x)$ is continuous. So we may take $a = e$.
  2. Again, by direct computation: \begin{align*} \lim_{x \to \infty}(1+x)^\frac{1}{x} &= \lim_{x \to \infty}\exp \left(\dfrac{\ln(1+x)}{x}\right) \\ & \overset{(*)}{=} \exp \left(\lim_{x \to \infty}\dfrac{\ln(1+x)}{x}\right) \\ & \overset{L'H}{=} \exp \left(\lim_{x \to \infty}\dfrac{\dfrac{1}{1+x}}{1}\right) \\ & = \exp(0) = 1 \end{align*} $(*)$ holds since $\exp(x)$ is continuous.
  3. This is just direct computation: \begin{align*} \dfrac{\mathrm{d}}{\mathrm{d}x}(1+x)^{\frac{1}{x}} &= \dfrac{\mathrm{d}}{\mathrm{d}x}\exp\left(\dfrac{\ln(1+x)}{x}\right) \\ &= \exp\left(\dfrac{\ln(1+x)}{x}\right) \cdot \left[\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\ln(1+x)}{x}\right)\right] \\ &= (1+x)^\frac{1}{x} \cdot \dfrac{\dfrac{1}{1+x} \cdot x-\ln(1+x) \cdot 1}{x^2} \\ &= (1+x)^{\frac{1}{x}-1}\cdot \dfrac{x-(1+x)\ln(1+x)}{x^2} \end{align*}
  4. This is just direct computation: \begin{align*} f'(0) &= \lim_{x \to 0}\dfrac{f(x)-f(0)}{x-0} \\ &= \lim_{x \to 0}\dfrac{(1+x)^{\frac{1}{x}}-e}{x} \\ &\overset{L'H}{=} \lim_{x \to 0} (1+x)^{\frac{1}{x}-1}\cdot \dfrac{x-(1+x)\ln(1+x)}{x^2} \\ \end{align*} Note that we have:
    5.     \begin{align*} \lim_{x \to 0} \dfrac{x-(1+x)\ln(1+x)}{x^2} &\overset{L'H}{=} \lim_{x \to 0} \dfrac{1-1\cdot \ln(1+x) - (1+x) \cdot \dfrac{1}{1+x}}{2x} \\ &\overset{L'H}{=} \lim_{x \to 0}\dfrac{-\dfrac{1}{1+x}}{2} \\ &= -\dfrac{1}{2} \end{align*} Last, we have: \begin{align*} \dfrac{\mathrm{d}}{\mathrm{d}x}(1+x)^{\frac{1}{x}} &= \lim_{x \to 0}(1+x)^\frac{1}{x} \cdot \lim_{x \to 0}\dfrac{1}{1+x} \cdot \lim_{x \to 1} \dfrac{x-(1+x)\ln(1+x)}{x^2} \\ &= e \cdot \dfrac{1}{1} \cdot \dfrac{-1}{2} = -\dfrac{1}{2}e \end{align*}
      $\square$
Exercise 3.2. (106.上微積分甲統一教學.第4題)
    Let $f(x) = \begin{cases} x^{\frac{4}{3}}\cos \left(\dfrac{1}{x}\right) &, x \neq 0\\ 0 &, x = 0\end{cases}$
  1. Is $f(x)$ continuous at $x = 0$?
  2. Compute $f'(x)$, for $x \neq 0$ and $f'(0)$.
  3. Is $f'(x)$ continuous at $x=0$?

Part 4: Proof

  • Reference.
    1. 老師的授課筆記與我的講義。
Problem 4.1. (105.上微積分甲統一教學一組.第6題)
    Suppose that $f$ is a differentiable function. If $f'(a) > 0$ and $f'(b)<0$, explain that there exists $c \in (a, b)$ such that $f'(c) = 0$. (Note that $f'(x)$ may not be continuous.)
Solution.
    Note that by EVT$^1$, there exist global maximum and minimum of $f(x)$ on $[a,b]$. We claim the following: Claim: $f(a)$ and $f(b)$ can't be global maximum.
      Note that $f'(a)>0$, that is, $\displaystyle \lim_{x \to a^+}\dfrac{f(x)-f(a)}{x-a} = f'(a)>0$, let $\epsilon = f'(a)>0$, then there exist $\delta >0$ such that for all $x \in (a,a+\delta)$, we have: $$\left|\dfrac{f(x)-f(a)}{x-a}-f'(a)\right| < \epsilon \Longleftrightarrow 0 < \dfrac{f(x)-f(a)}{x-a}<2f'(a)$$, hence, for all $x \in (a,a+\delta)$, $f(a)<f(x)$. ($f(a)$ is a local minimum) So $f(a)$ can't be global maximum. The case dealing $f(b)$ is similar.
    Hence, global maximum must appear in $(a,b)$, for such $c \in (a,b)$ with $f(c)$ being the global maximum, we have $f'(c) = 0$ and hence the proof.
($^1$: EVT: Extreme Value Theorem, every continous function on a closed and bounded subset of $\mathbb{R}$ must achieves its global maximum and minimum.)

 Exercise 4.2.
    A number $x_0$ is called a fixed point of a function $f(x)$ if $f(x_0) = x_0$. Prove that if $f \colon [0, 1] \to [0, 1]$ is differentiable, and $f'(x) \neq 1$ for all $x \in [0, 1]$, then $f$ has a unique fixed point $x_0 \in [0, 1]$. (Hint: You need to prove existence and uniqueness. Use the intermediate value theorem to prove the existence part, and mean value theorem for the uniqueness part)
Exercise 4.3. (106.上微積分甲統一教學一組.第9題)
    $f(x)$ is a differentiable function defined on $\mathbb{R}$. Let $g(x) = f(x) \cdot |f(x)|$.
  1. Find the domain of $g'(x)$ and compute $g'(x)$. (Hint: To compute $g'(x_0)$ you may need to discuss the cases $f(x_0) > 0$, $f(x_0) < 0$, and $f(x_0) = 0$ seperately.)
  2. Suppose that $f'(x)>0$ on the interval $(a, b)$. Show that $g(x)$ has at most one critical point on $(a, b)$.
  3. Suppose that $f'(x) > 0$ on the interval $(a, b)$. Show that $g(x_1) < g(x_2)$ for all $a \leqslant x_1 < x_2 \leqslant b$.

留言

張貼留言

這個網誌中的熱門文章

108學年度｜微積分模組班｜Class 03｜Homework 6.

-
圖片
Homework 6. Date: 108/10/31 Place: 共同教學館 102 [J] 代表微積分用書 Calculus: Early Transcendetals (8E), James Stewart. Problem 1. WeBWorK: homework 4.3, 4.4, 4.5, 4.7 Problem 2. ([J], Section 4.3, Problem 33., Page 301) Suppose $f$ is a continuous function where $f(x)>0$ for all $x$, $f(0) = 4$, $f'(x)>0$ if $x<0$ or $x>2$, $f'(x)<0$ is $0<x<2$, $f''(-1) = f''(1) = 0$, $f''(x)>0$ if $x<-1$ or $x>1$, $f''(x)<0$ if $-1<x<1$. Can $f$ have an absolute maximum? If so, sketch a possible graph of $f$. If not, explain why. Can $f$ have an absolute minimum? If so, sketch a possible graph of $f$. If not, explain why. Sketch a possible graph for $f$ that does not achieve an absolute minimum. Solution. Impossible, since $f'(x)>0$, $f''(x)>0$ for $x>>0$. (In particular, this is true for $x>2$), so $\displaystyle \lim_{x \to \infty }f(x) = \infty$, there is no absolute ma...
閱讀完整內容

賦距空間 (Metric Space)

-
Definition 1.  距離函數 (Metric) 與賦距空間 (Metric space) 給定一個集合 $M$，一個距離函數 $d \colon M \times M \to \mathbb{R}$ 指的是滿足以下三個性質的函數： (正定性 Positive definite) $d(x,y) \geqslant 0$, $\forall$ $x,y \in M$, 並且 $d(x,y)=0 \Leftrightarrow x=y$ (對稱性 Symmetric) $d(x,y)=d(y,x)$, $\forall$ $x,y \in M$ (三角不等式 Triangular inequality) $d(x,y) \leqslant d(x,z)+d(z,y)$, $\forall$ $x,y,z \in M$ 一個集合 $M$ 與定義在 $M$ 之上的距離函數 $d$，我們稱之為賦距空間 $(M,d)$ Example 1 (Metric space) $(\mathbb{R},d_2)$, $d_2(x,y)=|x-y|$ $(\mathbb{R^n},d_2)$, $\displaystyle d_2(\mathbf{x},\mathbf{y})=\sqrt{\sum_{i=1}^n(x_i-y_i)^2}$, 其中 $\mathbf{x}=(x_1,x_2,...,x_n)$, $\mathbf{y}=(y_1,y_2,...,y_n)$ $(\mathbb{R^n},d_{max})$, $d_{max}(\mathbf{x},\mathbf{y})=\max\{|x_i-y_i|,1 \leqslant i \leqslant n\}$ , 其中 $\mathbf{x}=(x_1,x_2,...,x_n)$, $\mathbf{y}=(y_1,y_2,...,y_n)$ $(M,d)$, $M$ 是任意的集合，$d$ 是離散距離 (discrete metric)，定義為： $d_M(x,y)=\begin{cases} 1 &,x \neq y\\ 0 &,x=y\end{cases}, \forall x,y \in M$ Proposition 1....
閱讀完整內容