108學年度｜微積分模組班｜Class 03｜Homework 6.

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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 $-1<x<1$.
  1. Can $f$ have an absolute maximum? If so, sketch a possible graph of $f$. If not, explain why.
  2. Can $f$ have an absolute minimum? If so, sketch a possible graph of $f$. If not, explain why.
  3. Sketch a possible graph for $f$ that does not achieve an absolute minimum.
Solution.
  1. 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 maximum.
  2. Possible, in the following (Green curve), we have absolute minimum be $f(2)$.
  3. As following (Red curve):

Problem 3. ([J], Section 4.3, Problem 80., Page 303)
  1. If $f$ and $g$ are concave upward on $I$, show that $f+g$ is concave upward on $I$.
  2. If $f$ is positive and concave upward on $I$, show that the function $g(x)=[f(x)]^2$ is concave upward on $I$.
Solution.
  1. Note that $f''(x)>0$ and $g''(x)>0$ on $I$, so $(f+g)'(x) = f'(x)+g'(x)>0$ on $I$, hence, $f+g$ is concave upward on $I$.
  2. Note that $g'(x) = 2f(x)f'(x) > 0$ by assumption, so $g(x)$ is concave upward on $I$.
Problem 4. ([J], Section 4.3, Problem 90., Page 304)
  • Suppose that $f'''$ is continuous and $f'(c) = f''(c) = 0$, but $f'''(c)>0$. Does $f$ have a local maximum or minimum at $c$? Does $f$ have a point of inflection at $c$?
Solution.
  • $f$ has an inflection point at $x=c$. Note that continuity of $f'''$ and $f'''(c)>0$ gives that  $f'$ has a local minimum at $x=c$. That is, there is a small interval $(c-\delta,c+\delta)$ such that for all $x \in (c-\delta, c+\delta)$, $f'(x) \geqslant f'(c) = 0$. So $f$ is increasing on $(c-\delta, c+\delta)$, which makes $x = c$ an inflection point.
Problem 5. ([J], Section 4.4, Problem 51., Page 312)
  • Find the limit. Use l'Hospital's Rule where appropriate. $$\lim_{x \to 1}\left(\dfrac{x}{x-1}- \dfrac{1}{\ln x}\right)$$
Solution.
  • By direct computation: \begin{align*} \lim_{x \to 1}\left(\dfrac{x}{x-1}- \dfrac{1}{\ln x}\right) &= \lim_{x \to 1} \dfrac{x\ln x - (x-1)}{(x-1)\ln x} \\&\overset{L'H}{=} \lim_{x \to 1}\dfrac{\ln x + 1 - 1}{\ln x + \dfrac{x-1}{x}} \\ &= \lim_{x \to 1}\dfrac{x\ln x}{x\ln x + x-1} \\ &\overset{L'H}{=} \lim_{x \to 1}\dfrac{\ln x + 1}{\ln x +1 + 1} = \dfrac{1}{2} \end{align*} Please remember to check if using l'Hospital's Rule is appropriate.
Problem 6. ([J], Section 4.4, Problem 62., Page 312)
  • Find the limit. Use l'Hospital's Rule where appropriate. $$\lim_{x \to \infty}x^{\frac{\ln 2}{1+\ln x}}$$
Solution.
  • By direct computation: \begin{align*} \lim_{x \to \infty}x^{\frac{\ln 2}{1+\ln x}} &= \lim_{x \to \infty} \exp \left(\dfrac{\ln 2\ln x}{1+\ln x}\right) \\ &\overset{(*)}{=} \exp \left( \lim_{x \to \infty} \dfrac{\ln 2 \ln x}{1+\ln x}\right) \\ & = \exp(\ln 2) = 2 \end{align*} $(*)$ holds since $f(x) = \exp(x)$ is continuous.
Problem 7. ([J], Section 4.4, Problem 65., Page 312)
  • Find the limit. Use l'Hospital's Rule where appropriate. $$\lim_{x \to 0^+}(4x+1)^{\cot x}$$
Solution.
  • By direct computation: \begin{align*} \lim_{x \to  0^+}(4x+1)^{\cot x} &= \lim_{x \to 0^+} \exp \left(\cot x\ln (4x+1)\right) \\ &\overset{(*)}{=} \exp \left( \lim_{x \to 0^+} \dfrac{\ln (4x+1)}{\tan x}\right) \\ &\overset{L'H}{=} \exp \left( \lim_{x \to 0^+} \dfrac{\dfrac{4}{4x+1}}{\sec^2 x}\right) \\ &= \exp\left(\dfrac{4}{1^2}\right) = e^4 \end{align*} $(*)$ holds since $f(x) = \exp(x)$ is continuous. Please remember to check if using l'Hospital's Rule is appropriate.
Problem 8. ([J], Section 4.4, Problem 85., Page 313)
  • Evaluate $$\lim_{x \to \infty}\left[ x- x^2\ln\left(\dfrac{1+x}{x}\right)\right]$$
Solution.
  • By direct computation: \begin{align*} \lim_{x \to \infty}\left[ x- x^2\ln\left(\dfrac{1+x}{x}\right)\right] &\overset{y=\frac{1}{x}}{=} \lim_{y \to 0^+} \dfrac{y-\ln(1+y)}{y^2} \\ &\overset{L'H}{=}\lim_{y \to 0^+} \dfrac{1-\dfrac{1}{1+y}}{2y} \\ &=\lim_{y \to 0^+} \dfrac{y}{2y(1+y)} = \dfrac{1}{2} \end{align*} Please remember to check if using l'Hospital's Rule is appropriate.
Problem 9. ([J], Section 4.5, Problem 30., Page 322)
  • Sketch the curve. $$y= x^\frac{5}{3}-5x^{\frac{2}{3}}$$
Solution.
  • We have the following:
  1. The domain of $y = f(x)$ is $\mathbb{R}$. (In particular, it is continuous on $\mathbb{R}$). Moreover, we have: $$y = f(x) = x^{\frac{2}{3}}(x-5)$$ $f(x) \leqslant 0$ on $(-\infty, 5)$ and $f(x) > 0$ on $(5,\infty)$.
  2. Note that the derovateves: $$f'(x) = \dfrac{5}{3}x^{\frac{2}{3}} - \dfrac{10}{3}x^{-\frac{1}{3}} = \dfrac{5(x-2)}{3x^{\frac{1}{3}}}$$ $$f''(x) = \dfrac{10}{9}x^{-\frac{1}{3}}+ \dfrac{10}{9}x^{-\frac{4}{3}} = \dfrac{10(x+1)}{9x^{\frac{4}{3}}}$$, hence we have the following table:

    3.
    $f(x) = x^\frac{5}{3}-5x^{\frac{2}{3}}$
    $(-\infty, -1)$
    $-1$
    $(-1,0)$
    $0$
    $(0,2)$
    $2$
    $(2,5)$
    $5$
    $(5,\infty)$
    $f(x)$
    $-$
    $-$
    $-$
    $-$
    $-$
    $-$
    $-$
    $0$
    $+$
    $f'(x)$
    $+$
    $+$
    $+$
    DNE
    $-$
    $0$
    $+$
    $+$
    $+$
    $f''(x)$
    $-$
    0
    $+$
    DNE
    $+$
    $+$
    $+$
    $+$
    $+$
  3. Note that $\displaystyle \lim_{x \to \infty} f(x) = \infty$ and $\displaystyle \lim_{x \to -\infty}f(x) = -\infty$, hence there is no horizontal asymptote. And there is no vertical asymptote since $f(x)$ is continuous. Similarly, $\displaystyle \lim_{x \to \infty}\dfrac{f(x)}{x} = \infty$ and $\displaystyle \lim_{x \to -\infty}\dfrac{f(x)}{x} = \infty$, so there are no slant asymptote.
  4. The graph as below:
Graph of $f(x) = x^\frac{5}{3}-5x^{\frac{2}{3}}$

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