Exercise 7.1
Solution:
(a) Note that\[\sqrt{x+3}-\sqrt{c+3}=\frac{(x+3)-(c+3)}{\sqrt{x+3}+\sqrt{c+3}}=\frac{x-c}{\sqrt{x+3}+\sqrt{c+3}}.\]We have\begin{align*}f'(c)&=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}\\ &=\lim_{x\to c}\frac{\sqrt{x+3}-\sqrt{c+3}}{x-c}\\ &=\lim_{x\to c}\frac{\sqrt{x+3}-\sqrt{c+3}}{x-c}\\&=\lim_{x\to c}\frac{1}{x-c}\frac{x-c}{\sqrt{x+3}+\sqrt{c+3}} \\ &=\lim_{x\to c}\frac{1}{\sqrt{x+3}+\sqrt{c+3}}\\&=\frac{1}{2\sqrt{c+3}}. \end{align*}(b) Note that\[\frac{1}{x}-\frac{1}{c}=\frac{c-x}{xc}.\]We have\begin{align*}g'(c)&=\lim_{x\to c}\frac{g(x)-g(c)}{x-c}\\ &=\lim_{x\to c}\frac{\frac{1}{x}-\frac{1}{c}}{x-c}\\ &= \lim_{x\to c}\frac{\frac{c-x}{xc}}{x-c}\\ &=\lim_{x\to c}\frac{-1}{xc}=-\frac{1}{c^2}.\end{align*}(c) Note that $x^2-c^2=(x-c)(x+c)$ and\[\frac{1}{x^2}-\frac{1}{c^2}=\frac{c^2-x^2}{x^2c^2}=\frac{(c-x)(c+x)}{x^2c^2}.\]We have\begin{align*}h'(c)&=\lim_{x\to c}\frac{h(x)-h(c)}{x-c}\\ &=\lim_{x\to c}\frac{4x^2+\frac{1}{x^2}-4c^2-\frac{1}{c^2}}{x-c}\\ &= \lim_{x\to c}\left(\frac{4(x^2-c^2)}{x-c}+\frac{\frac{1}{x^2}-\frac{1}{c^2}}{x-c}\right)\\ &= \lim_{x\to c}\left(\frac{4(x^2-c^2)}{x-c}+\frac{\frac{(c-x)(c+x)}{x^2c^2}}{x-c}\right)\\ &=\lim_{x\to c}\left(4(x+c)-\frac{x+c}{x^2c^2}\right)=8c-\frac{2}{c^3}.\end{align*}
Exercise 7.2
Solution: Suppose there exist differentiable functions $f$ and $g$ such that $x=f(x)g(x)$ and $f(0)=g(0)=0$. Differentiating both sides, we obtain\[(x)’=(f(x)g(x))’\Longrightarrow 1=f'(x)g(x)+f(x)g'(x).\]Let $x=0$, we have\[1=f'(0)g(0)+f(0)g'(0)=0,\]which is impossible. Hence we are done.
Exercise 7.3
Solution: We have\begin{align*}\left(\frac{f}{g}\right)'(c)&=\lim_{x\to c}\frac{\frac{f(x)}{g(x)}-\frac{f(c)}{g(c)}}{x-c}\\ &=\lim_{x\to c}\frac{f(x)g(c)-f(c)g(x)}{(x-c)g(x)g(c)}\\&=\lim_{x\to c}\frac{(f(x)g(c)-f(c)g(c))-(f(c)g(x)-f(c)g(c))}{(x-c)g(x)g(c)}\\&=\lim_{x\to c}\frac{(f(x)g(c)-f(c)g(c))}{(x-c)g(x)g(c)}-\lim_{x\to c}\frac{(f(c)g(x)-f(c)g(c))}{(x-c)g(x)g(c)}\\&= \lim_{x\to c}\frac{f(x)-f(c)}{x-c}\frac{g(c)}{g(x)g(c)}-\lim_{x\to c}\frac{g(x)-g(c)}{x-c}\frac{f(c)}{g(x)g(c)}\\&=\frac{f'(c)}{g(c)}-\frac{g'(c)f(c)}{(g(c))^2}=\frac{f'(c)g(c)-g'(c)f(c)}{(g(c))^2}.\end{align*}
Exercise 7.4
Solution: We show that the limit $\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}$ does not exist.
Clearly, we have $f(0)=0$. Let $(a_n)$ be a sequence of irrational numbers converging to $0$, then\[\lim_{n\to\infty}\frac{f(a_n)-f(0)}{a_n-0}=\lim_{n\to \infty}\frac{a_n-0}{a_n-0}=1.\]Let $(b_n)$ be a sequence of rational numbers converging to $0$, then\[\lim_{n\to\infty}\frac{f(b_n)-f(0)}{b_n-0}=\lim_{n\to \infty}\frac{\frac{1}{2}b_n-0}{b_n-0}=\frac{1}{2}.\]Since those two limits are different, we conclude that $\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}$ does not exist.
Exercise 7.5
Solution: We show it by induction on $n$. If $n=1$, this is clear. Then we suppose that $(x^n)’=nx^{n-1}$. For $n+1$, we use the product rule,$$(f \cdot g)'(x)=f'(x)g(x)+f(x)g'(x).$$Let $f(x)=x^n$ and $g(x)=x$, then we have\begin{align*}&\ (x^{n+1})’=(x^n\cdot x)’\\=&\ (x^n)’\cdot x+x^n \cdot (x)’\\ =&\ nx^{n-1} \cdot x+x^n\cdot 1\\ =&\ nx^n+x^n=(n+1)x^n. \end{align*}Hence it also holds for $n+1$. We are done.
Exercise 7.6
Solution: Since $f’$ is bounded, there exists $m>0$ such that $|f'(a)|< m$ for all $a\in I$. By Theorem 7.22, for any $[x_1,x_2]\subset I $, we have\begin{align*}\frac{f(x_1)-f(x_2)}{x_1-x_2}=f'(x_0)\end{align*}for some $x_0\in (x_1,x_2)$.
For $\e >0$, take $\delta=\frac{\e}{m}$. Then for any $x_1 ,x_2\in I$ such that $|x_1-x_2| < \delta$, we have\begin{align*}|f(x_1)-f(x_2)|&=|f'(x_0)||x_1-x_2|\\ & < m \frac{\e}{m}=\e. \end{align*}Hence $f$ is uniformly continuous.
Exercise 7.7
Solution:
(a) We require that $\lim_{x\to 0} x^a=0$. It is clear this happens if and only if $a >0$.
(b) If $f$ is differentiable, then we require that $\lim_{x\to 0}\frac{f_a(x)}{x}$ exists. This means $$\lim_{x\to 0^+}\frac{f_a(x)-f_a(0)}{x-0}=\lim_{x\to 0^-}\frac{f_a(x)-f_a(0)}{x-0}.$$ Namely, $$\lim_{x\to 0^+}\frac{x^a}{x}=0 \iff \lim_{x\to 0^+} x^{a-1}=0.$$Hence it follows from part (a) that this happens if and only if $a >1$.
Exercise 7.8
Solution: Since $f’$ is continuous on $[a,b]$, it follows from Corollary 6.30 that $f’$ is bounded. Hence there exist $C >0$ such that $|f'(\alpha)| < C$ for all $\alpha\in[a,b]$.
For all $ x ,y \in [a,b]$, by Theorem 7.22, we have\[f(x)-f(y)=f'(\alpha)(x-y)\]for some $\alpha$ in between of $x$ and $y$. Hence\[|f(x)-f(y)|\leqslant C|x-y|\]since $|f'(\alpha)| < C$ for all $\alpha\in[a,b]$. Therefore $f$ is Lipschitz.
Exercise 7.9
Solution: Since $f’$ is bounded on $A$, there exist $C >0$ such that $|f'(\alpha)| < C$ for all $\alpha\in A$.
For all $ x ,y \in A$, by Theorem 7.22, we have\[f(x)-f(y)=f'(\alpha)(x-y)\]for some $\alpha$ in between of $x$ and $y$. Hence\[|f(x)-f(y)|\leqslant C|x-y|\]since $|f'(\alpha)| < C$ for all $\alpha\in A$. Therefore $f$ is Lipschitz.
Exercise 7.10
Solution:
(a) Let $f(x)=x$, then $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. But $f'(c)\equiv 1$.
(b) Let $f(x)=\begin{cases}x, &\text{for }0\leqslant x\leqslant 1,\\ 2-x, &\text{for }1\leqslant x\leqslant 2\end{cases}$. Put $a=0$ and $b=2$. Then you can prove that $f'(c)=1 $ or $-1$, but never zero. In particular, it is not differentiable at $x=1$.
(c) Let $f(x)=\begin{cases}1, &\text{for }x=0,\\ 1/x, &\text{otherwise }\end{cases}$. Put $a=0$ and $b=1$.
Exercise 7.11
Solution: ??? No such values…
Exercise 7.12
Solution: Let $x,y$ be the length and width respectively. Suppose $p$ is the perimeter, which is a fixed parameter. Then \[2x+2y=p\Longrightarrow x=\frac{p-2x}{2}.\]The area is given by $$A(x)=xy=\frac{x(p-2x)}{2}.$$We compute the derivative of $A(x)$,\[A'(x)=\frac{p-4x}{2}.\]If $0 < x< p/4$, $A'(x)>0$ and hence $A(x)$ is increasing on $(0,p/4)$ by Corollary 7.25.
If $p/4 < x< p$, $A'(x)<0$ and hence $A(x)$ is decreasing on $(p/4,p/2)$ by Corollary 7.25.
Therefore, $A(x)$ takes the maximal value at $x=p/4$. In particular, at this time, $y=\frac{p-2x}{2}=p/4$ as well. Hence it is a square. This shows that the square has the maximal area.
Exercise 7.13
Solution: We would like to show that if $x >0$, then $x+\frac{1}{x}\geqslant 2$.
Let $f(x)=x+\frac{1}{x}$, then $f'(x)=1-\frac{1}{x^2}=\frac{x^2-1}{x^2}$.
If $ 0 < x< 1$, then $f'(x) < 0$ and hence by Corollary 7.25 that $f(x)$ is decreasing on $(0,1)$.
If $ 1 < x$, then $f'(x) > 0$ and hence by Corollary 7.25 that $f(x)$ is increasing on $(1,\infty)$.
Hence $f(x)$ takes the minimal value at $x=1$, where $f(1)=2$. Therefore $x+\frac{1}{x}\geqslant 2$ for all $x >0$.
Exercise 7.14
Solution: Define a new function\[h(x)=(f(b)-f(a))(g(x)-g(a))-(g(b)-g(a))(f(x)-f(a)).\]Then $h$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Moreover, we have $h(a)=0$ and $$h(b)=(f(b)-f(a))(g(b)-g(a))-(g(b)-g(a))(f(b)-f(a))=0.$$Using Rolle’s Theorem, there exists $x\in (a,b)$ such that $h'(x)=0$. Namely,\[(f(b)-f(a))g'(x)-(g(b)-g(a))f'(x)=0.\]
Exercise 7.15
Solution: Note that $\lim_{x\to a}\frac{f'(x)}{g'(x)}=L$, for any $\e>0$, there exist $\delta>0$ such that $$\left|\frac{f'(x)}{g'(x)}-L\right|<\e$$ for all $0<|x-a|<\delta$.
For $x$ such that $0<|x-a|<\delta$. Since $f(a)=g(a)=0$, by Exercise 7.14, we have\[(f(x)-f(a))g'(x_0)=(g(x)-g(a))f'(x_0)\]for some $x_0$ between $a$ and $x$. Hence $f(x)g'(x_0)=g(x)f'(x_0)$ for some $x_0$ such that $0< |x_0-a|<\delta$. Therefore,\[\left|\frac{f(x)}{g(x)}-L\right|=\left|\frac{f'(x_0)}{g'(x_0)}-L\right|<\e.\] Therefore $\lim_{x\to a}\frac{f(x)}{g(x)}=L$.
Exercise 7.16
Solution: See Proof of L’Hopital’s Rule.
