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\flushleft\underline{Homework 6 --- MATH 4590 Spring 2018} \\
\begin{enumerate}[1.)]
\item Chapter 3, \# 6 from the text: If $f \colon (a,b) \rightarrow \mathbb{R}$ assumes a maximum or minimum at some $\theta \in (a,b)$, prove that $f'(\theta)=0$. \\
\textit{Proof}: We need to show that $f'(\theta)=0$, i.e. that 
$$f'(\theta) \stackrel{\text{def}}{=} \displaystyle\lim_{t \rightarrow \theta} \dfrac{f(t)-f(\theta)}{t-\theta}  = 0.$$
Assume without loss of generality that $\theta$ is the location of a local minimum (i.e. there is a $\delta$ such that for all $t \in (\theta-\delta, \theta+\delta)$, $f(\theta) \leq f(t)$. Hence for all such $t$, $f(t)-f(\theta) \geq 0$. Assume that $t \in (\theta-\delta, \theta)$, which means that $t<\theta$. Then $t-\theta < 0$ and so
$$\dfrac{f(t)-f(\theta)}{t-\theta} < 0.$$
On the other hand, if $t \in (\theta, \theta+\delta)$, then $t-\theta > 0$ and so
$$\dfrac{f(t)-f(\theta)}{t-\theta} > 0.$$
Combining these formulas yields the following for $t \in (\theta-\delta, \theta+\delta)$:
$$0 < \dfrac{f(t)-f(\theta)}{t-\theta} < 0.$$
Taking the limit as $t \rightarrow \theta$ and using the fact that the limit in the middle exists since $f$ is differentiable at $\theta$ forces us to conclude that
$$0 \leq \displaystyle\lim_{t \rightarrow \theta} \dfrac{f(t)-f(\theta)}{t-\theta} = f'(\theta) \leq 0,$$
in other words $f'(\theta) = 0$, as was to be shown. The proof when $\theta$ is a local maximum is very similar. $\blacksquare$

\item Prove using induction and the rules of differentiation (or directly, if you want) that the derivative of $x^n$ is $n x^{n-1}$.  \\
\textit{Solution}: The case $n=0$ is the formula $\dfrac{\mathrm{d}}{\mathrm{d}x} x = 1$ which we prove:
$$\displaystyle\lim_{t \rightarrow x} \dfrac{x-t}{x-t} = \displaystyle\lim_{t \rightarrow x} 1 = 1 = 1x^{1-1}.$$
Now assume the formula holds for $n=N-1$. We now prove it for $n=N$:
$$\dfrac{\mathrm{d}}{\mathrm{d}x} x^{N} = \dfrac{\mathrm{d}}{\mathrm{d}x} x \cdot x^{N-1} \stackrel{\text{prod rule}}{=} 1 \cdot x^{N-1} + (N-1) x \cdot x^{N-2} =(N)x^{N-1},$$
completing the proof.

\item The notion of an ``absolute derivative" in metric spaces was recently defined: let $(M,d_1)$ and $(N,d_2)$ be metric spaces and let $f \colon M \rightarrow N$ be a function. The absolute derivative of $f$, $f^{|'|} \colon M \rightarrow [0,\infty)$, is defined by 
$$f^{|'|}(x) = \displaystyle\lim_{t \rightarrow x} \dfrac{d_2(f(x),f(t))}{d_1(x,t)}.$$
Fix a point $y \in N$ and let $f$ be the constant function $f(x)=y$. Prove that $f^{|'|}(x)=0$. \\
\textit{Solution}: Since $f$ is the constant function $f(x)y$, compute
$$f^{|'|}(x)=\displaystyle\lim_{t \rightarrow x} \dfrac{d_2(f(x),f(t))}{d_1(x,t)} = \displaystyle\lim_{t \rightarrow x} \dfrac{d_2(y,y)}{d_2(x,t)} = \displaystyle\lim_{t \rightarrow x} 0 = 0.$$

\item Prove that if $f \colon \mathbb{R} \rightarrow \mathbb{R}$ (taking the usual metric in each case) is differentiable, then $f^{|'|}(x)=|f'(x)|$. $\Big($\textit{hint: you may find the fact that, for a function $g$, $\displaystyle\lim_{t \rightarrow x} |g(x)| = \left| \displaystyle\lim_{t \rightarrow x} g(x) \right|$ useful}$\Big)$ \\
\textit{Solution}: Compute
$$\begin{array}{ll}
f^{|'|}(x) &= \displaystyle\lim_{t \rightarrow x} \dfrac{d_2(f(x),f(t))}{d_1(x,t)} \\
&\hspace*{-18pt}\stackrel{\text{usual metric}}{=} \displaystyle\lim_{t \rightarrow x} \left| \dfrac{f(x)-f(t)}{x-t} \right| \\
&\hspace*{-5pt}\stackrel{hint}{=} \left| \displaystyle\lim_{t \rightarrow x} \dfrac{f(x)-f(t)}{x-t} \right| \\
&= \left| f'(x) \right|.
\end{array}$$
\end{enumerate}

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