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1. \(f(x)=x\mathrm{sin}x\).
2. \(f(x)=\frac{1}{x^{2}+1}\).
11. Prove that if \(f(x)\) is a cubic polynomial, then \(f\) has a real root. That is, there is an \(x_{0}\) such that \(f(x_{0})=0\).
12. Show that if \(f:\mathbb{R}^{n}\to\mathbb{R}^{m}\) is continuous on all of \(\mathbb{R}^{n}\) and \(K\subset\mathbb{R}^{n}\) is bounded, then \(f(K)\) is bounded.
13. Suppose \(f\) is continuous on \([0,1]\), and set \(I_{k}=\left[\frac{k-1}{2^{n}},\frac{k}{2^{n}}\right]\) for \(k=1,2,\ldots,2^{n}\). Show that given \(\epsilon>0\) there is a natural number \(N\) such that \(n\geq N\) implies that
\[\sup_{x\in I_{k}}f(x)-\inf_{x\in I_{k}}f(x)<\epsilon,\quad k=1,2,\ldots,2^{n}.\]
14. A sequence of functions \(\{f_{n}\}\) defined on a set \(A\subset\mathbb{R}\) is called _equicontinuous_ if for every \(\epsilon>0\) there exists a \(\delta>0\) such that \(|f_{n}(x)-f_{n}(y)|<\epsilon\) for all \(n\in\mathbb{N}\) and \(|x-y|<\delta\) in \(A\). Let \(g_{n}(x)=x^{n}\) explain why \(g_{n}(x)\) is not equicontinuous on \([0,1]\). Is each \(g_{n}\) uniformly continuous on [0,1]?
### 1.7 Differentiability on \(\mathbb{R}\)
**Definition 34**.: Let \(f\) be a function defined on an open set \(I\) containing the point \(a\). We say \(f\) is _differentiable_ at \(a\) with derivative \(f^{\prime}(a)\) if the limit
\[f^{\prime}(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\]
exists. We say \(f\) is _differentiable_ if it is differentiable at each point \(a\) of its domain.
Note that an equivalent way to define the derivative of \(f\) at \(a\) is to claim
\[f^{\prime}(a):=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\]
exists. The assumption that \(f\) is defined on an open interval containing \(a\) is made so that the above quotient is defined for all \(h\neq 0\).
**Example 46**.: We now test this definition on some functions.
1. Let \(f(x)=c\), where \(c\) is constant; let us find \(f^{\prime}(3)\). \[f^{\prime}(3)=\lim_{x\to 3}\frac{f(x)-f(3)}{x-3}=\lim_{x\to 3}\frac{c-c}{x-3}=0.\]
2. Let \(f(x)=x^{2}\). To find \(f^{\prime}(3)\), set \[f^{\prime}(3)=\lim_{x\to 3}\frac{f(x)-f(3)}{x-3}=\lim_{x\to 3}\frac{x^{2}-9}{x-3}= \lim_{x\to 3}(x+3)=6.\]* Consider the function \[f(x)=\left\{\begin{array}{ll}1&:x\in\mathbb{Q}\\ 0&:x\notin\mathbb{Q}.\end{array}\right.\] Since the limit \(\lim_{x\to 0}\dfrac{f(x)}{x}\) fails to exist, \(f\) is not differentiable at \(x=0\).
* To check differentiability for the function \[f(x)=\left\{\begin{array}{ll}x^{3}&:x\geq 0\\ 0&:x<0\end{array}\right.\] at \(x=0\), we form \[\dfrac{f(x)}{x}=\left\{\begin{array}{ll}x^{2}&:x\geq 0\\ 0&:x<0\end{array}\right.\] and so \(f^{\prime}(0)=\lim_{x\to 0}\dfrac{f(x)}{x}=0\).
The derivative is defined in terms of a limit of a particular quotient, but there are several other views of the derivative. For example, the derivative of \(f\) at \(x=a\) is referred to as the _instantaneous rate of change_ of \(f\) when \(x=a\). In elementary calculus \(f^{\prime}(a)\) is sometimes described as the _slope_ of the tangent line to the graph of \(f\) at \(a\). Clearly, the quotient \(\dfrac{f(x)-f(a)}{x-a}\) is the slope of a certain _secant line_, which passes through the points \((x,f(x))\) and \((a,f(a))\). If the derivative exists, then the limiting value becomes _slope of the tangent line_ to the graph at \(x=a\). If the function \(f\) is differentiable at \(x=a\), then the _linear approximation_ to \(f\) at \(a\) is the function
\[L(x)=f(a)+f^{\prime}(a)(x-a).\]
For example, if \(f(x)=x^{2}\), then linear approximation to this function at \(a=2\) is \(L(x)=f(2)+f^{\prime}(2)(x-2)=4+4(x-2)=4x-4.\) The following second characterization of differentiability, in terms of linear approximation, is useful in understanding differentiability in higher dimensions.
**Proposition 11**.: _Let \(f:\mathbb{R}\rightarrow\mathbb{R}\). Then \(f\) is differentiable if and only if there is a function \(L(x):=mx\) such that_
\[\lim_{h\to 0}\dfrac{|f(a+h)-f(a)-L(h)|}{|h|}=0.\]
Applying the sequential criterion for limit, we can obtain the following _sequential characterization_ of derivatives. This characterization is often useful when trying to show that a given function is not differentiable at a particular point.
**Proposition 12**.: _Let \(S\) be an open set containing the point \(a\), and suppose \(f:S\to\mathbb{R}\). Then \(f\) is differentiable at \(a\) if and only if for every sequence \(\{x_{n}\}\in S\) that converges to \(a\) with \(x_{n}\neq a\) for all \(n\), the sequence_
\[\left\{\frac{f(x_{n})-f(a)}{x_{n}-a}\right\}\]
_converges. Furthermore, if \(f\) is differentiable at \(a\), then the above sequence of quotients converges to \(f^{\prime}(a)\)._
The proofs of the above two propositions are left to the reader.
### Rules of Differentiation
The familiar rules of differentiation can be easily deduced from the definition that
\[f^{\prime}(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.\]
* The derivative of the sum of two differentiable functions is the sum of their derivatives.
* The derivative of the product \(f(x)g(x)\) is equal to \(f^{\prime}(x)g(x)+f(x)g^{\prime}(x)\) (_Product Rule_).
* The quotient \(\frac{f(x)}{g(x)}\) has derivative \(\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g(x)^{2}}\) whenever \(g(x)\neq 0\) (_Quotient Rule_).
* The _Chain Rule_ for differentiation of the composition of \(h=f\circ g\) or \(h(x)=f(g(x))\) is more subtle. It turns out that \(h^{\prime}(x)=f^{\prime}(g(x))\,g^{\prime}(x)\).
Proofs of the above rules can be found in any calculus book.
### Properties of Differentiable Functions
The first important property is that differentiability implies continuity.
**Theorem 44**.: _If \(f\) is differentiable at \(a\), then \(f\) is continuous at \(a\)._
Proof
: To show \(f\) is continuous at \(a\), we need to show \(\lim_{x\to a}f(x)=f(a)\), or equivalently \(\lim_{x\to a}(f(x)-f(a))=0\). Now consider
\[\begin{split}\lim_{x\to a}(f(x)-f(a))&=\lim_{x\to a }\frac{f(x)-f(a)}{x-a}(x-a)\\ &=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\lim_{x\to a}(x-a)\\ &=f^{\prime}(a)\,0=0\end{split} \tag{1.3} \]\]
as desired. Note that going from the first equality to the second is possible because both limits exist.
**Remark 24**.: Continuity does not imply differentiability. For example, \(f(x)=|x|\) is continuous at \(x=0\) but not differentiable there. Since \(x\to 0\) implies \(|x|\to 0\), \(f\) is continuous at \(0\). On the other hand, notice when \(h>0\), \(|h|=h\), but when \(h<0\), \(|h|=-h\); thus when we form the difference quotient, we get two different limits as \(h\to 0^{+}\) or \(h\to 0^{-}\):
\[\lim_{h\to 0^{+}}\frac{f(h)-f(0)}{h}=1\quad\text{and}\quad\lim_{h\to 0^{-}} \frac{f(h)-f(0)}{h}=-1.\]
Since the limit exists if and only if its one-sided limits exist and are equal, the above limit does not exist when \(a=0\), and therefore \(f(x)=|x|\) is not differentiable at \(0\).
**Remark 25**.: The relationship between continuity and differentiability has led to many results. Of course as an absolute value function demonstrates a function can be continuous but not differentiable at some point. It is easy to construct a function which fails to be differentiable at a _finite set_. If we go one step further and ask the question is it possible to construct a function that is continuous for all of \(\mathbb{R}\) but fails to be differentiable at every rational points? Karl Weierstrass presented an example of a continuous function that is not differentiable at _any_ point. For more about Weierstrass function, see Theorem 98 in Chapter 2.
**Example 47**.: If a function \(f\) is differentiable at a certain subset \(I\subset\mathbb{R}\), its derivative \(f^{\prime}(x)\) may or may not be continuous there. For example, the function
\[f(x)=\left\{\begin{array}{ll}x^{2}\sin(1/x)&\text{if}\;\;x\neq 0\\ 0&\text{if}\;\;x=0\end{array}\right.\]
is differentiable on \(\mathbb{R}\), but \(f^{\prime}(x)\) is not continuous on any interval which contains the origin. For differentiability at \(x=0\),
\[f^{\prime}(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\frac{x^{2}\sin(1/x)} {x-0}=\lim_{x\to 0}x\sin(1/x),\]
which we have already showed by the squeeze principle is zero. Moreover,
\[f^{\prime}(x)=2x\sin(1/x)-\cos(1/x),\]
and \(\lim_{x\to 0}f^{\prime}(x)\) does not exist. A function with a continuous derivative is called a \(\mathcal{C}^{1}\)-function. Thus the function given in this example is not a \(\mathcal{C}^{1}\)-function.
Note that if a function has continuous derivatives to a certain order, say \(k\), we say that function is a \(\mathcal{C}^{k}\) function and infinitely differentiable functions are referred as \(\mathcal{C}^{\infty}\) or _smooth_ functions. For example, \(f(x)=e^{x}\) is a smooth function.
If a function \(f\) is defined on an open set \(S\) containing the point \(a\), we say that it has a _local maximum_ at \(a\) if \(f(x)\leq f(a)\) for all \(x\) in the neighborhood
\[V_{\delta}(a)=\{x\in\mathbb{R}:\;|x-a|<\delta\}\subset S.\]
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