Unlock the Secret to Local Maxima and Minima: The Power of Theorem 45 Revealed!

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A _local minimum_ is defined similarly.

**Theorem 45**.: _Let \(f\) be a function defined on some open set \(S\), and suppose it is differentiable at \(a\in S\). If \(f\) has a local maximum or a local minimum at \(a\in S\), then \(f^{\prime}(a)=0\)._

Proof

: Suppose \(f\) has a local minimum at \(a\). By hypothesis, \(f(a)\leq f(x)\) for all \(x\) in some neighborhood \(V_{\delta}(a)\). This implies that the difference quotient

\[g(x)=\frac{f(x)-f(a)}{x-a}\quad\text{for}\quad x\neq a\]

defined in \(V_{\delta}(a)\) except at \(a\) satisfies the inequalities:

\[g(x)\geq 0\quad\text{for}\quad x>a\]

\[g(x)\leq 0\quad\text{for}\quad x

Since \(f\) is differentiable at \(a\), it follows that

\[f^{\prime}(a)=\lim_{x\to a^{+}}g(x)\geq 0\quad\text{and}\quad f^{\prime}(a)= \lim_{x\to a^{-}}g(x)\leq 0.\]

Hence \(f^{\prime}(x)=0\) as claimed.

### The Mean Value Theorem

The Mean Value Theorem is an important and fundamental property of differentiable functions. It is useful in the same way the Intermediate Value Theorem or the Extreme Value Theorem are to continuous functions. The basic ingredient in the Mean Value Theorem is the following:

**Theorem 46** (Rolle’s Theorem).: _If a function is continuous in an interval \([a,b]\) and differentiable in the open interval \((a,b)\) and if \(f(a)=f(b)\), then \(f^{\prime}(c)=0\) for some \(c\in(a,b)\) (Figure 1.39)._

Proof

: Suppose \(f(x)=k\) for some constant \(k\) on \([a,b]\); then \(f^{\prime}(x)=0\) in \((a,b)\), and thus there is nothing to prove. Assume \(f\) is not constant; then \(f\) attains a maximum or a minimum at some point \(c\in(a,b)\), and it follows from the previous theorem that \(f^{\prime}(c)=0\).

Figure 1.39: Rolle’s theorem

**Theorem 47** (Mean Value Theorem).: _If a function \(f\) is continuous in an interval \([a,b]\) and differentiable in the open interval \((a,b)\), then (Figure 1.40)_

\[f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\quad\text{for some}\ \ c\in(a,b).\]

Proof

: Define the linear function

\[M(x)=\frac{f(b)-f(a)}{b-a}(x-a)+f(a).\]

Notice that \(M(a)=f(a)\) and \(M(b)=f(b)\). Then apply Rolle’s theorem to the function \(g(x)=f(x)-M(x)\).

The Mean Value Theorem can be applied to show that a function with zero derivative throughout an interval must be a constant function. A more general statement is the following:

**Corollary 10**.: _If the derivative of two functions coincides in some interval, the two functions must differ by a constant._

**Corollary 11**.: _A function with positive derivative in an interval is strictly increasing. Similarly, a function with negative derivative is strictly decreasing._

**Remark 26**.: In the following we point out a few things about the Mean Value Theorem. For more about differentiation on \(\mathbb{R}\), we refer the reader to [52].

* _The Generalized Mean Value Theorem_, sometimes referred to as _Cauchy’s Mean Value Theorem_, states that if \(f\) and \(g\) are both continuous on \([a,b]\) and differentiable on \((a,b)\), then there is a point \(c\in(a,b)\) such that \[f^{\prime}(c)[g(b)-g(a)]=g^{\prime}(c)[f(b)-f(a)].\] This theorem is useful when comparing derivatives of two functions, and it can be proved by setting \(h(x)=f(x)[g(b)-g(a)]=g(x)[f(b)-f(a)]\) and applying Rolle’s theorem to \(h(x)\).

Figure 1.40: The Mean Value Theorem* One can extract information about the function \(f\) from \(f^{\prime}\) using the Mean Value Theorem. For example, suppose you are asked to prove the inequality \(\ln(1+x)\leq x\) for any \(x\geq 0\). First assume \(x>0\). The Mean Value Theorem applied to \(\ln(1+x)\) on the interval \([0,x]\) implies the existence of \(c\in(0,x)\) such that \[\frac{\ln(1+x)-\ln(1+0)}{x-0}=\frac{1}{1+c}\;,\] or \(\ln(1+x)=\frac{x}{1+c}\). Since \(0<\frac{1}{1+c}<1\), we get \(\ln(1+x)

### Inverse Function Theorem

There are various ways of forming new functions from old functions by using addition, multiplication, division, composition, etc. Now we will see another way to construct functions which might double the list of functions we might have. Recall that a function is _one-to-one_ or _injective_ if \(f(a)\neq f(b)\) whenever \(a\neq b\). The identity function \(I\) is one-to-one, but the function \(f(x)=x^{2}\) is not so, since \(f(-1)=f(1)\). However if we restrict the domain of \(f(x)=x^{2}\) to the set of \(x\geq 0\), then \(f\) becomes one-to-one. Recall also that a function is a collection of pairs of numbers subject to some constraints.

**Definition 35**.: For any function \(f\), the inverse of \(f\), denoted by \(f^{-1}\), is the set of all pairs \((a,b)\) for which the pair \((b,a)\) is in \(f\).

**Proposition 13**.: \(f^{-1}\) _is a function if and only if \(f\) is one-to-one._

Proof

: Suppose \(f^{-1}\) is a function. If \(f(b)=f(c)\), then \(f\) contains the pairs \((b,f(b))\) and \((c,f(c))\) and \((c,f(c))=(c,f(b))\), so \((f(b),b)\) and \((f(b),c)\) are in \(f^{-1}\). Since \(f^{-1}\) is a function, this implies that \(b=c\). Thus \(f\) is one-to-one. Conversely suppose \(f\) is one-to-one. Let \((a,b)\) and \((a,c)\) be two pairs in \(f^{-1}\). Then \((b,a)\) and \((c,a)\) are in \(f\), so \(a=f(b)\) and \(a=f(c)\), but \(f\) is one-to-one, so \(f(b)=f(c)\) implies \(b=c\). Thus \(f^{-1}\) is a function.

Since the points \((a,b)\) and \((b,a)\) are reflections of each other through the graph of \(I(x)=x\) (this is called the _diagonal_), to obtain the graph of \(f^{-1}\), one can reflect the graph of \(f\) through the line \(I(x)=x\) as seen in Figure 1.41.

Figure 1.41: The inverse of a function \(f\)

Note that \(f^{-1}(b)\) is the unique number \(a\) such that \(f(a)=b\); this gives us another way to define the inverse. Observe that

\[f(f^{-1}(x))=x\quad\text{for all}\ \ x\ \text{in the domain of}\ f^{-1}\,\ \text{and}\]

\[f^{-1}(f(x))=x\quad\text{for all}\ \ x\ \text{in the domain of}\ f.\]

These two equations can be expressed as

\[f\circ f^{-1}=I\quad\text{and}\quad f^{-1}\circ f=I.\]

Since many standard functions can be defined as the inverse of other functions, it becomes clear to us that we must identify functions which are one-to-one, and it will be useful to know how the properties of \(f\) and \(f^{-1}\) are related. It is not difficult to see that if \(f\) is increasing, then \(f^{-1}\) is also increasing, and if \(f\) is decreasing, \(f^{-1}\) is decreasing (try to prove it). It is not true that every one-to-one function is either increasing or decreasing as seen in the following example (Figure 1.42):

\[f(x)=\left\{\begin{array}{ll}x&\text{if}\ \,x\neq 1,2\\ 1&\text{if}\ \,x=2.\\ 2&\text{if}\ \,x=1.\\ \end{array}\right.\]

However, every continuous one-to-one function defined on an interval is either increasing or decreasing.

**Theorem 48**.: _If \(f\) is continuous and one-to-one on a closed and bounded interval \([a,b]\), then \(f\) is either increasing or decreasing on that interval._

For the proof we refer the reader to [52], p. 125. Notice that when \(f\) is a continuous and increasing function on a closed interval \([a,b]\), then by the Intermediate Value Theorem \(f\) takes every value between \(f(a)\) and \(f(b)\). Therefore, the domain of \(f^{-1}\) is the closed interval \([f(a),f(b)]\). Similarly if \(f\) is continuous and decreasing on \([a,b]\), then the domain of \(f^{-1}\) is \([f(b),f(a)]\). When using this theorem, you can assume \(f\) is increasing and use the standard trick of considering \(-f\) to move to decreasing functions.

Figure 1.42: A function that is neither increasing nor decreasing

**Proposition 14**.: _If \(f\) is continuous and one-to-one on an interval, then \(f^{-1}\) is also continuous._

Proof

: We know by the previous theorem that \(f\) is either increasing or decreasing. Without loss of generality assume that \(f\) is increasing. To show \(f^{-1}\) is continuous, we must show for every \(b\) in the domain of \(f^{-1}\) we have \(\lim\limits_{x\to b}f^{-1}(x)=f^{-1}(b)\). Note that such a number \(b\) has the form \(f(a)\); thus using the \(\epsilon-\delta\) definition of the limit, for any \(\epsilon>0\), we want to find \(\delta\) such that

\[\text{if}\quad|x-f(a)|<\delta\quad\text{then}\quad|f^{-1}(x)-a|<\epsilon.\]

Now since \(a\in(a-\epsilon,a+\epsilon)\), it follows that \(f(a)\in(f(a-\epsilon),f(a+\epsilon))\). Thus the choice of

\[\delta=\min\{f(a+\epsilon)-f(a),f(a)-f(a-\epsilon)\}\]

works (see Figure 1.43).