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# 数学代写|复分析代写Complex analysis代考|MATH3711 The Mean value theorem

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## 数学代写|复分析代写Complex analysis代考|The Mean value theorem

In this section the domains and codomains of all functions are subsets of $\mathbb{R}$.
Theorem 6.3.1. Let $f:[a, b] \rightarrow \mathbb{R}$, and let $c \in[a, b]$ such that $f$ achieves an extreme value at $c$ (i.e., either for all $x \in[a, b], f(c) \leq f(x)$ or for all $x \in[a, b], f(c) \geq f(x))$. Then at least one of the following holds:
(1) $c=a$;
(2) $c=b$;
(3) $f$ is not continuous at $c$;
(4) $f$ is not differentiable at $c$;

(5) $f$ is differentiable at $c$ and $f^{\prime}(c)=0$.
Proof. It suffices to prove that if the first four conditions do not hold, then the fifth one has to hold. So we assume that $c \neq a, c \neq b$, and that $f$ is differentiable at $c$.

Suppose that $f^{\prime}(c)>0$. By the definition of derivative, $f^{\prime}(c)=\lim _{x \rightarrow c} \frac{f(x)-f(c)}{x-c}$. Thus for all $x$ very near $c$ but larger than $c, \frac{f(x)-f(c)}{x-c}>0$, so that $f(x)-f(c)>0$, so that $f$ does not achieve its maximum at $c$. Also, for all $x$ very near $c$ but smaller than $c$, $\frac{f(x)-f(c)}{x-c}>0$, so that $f(x)-f(c)<0$, so that $f$ does not achieve its minimum at $c$. This is a contradiction, so that $f^{\prime}(c)$ cannot be positive. Similarly, $f^{\prime}(c)$ cannot be negative. Thus $f^{\prime}(c)=0$.

Thus to find extreme values of a function, one only has to check if extreme values occur at the endpoints of the domain, at points where the function is not continuous or non-differentiable, or where it is differentiable and the derivative is 0 . One should be aware that just because any of the five conditions is satisfied, we need not have an extreme value of the function. Here are some examples:
(1) The function $f:[-1,1] \rightarrow \mathbb{R}$ given by $f(x)=x^3-x$ has neither the maximum nor the minimum at the endpoints.
(2) Let $f:[-1,1] \rightarrow \mathbb{R}$ be given by $f(x)=\left{\begin{array}{ll}x, & \text { if } x>0 ; \ 1 / 2, & \text { if } x \leq 0 .\end{array}\right.$ Then $f$ is not continuous at 0 but $f$ does not have a minimum or maximum at 0 .
(3) Let $f:[-1,1] \rightarrow \mathbb{R}$ be given by $f(x)=\left{\begin{array}{ll}x, & \text { if } x>0 ; \ 2 x, & \text { if } x \leq 0 .\end{array}\right.$ Then $f$ is continuous and not differentiable at 0 , yet $f$ does not have a minimum or maximum at 0 .
(4) Let $f:[-1,1] \rightarrow \mathbb{R}$ be given by $f(x)=x^3$. Then $f$ is differentiable, $f^{\prime}(0)=0$, but $f$ does not have a minimum or maximum at 0 .

## 数学代写|复分析代写Complex analysis代考|L’Hôpital’s rule

For L’Hôpital’s rule we pass from $\frac{f}{g}$ to $\frac{f^{\prime}}{g^{\prime}}$; it is worth reviewing that the quotient rule for derivatives is different:
$$\left(\frac{f}{g}\right)^{\prime}=\frac{f^{\prime} g-f g^{\prime}}{g^2} .$$
Theorem 6.4.1. (L’Hôpital’s rule, easiest version) Let $A \subset \mathbb{C}$, let $a \in A$ be a limit point of $A$, and $f, g: A \rightarrow \mathbb{C}$ such that
(1) $f(a)=g(a)=0$.
(2) $f$ and $g$ are differentiable at $a$.
(3) $g^{\prime}(a) \neq 0$.
Then $\lim {x \rightarrow a} \frac{f(x)}{g(x)}=\frac{f^{\prime}(a)}{g^{\prime}(a)}$. Proof. $\quad \lim {x \rightarrow a} \frac{f(x)}{g(x)}=\lim {x \rightarrow a} \frac{f(x)-f(a)}{g(x)-g(a)}$ (since $f(a)=g(a)=0$ ) \begin{aligned} &=\lim {x \rightarrow a} \frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}} \ &=\frac{f^{\prime}(a)}{g^{\prime}(a)} \text { (by the quotient rule for limits since } g^{\prime}(a) \neq 0 \text { ). } \end{aligned}
The rest of the versions of L’Hôpital’s rule in this section only work for real-valued functions on domains that are subsets of $\mathbb{R}$ because the proofs invoke the Cauchy’s mean value theorem (Theorem 6.3.8).

## 数学代写|复分析代写Complex analysis代考|The Mean value theorem

(1) $c=a$;
(2) $c=b$;
(3) f不连续c;
(4) $f$ 不可微分 $c$;
（5） $f$ 可微分于 $c$ 和 $f^{\prime}(c)=0$.

\left(\frac{f}{g}\right)^{\prime}=\frac{f^{\prime} g-f g^{\prime}}{g^2}


(1) $f(a)=g(a)=0$
(2) $f$ 和 $g$ 可微分于 $a$.
(3) $g^{\prime}(a) \neq 0$.

$=\lim x \rightarrow a \frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}} \quad=\frac{f^{\prime}(a)}{g^{\prime}(a)}$ (by the quotient rule for limits since $\left.g^{\prime}(a) \neq 0\right)$.

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