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# 数学代写|实分析代写Real analysis代考|MTH3140 Intermediate Value Theorem for Derivatives

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## 数学代写|实分析代写Real analysis代考|Intermediate Value Theorem for Derivatives

Our second important result of this section, due to Jean Gaston Darboux (1842-1917), is the intermediate value theorem for derivatives. The remarkable aspect of this theorem is that the hypothesis does not require continuity of the derivative. If the derivative were continuous, then the result would follow from Theorem 4.2.11 applied to $f^{\prime}$.

THEOREM 5.2.13 (Intermediate Value Theorem for Derivatives) Suppose $I \subset \mathbb{R}$ is an interval and $f: I \rightarrow \mathbb{R}$ is differentiable on $I$. Then given $a, b$ in $I$ with $a<b$ and a real number $\lambda$ between $f^{\prime}(a)$ and $f^{\prime}(b)$, there exists $c \in(a, b)$ such that $f^{\prime}(c)=\lambda$.

Proof. Define $g$ by $g(x)=f(x)-\lambda x$. Then $g$ is differentiable on $I$ with $g^{\prime}(x)=f^{\prime}(x)-\lambda$.

Suppose $f^{\prime}(a)<\lambda0$. As in the remark following Theorem 5.2.9, since $g^{\prime}(a)<0$ there exists an $x_{1}>a$ such that $g\left(x_{1}\right)0$, there exists an $x_{2}<b$ such that $g\left(x_{2}\right)<g(b)$. As a consequence, $g$ has an absolute minimum at some point $c \in(a, b)$. But then
$$g^{\prime}(c)=f^{\prime}(c)-\lambda=0,$$
i.e., $f^{\prime}(c)=\lambda$.
The previous theorem is often used in calculus to determine where a function is increasing or decreasing. Suppose it has been determined that the derivative $f^{\prime}$ is zero at $c_{1}$ and $c_{2}$ with $c_{1}<c_{2}$, and that $f^{\prime}(x) \neq 0$ for all $x \in\left(c_{1}, c_{2}\right)$. Then by the previous theorem, it suffices to check the sign of the derivative at a single point in the interval $\left(c_{1}, c_{2}\right)$ to determine whether $f^{\prime}$ is positive or negative on the whole interval $\left(c_{1}, c_{2}\right)$. Theorem $5.2 .9$ then allows us to determine whether $f$ is increasing or decreasing on $\left(c_{1}, c_{2}\right)$.

## 数学代写|实分析代写Real analysis代考|Inverse Function Theorem

We conclude this section with the following version of the inverse function theorem.

THEOREM 5.2.14 (Inverse Function Theorem) Suppose $I \subset \mathbb{R}$ is an interval and $f: I \rightarrow \mathbb{R}$ is differentiable on $I$ with $f^{\prime}(x) \neq 0$ for all $x \in I$. Then $f$ is one-to-one on $I$, the inverse function $f^{-1}$ is continuous and differentiable on $J=f(I)$ with
$$\left(f^{-1}\right)^{\prime}(f(x))=\frac{1}{f^{\prime}(x)}$$
for all $x \in I$.
Proof. Since $f^{\prime}(x) \neq 0$ for all $x \in I$, by Theorem 5.2.13, $f^{\prime}$ is either positive on $I$, or negative on $I$. Assume that $f^{\prime}(x)>0$ for all $x \in I$. Then by Theorem $5.2 .9, f$ is strictly increasing on $I$ and by Theorem 4.4.12 $f^{-1}$ is continuous on $J=f(I)$.

It remains to be shown that $f^{-1}$ is differentiable on $J$. Let $y_{o} \in J$, and let $\left{y_{n}\right}$ be any sequence in $J$ with $y_{n} \rightarrow y_{o}$, and $y_{n} \neq y_{o}$ for all $n$. For each $n$, there exists $x_{n} \in I$ such that $f\left(x_{n}\right)=y_{n}$. Since $f^{-1}$ is continuous, $x_{n} \rightarrow x_{o}=f^{-1}\left(y_{o}\right)$. Hence
\begin{aligned} \lim {n \rightarrow \infty} \frac{f^{-1}\left(y{n}\right)-f^{-1}\left(y_{o}\right)}{y_{n}-y_{o}} &=\lim {n \rightarrow \infty} \frac{x{n}-x_{o}}{f\left(x_{n}\right)-f\left(x_{o}\right)} \ &=\frac{1}{f^{\prime}\left(x_{o}\right)} \end{aligned}

Since this holds for any sequence $\left{y_{n}\right}$ with $y_{n} \rightarrow y_{o}, y_{n} \neq y_{o}$, by Theorem 4.1.3 and the definition of the derivative
$$\left(f^{-1}\right)^{\prime}\left(y_{o}\right)=\frac{1}{f^{\prime}\left(x_{o}\right)} .$$
Remark. The hypothesis that $f^{\prime}(x) \neq 0$ for all $x \in I$ is crucial. For example, the function $f(x)=x^{3}$ is strictly increasing on $[-1,1]$ with $f^{\prime}(0)=0$. The inverse function $f^{-1}(y)=y^{1 / 3}$ however is not differentiable at $y=0$.

## 数学代写|实分析代写Real analysis代考|Intermediate Value Theorem for Derivatives

$$g^{\prime}(c)=f^{\prime}(c)-\lambda=0$$
$\mathrm{IE}{\text {。 }} f^{\prime}(c)=\lambda$ 前面的定理经常在微积分中用于确定函数在哪里增加或减少。假设已经确定导数 $f^{\prime}$ 为雪时 $c{1}$ 和 $c_{2}$ 和 $c_{1}<c_{2}$ ，然后 $f^{\prime}(x) \neq 0$ 对所 有人 $x \in\left(c_{1}, c_{2}\right)$. 然后根据前面的定理，在区间中的单个点检育导数的符号就足够了 $\left(c_{1}, c_{2}\right)$ 来确定是否 $f^{\prime}$ 在整个区间上为正或负
$\left(c_{1}, c_{2}\right)$. 定理 $5.2 .9$ 然后允许我们确定是否 $f$ 㘿加或减少 $\left(c_{1}, c_{2}\right)$.

## 数学代写|实分析代写Real analysis代考|Inverse Function Theorem

$$\left(f^{-1}\right)^{\prime}(f(x))=\frac{1}{f^{\prime}(x)}$$

$$\lim n \rightarrow \infty \frac{f^{-1}(y n)-f^{-1}\left(y_{o}\right)}{y_{n}-y_{o}}=\lim n \rightarrow \infty \frac{x n-x_{o}}{f\left(x_{n}\right)-f\left(x_{o}\right)} \quad=\frac{1}{f^{\prime}\left(x_{o}\right)}$$

$$\left(f^{-1}\right)^{\prime}\left(y_{o}\right)=\frac{1}{f^{\prime}\left(x_{o}\right)}$$

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