# 数学代写|微积分代写Calculus代写|MTH125

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## 数学代写|微积分代写Calculus代写|Mapping Properties

To differentiate roots, we need to know how derivatives of inverses behave. But continuous functions are invertible iff they are strictly monotone ( $\S 2.3)$, so, we begin by using the derivative to identify monotonicity.

Theorem 3.2.1. Let $f:(a, b) \rightarrow \mathbf{R}$ be differentiable. If $f^{\prime}(x) \neq 0$ for $a0$ on $(a, b)$ or $f^{\prime}(x)<0$ on $(a, b)$. Moreover, $f^{\prime}(x) \geq 0$ on $(a, b)$ iff $f$ is increasing, and $f^{\prime}(x) \leq 0$ on $(a, b)$ iff $f$ is decreasing.

By the mean value theorem, given $ac$, so,
$$f^{\prime}(c)=\lim _{x \rightarrow c+} \frac{f(x)-f(c)}{x-c} \geq 0,$$
for all $a0$ on $(a, b)$ or $f^{\prime}<0$ on $(a, b)$.

It is not, generally, true that strict monotonicity implies the nonvanishing of $f^{\prime}$. For example, $f(x)=x^3$ is strictly increasing on $\mathbf{R}$ but $f^{\prime}(0)=0$.
Since its derivative was computed in the previous section, the function
$$f(x)=\frac{x^2-1}{x^2+1}$$
is strictly increasing on $(0, \infty)$ and strictly decreasing on $(-\infty, 0)$. Thus, the critical point $x=0$ is a minimum of $f$ on $\mathbf{R}$.

A useful consequence of this theorem is the following: If $f$ and $g$ are differentiable on $(a, b)$, continuous on $[a, b], f(a)=g(a)$, and $f^{\prime}(x) \geq g^{\prime}(x)$ on $(a, b)$, then, $f(x) \geq g(x)$ on $[a, b]$. This follows by applying the theorem to $h=f-g$.

Another consequence is that derivatives, although not necessarily continuous, satisfy the intermediate value property (Exercise $\mathbf{3 . 2 . 8}$ ).

Now we can state the inverse function theorem for differentiable functions.

## 数学代写|微积分代写Calculus代写|Graphing Techniques

Let $f$ be differentiable on $(a, b)$. If $f^{\prime}=d f / d x$ is differentiable on $(a, b)$, we denote its derivative by $f^{\prime \prime}=\left(f^{\prime}\right)^{\prime} ; f^{\prime \prime}$ is the second derivative of $f$. If $f^{\prime \prime}$ is differentiable on $(a, b), f^{\prime \prime \prime}=\left(f^{\prime \prime}\right)^{\prime}$ is the third derivative of $f$. In general, we let $f^{(n)}$ denote the $n$th derivative or the derivative of order $n$, where, by convention, we take $f^{(0)}=f$. If $f$ has all derivatives, $f^{\prime}, f^{\prime \prime}, f^{\prime \prime \prime}, \ldots$, we say $f$ is smooth on $(a, b)$.

An alternate and useful notation for higher derivatives is obtained by thinking of $f^{\prime}=d f / d x$ as the result of applying $d / d x$ to $f$, i.e., $d f / d x=$ $(d / d x) f$. From this point of view, $d / d x$ signifies the operation of differentiation. Thus, applying $d / d x$ twice, we obtain
$$f^{\prime \prime}=\left(\frac{d}{d x}\right)\left(\frac{d}{d x}\right) f=\left(\frac{d^2}{d x^2}\right) f=\frac{d^2 f}{d x^2} .$$
Similarly, third derivatives may be denoted
$$f^{\prime \prime \prime}=\left(\frac{d}{d x}\right)\left(\frac{d^2 f}{d x^2}\right)=\frac{d^3 f}{d x^3} .$$
For example, $f(x)=x^2$ has $f^{\prime}(x)=2 x, f^{\prime \prime}(x)=2$, and $f^{(n)}(x)=0$ for $n \geq 3$. More generally, by induction, $f(x)=x^n, n \geq 0$, has derivatives
$$f^{(k)}(x)=\frac{d^k f}{d x^k}= \begin{cases}\frac{n !}{(n-k) !} x^{n-k}, & 0 \leq k \leq n \ 0, & k>n\end{cases}$$
so, $f(x)=x^n$ is smooth. By the arithmetic properties of derivatives, it follows that rational functions are smooth wherever they are defined.

Not all functions are smooth. The function $f(x)=|x|$ is not differentiable at zero. Using this, one can show that $f(x)=x^n|x|$ is $n$ times differentiable on $\mathbf{R}$, but $f^{(n)}$ is not differentiable at zero. More generally, for $f, g$ differentiable, we do not expect $\max (f, g)$ to be differentiable. However, since $f(x)=x^{1 / n}$ is smooth on $(0, \infty)$, algebraic functions are smooth on any open interval of definition. Also, the functions $x^a, a^x$, and $\log _a x$ are smooth on $(0, \infty)$.

We know the sign of $f^{\prime}$ determines the monotonicity of $f$, in the sense that $f^{\prime} \geq 0$ iff $f$ is increasing and $f^{\prime} \leq 0$ iff $f$ is decreasing. How is the sign of $f^{\prime \prime}$ reflected in the graph of $f$ ? Since $f^{\prime \prime}=\left(f^{\prime}\right)^{\prime}$, we see that $f^{\prime \prime} \geq 0$ iff $f^{\prime}$ is increasing and $f^{\prime \prime} \leq 0$ iff $f^{\prime}$ is decreasing.
More precisely, we say $f$ is convex on $(a, b)$ if, for all $a<x<y<b$,
$$f((1-t) x+t y) \leq(1-t) f(x)+t f(y), \quad 0 \leq t \leq 1 .$$

# 微积分代考

## 数学代写|微积分代写Calculus代写|Mapping Properties

.

3.2.1.

$$f^{\prime}(c)=\lim _{x \rightarrow c+} \frac{f(x)-f(c)}{x-c} \geq 0,$$

$$f(x)=\frac{x^2-1}{x^2+1}$$

## 数学代写|微积分代写Calculus代写| graphingtechniques

.绘图技术 .绘图技术 .绘图技术数学代写|微积分代写|

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