这是一份northeastern东北大学(美国) MATH 3150作业代写的成功案例
Proof. Write $h=g \circ f .$ By 8.1.5, there exists a function $A: S \rightarrow \mathbb{R}$, continuous at $c$, such that
$$
f(x)-f(c)=A(x)(x-c) \text { for all } x \in \mathbf{S}
$$
Similarly, there is a function $B: T \rightarrow \mathbb{R}$, continuous at $f(c)$, such that
(2) $g(y)-g(f(c))=B(y)(y-f(c))$ for all $y \in \mathrm{T}$.
If $s \in \mathrm{S}$ then $f(x) \in \mathrm{T}$; putting $y=f(x)$ in (2), we have
$$
\begin{aligned}
g(f(x))-g(f(c)) &=B(f(x)) \cdot(f(x)-f(c)) \
&=B(f(x)) \cdot A(x)(x-c)
\end{aligned}
$$
MATH 3150COURSE NOTES ：
$$
\sigma={a<b}, \quad \tau \doteq{a<c<b}
$$
Writing $M=\sup f$ as before, and
$$
\begin{aligned}
M^{\prime} &=\sup {f(x): a \leq x \leq c}, \
M^{\prime \prime} &=\sup {f(x): c \leq x \leq b},
\end{aligned}
$$
we have
$$
S(\sigma)=M(b-a) \quad \text { and } \quad S(\tau)=M^{\prime}(c-a)+M^{\prime \prime}(b-c)
$$