In proof of Straddle Lemma, how $ \left \lvert f(u)-f(c) - (u-c)f^\prime(c) \right\rvert \leq \varepsilon \lvert u-c \rvert = \varepsilon ( c-u )$?

Question

Saqqib Mahmood was questioning Bartle, Introduction to Real Analysis (2011 4 ed), Section 6.1, Exercise 17, p 171. How did he deduce (2*) and (2**) below? I'm flustered by all these variables $c, \delta, u, v, x$.

Therefore we can conclude that $$ \left\lvert f(x)-f(c) - (x-c)f^\prime(c) \right\rvert \leq \varepsilon \lvert x - c \rvert \tag{2'} $$ for all $x \in I$ for which $c-\delta(\varepsilon) < x < c+\delta(\varepsilon)$.

From (2') we conclude that if $u, v \in I$ and $c-\delta(\varepsilon) < u \leq c \leq v < c + \delta(\varepsilon)$, then we have $$ \left\lvert f(u)-f(c) - (u-c)f^\prime(c) \right\rvert \leq \varepsilon \lvert u-c \rvert = \varepsilon ( c-u ) \tag{2*} $$ and also $$ \left\lvert f(v)-f(c) - (v-c)f^\prime(c) \right\rvert \leq \varepsilon \lvert v-c \rvert = \varepsilon (v-c), \tag{2**} $$

Answer 1

They're just rewriting (2') by considering the two cases $x \le c$ and $x \ge c$ separately (and renaming $x$ as $u$ and $v$ respectively).

For instance, if $c-\delta(\epsilon) < x < c + \delta(\epsilon)$ and $x \le c$, then (2') still applies, but you can replace $|x-c|$ with $(c-x)$. This gives (2*).