*17 (a) Show that if $|g'(x)|\leq M|x-a|^n$ for $|x-a|<\delta$, then $|g(x)-g(a)|\leq \frac{M|x-a|^{n+1}}{n+1}$ for $|x-a|<\delta$
Here is the solution in the solution manual
By hypothesis,
$$-M(x-a)^n\leq g'(x)\leq M(x-a)^n, \text{ for } x\geq a$$
It follows from the Mean Value Theorem that
$$\frac{-M(x-a)^{n+1}}{n+1}\leq g(x)-g(a)\leq \frac{M(x-a)^{n+1}}{n+1}$$
i.e., that $|g(x)-g(a)|\leq \frac{M(x-a)^{n+1}}{n+1}$. The case $x\leq a$ is treated similarly.
Sometimes the solutions are quite terse like this and skip over many steps. Here is my attempt at filling in the intermediate steps in the solution manual solution.
We know that $|g'(x)|\leq M|x-a|^n$ and therefore
$$-M|x-a|^n\leq g'(x)\leq M|x-a|^n$$
If $x\geq a$, then $|x-a|=x-a$ and
$$-M(x-a)^n\leq g'(x)\leq M(x-a)^n\tag{1}$$
The Mean Value Theorem says that there is some point $c\in (a,x)$ at which
$$g'(c)=\frac{g(x)-g(a)}{x-a}\tag{2}$$
Hence
$$-M(c-a)^n\leq g'(c)\leq M(c-a)^n$$
But since $x-a>c-a$, and using $(2)$ we have
$$-M(x-a)^n<-M(c-a)^n\leq \frac{g(x)-g(a)}{x-a}\leq M(c-a)^n<M(x-a)^n$$
$$-M(x-a)^{n+1}<g(x)-g(a)<M(x-a)^{n+1}$$
If this is correct so far, where does the factor $\frac{1}{n+1}$ come from?
EDIT: As per a comment below, it appears that $(1)$ is simply being integrated. Does the solution manual solution make sense at all? Ie, is there any path that uses MVT?
It looks like the proof wants to integrate $g'(x)$, but without knowing the integrability of $g'$, it is not possible to directly do that step. Instead for $a+\delta>x \geq a$, consider $F_1(x) = g(x)-g(a)-M(x-a)^{n+1}/n+1$. Then $F_1$ is differentiable on any $[a,x]$ for $x<a+\delta$, and $F_1'(x)\leq 0$. Then by the mean value theorem, $F_1(x)-F_1(a) = F_1(x) = F_1'(c)(x-a) \leq 0$. Using this kind of construction for the other cases you would have the proof.