Proving $|h(t)−h(0)−\frac{t^2h''(0)}{2}|\leq at^2$, for some $b>0$ and $t\leq b$ using extended mean value theorem Trying to prove: By the extended mean value theorem for functions of the real variable $t$, we have for $a>0$: $|h(t)−h(0)−\frac{t^2h''(0)}{2}|\leq at^2$, for some $b>0$ and $t\leq b$.  We are also given that $h'(0)=0$, $h'(t)$ exists for some $b$ such that $0\leq t\leq b$ and $h''(0)<0$.
The statement is made on p452 of Applied Complex Variables by J.W. Dettman as a lead up to introducing the method of steepest descents, a means of finding the asymptotic expansion of a particular form of complex integral with a parameter that tends to infinity. I think the statement can be validated by simply using Taylor's theorem for the range of values of t necessary, but I wanted to understand how Dettman justifies it with the extended MVT.
The textbook is the set book for a course I'm taking. The method (steepest descents) is described more clearly elsewhere in the course, but I'm just trying to understand all the steps in Dettman too and this statement is eluding me at the present.

Let $$f(t)=h(t)-h(0)-\frac{t^2h’’(0)}2,\quad g(t)=t^2,$$ defined on $[0,b_1]$ for some $b_1>0.$
Then by the extended mean value theorem, there exists $0<\xi<t$ such that $$\frac{f(t)-f(0)}{g(t)-g(0)}=\frac {f’(\xi)}{g’(\xi)},$$ where the latter equals $$\frac 12\cdot\left[\frac{h’(\xi)-h’(0)}{\xi-0}-h’’(0)\right].\quad (1)$$ Since $h’’(0)$ exists, for every $a>0$, there exists $b_2>0$ such that $0<|\xi|<b_2$ implies the absolute value of the expression in (1) is less than $a$.
Now choosing $b=\min(b_1,b_2)$, one has for $0<t\leq b$ $$\left|\frac{f(t)}{g(t)}\right|<a$$
$$\Rightarrow |f(t)|<at^2$$ or $$|f(t)|\leq at^2,0\leq t\leq b,$$ where in the above one uses $f(0)=g(0)=0$ and $h’(0)=0.$
