Need clarification on a Taylor polynomial question $$f(x) = 5 \ln(x)-x$$
second Taylor polynomial centered around $b=1$ is $-1 + 4(x-1) - (5/2)(x-1)^2$
let $a$ be a real number : $0 < a < 1$
let $J$ be closed interval $[1-a, 1+a]$
find upper bound for the error $|f(x)-T_2(x)|$ on interval $J$
answer in terms of a
so i got to the point where i have $|f(x)-T_2(x)| <= (10/6)(x-1)^3$ however the answer is 
$(5/3)(a/(1-a))^3$
How did the $a/(1-a)$ get there? If i sub in for the max $x$ which would be $1 + a$, then i 
would get $(5/3)(a)^3$. Why is it divided by $(1-a)$?

We are given:
$$f(x) = 5 \ln(x) - x$$
The second Taylor polynomial centered around $b=1$ is given by:
$$T_2(x) = -\frac{1}{2} 5 (x-1)^2+4 (x-1)-1$$
We are told to let $a$ be a real number such that $0 \lt a \lt 1$ and let $J$ be the closed interval $[1 −a, 1 +a]$.
We are then asked to use the Quadratic Approximation Error Bound to find an upper bound for the error $|f(x) − T_2(x)|$ on the interval $J$.
The error term is given by:
$$R_{n+1} = \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-b)^{n+1} \le \dfrac{M}{(n+1)!}(x-b)^{n+1}$$
We have $b=1, n= 2$, and have to find the max error for two items, thus:
$$\dfrac{d^3}{dx^3} (5 \ln(x)-x) = \dfrac{10}{x^3}$$
So the maximum of $f^{(3)}(x)$ is given by:
$$\displaystyle \max_{J} \left|f'''(x)\right| = \max_{ 1-a \le x \le 1+a} \left|\dfrac{10}{x^3}\right|$$
The max occurs at the left endpoint because $0 \lt a \lt 1$, so the maximum is given by:
$$\dfrac{10}{(1-a)^3}$$
Next, we have to repeat this and find the maximum of $(x-1)^3$. In this case, the maximum occurs at the rightmost endpoint, hence the maximum is:
$$((1+a) - 1)^3 = a^3$$
Putting this together yields:
$$\left|f(x)-T_2(x)\right| = R_{3} = \dfrac{f^{(3)}(c)}{3!}(x-1)^{3} \le \dfrac{M}{3!}(x-1)^{3} = \dfrac{10}{3!(1-a)^3}a^3 = \dfrac{5}{3} \left(\dfrac {a}{1-a}\right)^3$$
