# Taylor's Inequality - What is x?

In Taylor's Inequality,

$|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}$

What exactly goes inside the $|x-a|$ portion of the equation?

The problem I'm given is asking me what would give a bound on the remainder if $T_3(x)$ was used to approximate $f(x) = ln(x)$ on the interval $1\le x\le2.5$. (Where a = 2).

I know that $f^{4}(x)$ is $-6x^{-4}$ and since it's a decreasing function,

$|f^4(x)| = |f^4(1)| = 6\le M$

But I have been having so much trouble understanding what goes in the $|x-a|$ part of Taylor's inequality.

If you're looking to create an upper bound, you should make $|x-a|$ as big as possible.
Since $a=2$ and $1 \leq x \leq 2.5$, the absolute value is maximized when $x=1$, so $|x-a| \leq |1-2|=|-1|=1$.