To apply the Lagrange Remainder Formula, we first need to find an
such that
, that is,
.
Since
is an even function, we choose the positive root
without loss of generality.
By Taylor's Remainder theorem (as applied to Maclaurin polynomials),
![{\displaystyle T_{1}(x)+{\text{min}}\{R_{1}(x)\}\leq f(x)\leq T_{1}(x)+{\text{max}}\{R_{1}(x)\}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/bc8cd7081b693d1173455cf64ed070906e0bbae6)
where
and
denote the minimum and maximum values of
on the interval
, respectively.
Since we know that
from part (a), we simply consider:
![{\displaystyle {\text{min}}\{R_{1}(x)\}\leq \ln \left({\frac {8}{9}}\right)\leq {\text{max}}\{R_{1}(x)\}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9891e602de372791671c9c925bf9b1ec6295df70)
We now calculate the upper and lower bounds of the error. To do so, we substitute
into the Lagrange Remainder Formula:
![{\displaystyle R_{1}\left({\frac {1}{3}}\right)={\frac {f''(c)}{2!}}\left({\frac {1}{3}}\right)^{2}={\frac {f''(c)}{18}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/0714835b392ad9f952cba2542d9fa8bd6b280afb)
for some
in in the interval
. To find the error bounds, we maximize this function as a function of
. But observe that
![{\displaystyle f''(x)={\frac {-2(x^{2}+1)}{(1-x^{2})^{2}}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/df6477fcde47be2488e9d0803744a7fd084b4df3)
is monotonically decreasing on the interval
(the denominator is increasing while the numerator is decreasing (note the minus sign), so overall
is decreasing). Hence the minimum and maximum values of the error are attained at
and
, respectively.
Plugging this in we obtain the required lower and upper bound for the error: