Answer
$a)$ = $\frac{1}{4}$
$b)$ = $\frac{1}{12}$
Work Step by Step
$a)$
$f'(x)$ = $\frac{-2}{(x-1)^{3}}$
$f''(x)$ = $\frac{6}{(x-1)^{4}}$
$f''(2)$ = $6$
$f''(4)$ = $\frac{6}{81}$
$max|f''(x)|$ = $6$
$\frac{(b-a)^{3}}{12n^{2}}$$max|f''(x)|$ =$\frac{6(2)^{3}}{12(4)^{2}}$
= $\frac{1}{4}$
$b)$
$f^{3}(x)$ = $\frac{-24}{(x-1)^{5}}$
$f^{4}(x)$ = $\frac{120}{(x-1)^{6}}$
$f^{4}(2)$ = $120$
$f^{4}(4)$ = $\frac{120}{729}$
$max|f^{4}(x)|$ = $120$
$\frac{(b-a)^{5}}{180n^{4}}$$max|f^{4}(x)|$ = $\frac{120(2^{5})}{180(4)^{4}}$
= $\frac{1}{12}$
