Answer
(a)
$C(q)$ = $84+0.16q-0.0006q^{2}+0.000003q^{3}$
$C'(q)$ = $0.16-0.0012q+0.000009q^{2}$
$C'(100)$ = $0.16-0.0012(100)+0.000009(100)^{2}$ = $0.13$
(b)
$C(101)-C(100)$ $\approx$ $97.13-97$ $\approx$ $0.13$
This is close to the marginal cost from part (a)
Work Step by Step
(a)
$C(q)$ = $84+0.16q-0.0006q^{2}+0.000003q^{3}$
$C'(q)$ = $0.16-0.0012q+0.000009q^{2}$
$C'(100)$ = $0.16-0.0012(100)+0.000009(100)^{2}$ = $0.13$
(b)
$C(101)-C(100)$ $\approx$ $97.13-97$ $\approx$ $0.13$
This is close to the marginal cost from part (a)