Answer
50
Work Step by Step
We are given $s(t)=t^3+2t+9$
For $t = 1$ the instantaneous velocity is
$\lim\limits_{h \to 0}\frac{s(4+h)-s(14)}{h}$
$s(4+h)=(4+h)^3+2(4+h)+9=64+48h+12h^2+h^3+8+2h+9=h^3+12h^2+50h+81$
$s(4)=81$
The instantaneous rate of change is then
$\lim\limits_{h \to 0}\frac{h^3+12h^2+50h+81-81}{h}=\lim\limits_{h \to 0}\frac{h^3+12h^2+50h}{h}=\lim\limits_{h \to 0}h^2+12h+50=50$
