Answer
For t = -1 the instantaneous rate of change is $\lim\limits_{h \to 0} \frac{h(-1 + h) - h(-1)}{h}$
$g(-1+h)=1-(-1+h)^{2}=1-(1-2h+h^{2})=2h-h^{2}$
$g(-10) = 0$
The instantaneous rate of change is then: $\lim\limits_{h \to 0} \frac{g(2 + h) - g(2)}{h} = \lim\limits_{h \to 0} \frac{2h-h^{2}-0}{h}=\lim\limits_{h \to 0} \frac{h(2-h)}{h} = 2$
Work Step by Step
For t = -1 the instantaneous rate of change is $\lim\limits_{h \to 0} \frac{h(-1 + h) - h(-1)}{h}$
$g(-1+h)=1-(-1+h)^{2}=1-(1-2h+h^{2})=2h-h^{2}$
$g(-10) = 0$
The instantaneous rate of change is then: $\lim\limits_{h \to 0} \frac{g(2 + h) - g(2)}{h} = \lim\limits_{h \to 0} \frac{2h-h^{2}-0}{h}=\lim\limits_{h \to 0} \frac{h(2-h)}{h} = 2$
