ANSWER;
The average rate of change of the function over the given interval is 8(t+1)
EXPLANATION;
The average rate of change of a given function over the interval [a,b] can be represented as;
[tex]\frac{f(b)-f(a)}{b-a}[/tex]With respect to the question given, a = 1 and b = t
[tex]\begin{gathered} f(b)=f(t)=8(t)^2-3=8t^2-3 \\ f(a)=f(1)=8(1)^2-3\text{ = 8-3 = 5} \end{gathered}[/tex]Now, we substitute the expressions in the given rate of change relation above as follows;
[tex]\begin{gathered} \frac{8t^2-3-(5)}{t-1}\text{ = }\frac{8t^2-8}{t-1}\text{ = }\frac{8(t^2-1)}{t-1} \\ \\ =\text{ }\frac{8(t-1)(t+1)}{t-1}\text{ = 8(t+1)} \end{gathered}[/tex]