What is the average rate of change of the function #g(t)=(t+2)^2# on the interval [t-h, t]?

1 Answer
Steve M
Mar 19, 2017

# ( 8t-h^2+2ht+4h ) / ( h) #

Explanation:

The average rate of change of a continuous function,
#f(x)# , on a closed interval #[a,b]# is given by

# bar f(x) =(f(b)-f(a))/(b-a) #

So the average rate of change of the function #g(t)=(t+2)^2# on #[t-h,t]# is:

# bar(g) = ( g(t)-g(t-h) ) / ( t-(t-h) )#
# \ \ = ( (t+2)^2-(t-h+2)^2 ) / ( t-t+h) #
# \ \ = ( (t^2+4t+4)-(t^2+h^2-2ht-4h-4t+4) ) / ( h) #
# \ \ = ( t^2+4t+4-t^2-h^2+2ht+4h+4t-4 ) / ( h) #
# \ \ = ( 8t-h^2+2ht+4h ) / ( h) #

Note
If we take the limit as #h rarr 0# then the above becomes the derivative:

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