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

Answer 1

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

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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Answer 2

Scarlett Allison

The average rate of change of the function g(t) = (t + 2)^2 on the interval [t - h, t] can be found by evaluating g(t) at the endpoints of the interval and then calculating the difference in the function values divided by the difference in the input values. So, the average rate of change is given by:

[ \frac{g(t) - g(t - h)}{t - (t - h)} ]

[ = \frac{(t + 2)^2 - ((t - h) + 2)^2}{t - (t - h)} ]

[ = \frac{(t + 2)^2 - (t - h + 2)^2}{h} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.