Question

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)`

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: