Question

How do you find the average rate of change of `f(x)=2x^2+1` on [x,x+h]?

Answer 1

`4x+2h`

Explanation:

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

`(f(b)-f(a))/(b-a)`

So the average rate of change of the function `f(x)=2x^2+1` on `[x,x+h]` is:

`Aroc = ( f(x+h)-f(x) ) / ( (x+h)-(x) )`

`" " = ( f(x+h)-f(x) ) / ( h ) \ \ \ \ \ ..... [1]`

`" " = ( 2(x+h)^2+1-(2x^2+1) ) / ( h )`
`" " = ( 2(x^2+2xh+h^2)+1 -2x^2-1 ) / ( h )`
`" " = ( 2x^2+4xh+2h^2 -2x^2 ) / ( h )`
`" " = ( 4xh+2h^2 ) / ( h )`
`" " = 4x+2h`

Which is the required answer.

Additional Notes:

Note that this question is steered towards deriving the derivative `f'(x)` from first principles, as the definition of the derivative is:

`f'(x) = lim_(h rarr 0) (f(x+h)-f(x))/h`

This is the function we had in [1], so as we take the limit as `h rarr 0` we get the derivative `f'(x)` for any `x`, This:

`f'(x) = lim_(h rarr 0) 4x+2h`

`" " = 4x`