Question

How do you find the average rate of change of `f(x)=x^2+2` over [0,2]?

Answer 1

By calculus: average rate of change is 2

Explanation:

As you have used `f(x)` I am assuming you are using Calculus.

`color(blue)("Short cut method")`

By sight: `"rate of change "-> (dy)/(dx)=2x`

At `x=0" "(dy)/(dx)=2(0)=0`

At `x=2" "(dy)/(dx)=2(2)=4`

Thus the average rate of change is

`(0+4)/2=2`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("From first principles")`

Let `" "y=x^2+2`..........................(1)

Increment `x" by the very small amount of "deltax`

As `x` has change then `y` will change as well

Let the change in `y" be "deltay`

Now we have

`y+deltay=(x+deltax)^2+2`

`y+deltay=x^2+2xdeltax+2`.....................(2)

Subtract equation (1) from equation (2)

`y+deltay=x^2+2xdeltax+2`
`underline(y" "=x^2" "+2)` apply the subtraction
`" "deltay = 0""+2xdeltax+0`

Divide by `deltax`

`(deltay)/(deltax)=2x xx(deltax)/(deltax)`

but `(deltax)/(deltax)=1`

So

`lim_(deltaxto0) (deltay)/(deltax)= (dx)/(dy)=2x`

Then solve for average rate of change as above.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Note that`lim_(deltaxto0)` means consider the situation where

`deltax` gets so small it may as well be zero, but in reality it has not quite got there!

Answer 2

2

Explanation:

Average rate = `(intd/dxf(x)dx)/(intdx)`, between the limits.
= `(intdf(x))/(intdx)`, between the limits.
= `(f(2)-f(0))/((2-0)` = `(6-2)/2`.