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#
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#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.
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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!
