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

1 Answer
Steve M
Feb 28, 2017

#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#

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