How do you find the average rate of change of #h(x)=x^2+3x-1# over [x, x+h]?

1 Answer
Jim H
May 22, 2017

For a function #f#, the average rate of change of #f# over interval #[a,b]# is defined to be #(f(b)-f(a))/(b-a)#

Explanation:

This question uses #h# to mean two different things. To try to avoid confusion, let's rename the function #f#.

#f(x) = x^2+3x-1#

#f(x+h) = (x+h)^2+3(x+h) -1#

The average rate of change is

#(f(x+h)-f(x))/((x+h)-x) = (f(x+h)-f(x))/h#

#(f(x+h)-f(x))/h = (overbrace([(x+h)^2+3(x+h) -1])^(f(x+h))-overbrace([x^2+3x-1])^(f(x)))/h#

expand insude the brackets

# = ([x^2+2xh+h^2+3x+3h-1]-[x^2+3x-1])/h#

remove the brackets (distribute the #-# sign)

# = (x^2+2xh+h^2+3x+3h-1-x^2-3x+1)/h#

Simplify the numerator

# = (cancel(color(red)(x^2))+2xh+h^2+cancel(color(green)(3x))+3h-cancel(1)-cancel(color(red)(x^2))-cancel(color(green)(3x))+cancel(1))/h#

# = (2xh+h^2+3h)/h#

factor the #h# in the numerator and reduce the fraction

# = (cancel(h)(2x+h+3))/(cancel(h)1)#

Finish writing

# = 2x+h+3#

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