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

Question

Christopher Agresta · Accepted Answer

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)#

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 change's average rate 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#

Enlarge beyond 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#

Make the numerator simpler.

# = (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)#

Complete your writing.

# = 2x+h+3#

Emilia Aguilar · Answer

undefined To find the average rate of change of $ h(x) = x^2 + 3x - 1 $ over the interval $[x, x+h]$, you first need to find the value of $ h(x+h) $ and subtract $ h(x) $ from it. Then, divide the result by $ h $. The formula for the average rate of change is:

$$ \text{Average rate of change} = \frac{h(x+h) - h(x)}{h} $$

Substitute $ x $ and $ x+h $ into the function $ h(x) $, perform the calculations, and then apply the formula.