How do you find the average rate of change of #y=x^3+1# from #x=1# to #x=3#?

Question

Charles Arocha · Accepted Answer

#(f(x+h)-f(x))/h=(f(b)-f(a))/(b-a)#, where #a# is the lower bound and #b# is the upper bound. The mean rate of variation #slope=(f(b)-f(a))/(b-a)=(f(3)-f(1))/(3-1)=((3)^3+1-((1)^3+1))/(3-1)=((27+1)-(1+1))/(3-1)=(28-2)/(3-1)=26/2=13# Point: #(3,28)#
Point: #(1,2)# #y=mx+b# #2=13(1)+b# #2=13+b# #-11=b# #y=13x-11#, the secant line through the points #(3,28)# and #(1,2)#.

Cameron Alexander · Answer

undefined To find the average rate of change of $ y = x^3 + 1 $ from $ x = 1 $ to $ x = 3 $, first find the values of $ y $ at $ x = 1 $ and $ x = 3 $.

1. Substitute $ x = 1 $ into the equation: 
   $ y(1) = (1)^3 + 1 = 1 + 1 = 2 $.

2. Substitute $ x = 3 $ into the equation: 
   $ y(3) = (3)^3 + 1 = 27 + 1 = 28 $.

Then, calculate the change in $ y $ over the interval $ x = 1 $ to $ x = 3 $, which is $ \Delta y = y(3) - y(1) = 28 - 2 = 26 $.

The change in $ x $ over the same interval is $ \Delta x = 3 - 1 = 2 $.

Finally, divide the change in $ y $ by the change in $ x $ to find the average rate of change:
$$ \text{Average rate of change} = \frac{\Delta y}{\Delta x} = \frac{26}{2} = 13. $$