Given C(x)=500+10x+0.05x2
i) The average rate of change is the change in y value over the change in x value for two distinct points, hence if x changes from 100 to 105, then you may evaluate the average rate of change of C(x) such that: 105−100C(105)−C(100) .
Now, C(105)−C(100)=(500+10(105)+0.05(105)2)−(500+10(100)+0.05(100)2)=(1050−1000)+(551.25−500)=101.25
So, Average rate of change of C(x) from 100 to 105 is 5101.25=20.25.
ii) The instantaneous rate of change of C with respect to x=100 is C′(x)∣x=100 .
Now, given C(x)=500+10x+0.05x2
⟹C′(x)=10+0.1x
The instantaneous rate of change of C with respect to 100 is C′(100)=10+0.1(100)=20 .
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