MEAN VALUE THEOREM AND AVERAGE SLOPE

Question

Consider the function f(x)=8sqrt(x)+7 on the interval [36] . Find the average or mean slope of the function on this interval. By the Mean Value Theorem, we know there exists a c in the open interval (36) such that f(c) is equal to this mean slope. For this problem, there is only one c that works. Find it.

Answer 1

Remember the slope formula for the slope of a line in calculus? Given two points on a line, the slope m is:

 

m=[f(x2)-f(x1)]/(x2-x1)

 

We can find the average slope on an interval the same way, using the endpoints of the interval, in this case x=3 and x=6:

 

Average slope = [f(6)-f(3)]/(6-3) = [(8\sqrt(6)+7)-(8\sqrt(3)+7)]/3 = (8/3)[\sqrt(6)-\sqrt(3)] ≈ 1.91317

 

So that's the average slope. According to the mean value theorem, there is some c in the interval (3,6) such that

 

f'(c)=1.91316

 

That means first we have to find f'(c):

 

f'(c)=4c^(-1/2)

 

Now we calculate:

 

4c^(-1/2) = 1.9136

c^(-1/2) = 0.4782925

c^(1/2) = 1/(0.4782925)

c = 4.3713

 

Note that all of these calculations are approximations after we approximated the average slope itself.