Mean Value Theorum?

Question

Part A: Consider the function f(x)= 8sqrt[x] +4 on the interval. [1,7] Find the average slope of the function on this interval.

Part B: By the Mean Value Theorem, we know there exists a in the open interval (1,7) such that f'(c) is equal to this average slope. For this problem, there is only one that works. Find it.

Answer 1

The average slope is

1/(7-1) integral from 1 to 7 of f'(x) dx

f' = 4/√x, so we get that the average slope is

(1/6) integral from 1 to 7 of 4x^(-½) dx = (1/6) 8√x evaluated at 1 and 7 =(4/3)(√7 - 1)

(We could have avoided taking the derivative by noting that
the integral from 1 to 7 of f'(x) dx = f(7) - f(1), but you need the derivative for part B)

Part B is just solving 4/√c = (4/3)(√7 - 1) for c (confirm that it does lie in (1,7)). I leave that to you.

Answer 2

basically the mean value theroem states that f'(x)=[f(b)-f(a)]/(b-a) so to make it more clear. It is basically saying that sometime on this intervale if f(x) is continuous that the average must equal the slope of the secant line. If you think of this as running. At some time you are running you are going to be running at the same rate as the average speed. f'(x)=x/(sqrt(x^2+1)) f(8)=3 f(0)=1. x/(sqrt(x^2+1))=(3-1)/8 x/(sqrt(x^2+1))=1/(sqrt(2)) so use a graphing calcuator to determine what x is. make sure that the x you get is in the interval of [0, sqrt(8)]. When you do put into a graphing calculator you get x=1. This is in the interval of [0,sqrt(8)]. So there is a value of c that satisfies the mean value theroem in this interval. If we had gotten x=9 then we will have said no there isn't any value of c that satisfies the mean value thereom in the interval [0, sqrt(8)] because 9 isn't in that interval.