Mean Value Theorem

Question

Consider the function f(x)=4√x+4f(x)=4x+4 on the interval [3,5][3,5]. Find the average or mean slope of the function on this interval.

By the Mean Value Theorem, we know there exists a cc in the open interval (3,5)(3,5) such that f'(c)f′(c) is equal to this mean slope. For this problem, there is only one cc that works. Find it.

Answer 1

for y=4x+4, the mean or average slope = 4, for any c. the slope is always 4

for y = 4sqrx + 4

slope at x= 3 is

y' = 2/sqr(3) = 1.1547 = about 1.15

at x=5 slope is

y' = 2/sqr(5) = .894427 about 0.89

average slope = about 1.024564

y' = 2/sqr(c) = 1.024564

sqr(c) = 2/1.024564 = about 1.952

c = about 3.81

according to the MeanVAule theorem

f'(c) = (f(b) - f(a))/(b-a) = (4sqr5 - 4sqr3)/(5-3) = 2(sqr5-sqr3) = about 1.024564

f'(x) = 2/sqrx = 1.024564

sqrc = 2/1.024564 = about 1.952

c = (1.952)^2 = about 3.81

c= 3.81

3<c<5

y=4sqrx +4

y-4 = 4sqrx

(y-4)^2 = 16x

x = (1/16)(y-4)^2 is a rightward opening parabola with vertex = (0,4)