What are the coordinates for this point?

Question

The Mean Value Theorem guarantees the existence of a special point on the graph of y=sqrt(x) between (0, 0) and (4, 2). What are the coordinates of this point? 

 

Answer: (1, 1)

 

Please show all your work. 

Answer 1

The Mean Value Theorem guarantees the existence of a special point on the graph of y=sqrt(x) between (0, 0) and (4, 2). What are the coordinates of this point?

 

The Mean Value Theorem states that for a function f(x) defined on a closed interval, [a,b], there will exist a point c, a ≤ c ≤ b, such that:

 

f'(c) = (f(b)-f(a))/(b-a)            ( f' = df(x)/dx )

 

Basically it means that there will be at least one point between a and b where f(x) has the same slope as the secant line connecting f(a) to f(b).

 

For f(x) = x^1/2, the slope of the secant line between (0,0) and (4,2) is:

 

(2 - 0)/(4 - 0) = 1/2

 

To find the point c, find f'(x) and set it equal to 1/2.

 

f'(x) = (1/2)x^-1/2 = 1/2 when x = 1

 

To find the y value of point c, plug x=1 into f(x)

 

f(x=1) = (1)^1/2 = 1

 

The point c = (1,1) and it lies within the specified interval.

Answer 2

The Mean Value Theorem guarantees the existence of a special point on the graph of y=√(x) between (0,0) and (4,2). What are the coordinates of this point?

Answer: (1, 1)

Phillip has the right answer for THE Mean Value Theorem. The following answer is for the Mean Value Theorem for Integrals.

f(c)(b-a)=∫{a,b}(f(x) dx)

√(c) (4-0) = ∫{0,4}(x^(1/2) dx)

4√(c) = 2/3 [x^(3/2]{0,4}

= 2/3 [4^(3/2)]

= 2/3 [2^3]

= 16/3

√(c) = 16/3/4 = 4/3

c = 16/9

f(16/9) = 4/3

So the special point is (16/9,4/3),
not (1,1).