Mean Value Theorem?

Question

Verify that the function satisfies the hypotheses of The Mean Value Theorem on the given interval. Then find the number c that satisfy the conclusion of The Mean Value Theorem.
f(x)=x/(x+6), [0,1]

Answer 1

the hypotheses are:
1. f is continuous on the closed interval [0,1] and
2. f is differentiable on the open interval (0,1).

Clearly, the function we are dealing with is continuous on [0,1] since the only point of discontinuity is at x equal negative six. and the derivative of f is 6/(x+6)^2 and this derivative clearly exists for all x in (0,1). so both of the hypotheses are satisfied.
now we need only find the number c such that
f '(c)=[f(1)-f(0)] / [1-0]. so by substitution we get

6/(c+6)^2=[1/7 - 0] / 1 = 1/7
multiply both sides by 7 and (c+6)^2
to get (c+6)^2=42.
then take the square root of both sides to get c+6=sqrt(42). then subtract 6 from both sides to get c = sqrt(42)-6.

Answer 2

The function is continuos in the closed interval and the derivative exists in the open interval
so the MVT is applicable
[f(1)-f(0)]/(1-0) = f'(c)

f´(x) = 6/(x+6)^2 so 6/(c+6)^2= 1/7
(c+6)^2= 42 c+6 =sqrt42 ( positive sign as 0 c=sqrt42-6=0.4807

Answer 3

True