How do you determine all values of c that satisfy the mean value theorem on the interval [1, 4] for f(x)=1/sqrt(x)?

1 Answer
Aug 6, 2017

The value of c=2.08

Explanation:

The mean value theorem states that if a function f(x) is continuous on the interval [a,b] and differentiable on the interval (a,b), then

EE c in [a,b] such that

f'(c)=(f(b)-f(a))/(b-a)

Here,

f(x)=1/sqrtx

The domain of f(x) is D_f(x) in (0, +oo)

The interval [1,4] in D_f(x)

f'(x)=-x^(-3/2)/2

f'(c)=-c^(-3/2)/2

f(4)=1/sqrt4=1/2

f(1)=1/sqrt1=1

Therefore,

-c^(-3/2)/2=(1/2-1)/(4-1)=-1/6

c^(-3/2)=1/3

c^(3/2)=3

c^3=9

c=root(3)9=2.08

2.08 in [1,4]