mean value theorem

Question

find the value(s) of c that satisfy the mean value theorem for the given function and interval.

 

f'(x)= √(x-3)   [3,8]

Answer 1

“Find the value(s) of c that satisfy the mean value theorem for the given function and interval.

f'(x)= √(x-3) [3,8]”

It’s unclear which functions the mean value theorem is to be applied to:

1) g(x) = √(x-3)

or

2) ∫f’(x)dx = f(x) = (2/3)(x-3)^(3/2) + C

===

1) g(c) = (1/(8-3)) ∫{3,8}(√(x-3)dx) = √(c-3)

5√(c-3) = ∫{3,8}((x-3)^(1/2) dx)

= (2/3)[(x-3)^(3/2)]{3,8}

= (2/3)[(8-3)^(3/2)–(3-3)^(3/2)]

= (2/3)[(5)^(3/2)–(0)^(3/2)]

5(c-3)^(1/2) = (2/3)(5)^(3/2)

25(c-3) = (4/9)(5)^3

c-3 = 20/9

c = 20/9 + 27/9 = 47/9 ≈ 5.22222222222222

===

2) f(x) = (2/3)(x-3)^(3/2) + C

f(c) = (1/(8-3)) ∫{3,8}((2/3)(x-3)^(3/2) + C dx)

5f(c) = [(4/15)(x-3)^(5/2) + Cx]{3,8}

5f(c) = (4/15)(8-3)^(5/2) + 8C – (4/15)(3-3)^(5/2) – 3C

5f(c) = (4/15)(5)^(5/2) + 5C – (4/15)(0)^(5/2)

f(c) = (4/15)(5)^(3/2) + C

(2/3)(c-3)^(3/2) + C = (4/15)(5)^(3/2) + C

(2/3)(c-3)^(3/2) = (4/15)(5)^(3/2)

(c-3)^(3/2) = (2/5)(5)^(3/2)

c-3 = 5(2/5)^(2/3)

c = 5(2/5)^(2/3) + 3 ≈ 5.7144176165949