Answer:
Step-by-step explanation:
[tex]f'(x) =\frac{1}{2\sqrt{x-3}}[/tex]
By MVT, there exist a number c on [4, 12] such that
[tex]f'(c) = \frac{f(12)-f(4)}{12-4}[/tex]
[tex]f(12) = 3, and, f(4) = 1, hence f'(c) = \frac{1}{4}[/tex]
therefore,
[tex]\frac{1}{2\sqrt{x-3}} = \frac{1}{4}[/tex]
Solve this equation, you have x =1.
However, this is out of the given interval. Hence does not exist
