Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 44

The mean value theorem is applicable to the given function, since it is a polynomial function. All polynomial functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval.
The mean value theorem states:
f(b) - f(a) = f'(c)(b-a)
Replacing 2 for b and -7 for a, yields:
f(2) - f(-7) = f'(c)(2+7)
Evaluating f(2) and f(-7) yields:
f(2) =sqrt(2-2) = 0
f(-7) = sqrt(2+7)=3
You need to evaluate f'(c), using quotient rule:
f'(c) = (sqrt(2-c))' => f'(c) =((2-c)')/(2sqrt(2-c)) => f'(c) =-1/(2sqrt(2-c))
Replacing the found values in equation f(2) - f(-7) = f'(c)(2+7):
0 - 3= -9/(2sqrt(2-c)) =>3/(2sqrt(2-c)) = 1 =>2sqrt(2-c) = 3=> sqrt(2-c) = 3/2 => 2 - c = 9/4 => c = 2 - 9/4 => c = -1/4 in (-7,2)
Hence, in this case, the mean value theorem may be applied for c = -1/4.

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