For the mean value theorem to be valid, the function f(x) = sqrt x must satisfy the following conditions on the interval, such that:
f(x) is continuous over the interval [0,4] and it is because it is an elementary function.
f(x) is differentiable on (0,4).
If both conditions are satisfied, then, it exists a point c in (0,4) , such that:
f(4) - f(0) = f'(c)(4 - 0)
You need to evaluate f(4) and f(0), by replacing 4 and 0 for x in equation of the function:
sqrt 4 - sqrt 0 = (sqrt c)'(4-0)
2 - 0 = 4/(2sqrt c)
Reducing by 2 yields:
2 = 2/sqrt c => 2sqrt c = 2 => sqrt c = 1 => c = 1 in (0,4)
Hence, evaluating the number c that satisfies the mean value theorem yields c = 1.
Friday, May 27, 2016
Calculus: Early Transcendentals, Chapter 4, 4.2, Section 4.2, Problem 13
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