Question #38830

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Question #38830

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Answer 1 · 2017-07-16T22:43:40

Normally, in the real domain, $f (x) = \sqrt{25 - x^{2}}$ $f(x) = sqrt(25-x^2)$ is meant to be the positive value of the root.

Explanation:

The Mean Value Theorem applies to a single valued real function, so in this case, $f (3) = 4$ $f(3) = 4$ satisfies the MVT for the function $f (x) = ∣ ∣ \sqrt{25 - x^{2}} ∣ ∣$ $f(x) = abs sqrt(25-x^2)$ and $f (3) = - 4$ $f(3) = -4$ satisfies the MVT for the function $f (x) = - ∣ ∣ \sqrt{25 - x^{2}} ∣ ∣$ $f(x) = -abs sqrt(25-x^2)$

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Question #38830

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