Let $f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}},\ 1 \leq x \leq 26$ be real valued function, then $f^{'}(x)$ for $1

Question

Let $f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}},\ 1 \leq x \leq 26$ be real valued function, then $f^{'}(x)$ for $1 < x < 26$ is

Dr. Abubacker · Accepted Answer

(a)
From Rolle’s theorem in given interval, [1,26], $f(1) = (26) = 5$ function satisfy all the condition of Rolle’s theorem, therefore in [1,26], at least, there is one point for which $f^{'}(x) = 0$

Q. Let $f(x) = \sqrt{x - 1} + \sqrt{x + 24 - 10\sqrt{x - 1}},\ 1 \leq x \leq 26$ be real valued function, then $f^{'}(x)$ for $1 < x < 26$ is

Explanation

Reviewed by Dr. Abubacker

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