Question 230599
<font face="Garamond" size="+2">


That is a pretty cryptic question you have posed there.  I'm guessing that you want to find a critical point of the function *[tex \Large y\ =\ \sqrt{x}] on the closed interval [16, 25] using the Mean Value Theorem.


The Mean Value Theorem says:


Let *[tex \LARGE f:][<i>a, b</i>]*[tex \LARGE \rightarrow\R] be a continuous function on the closed interval [<i>a, b</i>] and differentiable on the open interval (<i>a, b</i>), where *[tex \LARGE a\ <\ b].  Then there exists some *[tex \LARGE c] in (<i>a, b</i>) such that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(c)\ =\ \frac{f(b)-f(a)}{b-a}]


The given interval is [16, 25] and *[tex \Large f(x)\ =\ \sqrt{x}], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{f(b)-f(a)}{b-a}\ =\ \frac{5-4}{25-16}\ =\ \frac{1}{9}]


But given *[tex \Large f(x)\ =\ \sqrt{x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x)\ =\ -\frac{1}{2\sqrt{x}}\ \ \ \ ] (using the power rule with *[tex \Large f(x)\ =\ x^{\frac{1}{2}}])


Therefore, if


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(c)\ =\ \frac{1}{9}]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{1}{2\sqrt{c}}\ =\ \frac{1}{9}]


Solve for *[tex \Large c]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\sqrt{c}\ =\ -9]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{c}\ =\ -\frac{9}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ = \frac{81}{4}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>