Hello, domo256!
I assume the function is: \(\displaystyle \:f(x) \:=\:11\,-\,x\,+\,2(25\,+\,x^2)^{\frac{1}{2}}\)
Then: \(\displaystyle \:f'(x)\;=\;-1\,+\,2\cdot\frac{1}{2}(25\,+\,x^2)^{-\frac{1}{2}}\cdot2x\;=\;-1\,+\,\frac{2x}{\sqrt{25\,+\,x^2}}\)
We have: \(\displaystyle \:-1\,+\,\frac{2x}{\sqrt{25\,+\,x^2}}\;=\;0\)
Multiply through by \(\displaystyle \sqrt{25\,+\,x^2}:\;\;-\sqrt{25\,+\,x^2}\, +\, 2x \;=\;0\;\;\Rightarrow\;\;2x\:=\:\sqrt{25\,+\,x^2}\)
Square both sides: \(\displaystyle \:4x^2\:=\:25\,+\,x^2\;\;\Rightarrow\;\;3x^2\:=\:25\;\;\Rightarrow\;\;x^2\:=\:\frac{25}{3}\)
Hence: \(\displaystyle \:x\:=\:\pm\sqrt{\frac{25}{3}} \:=\:\pm\frac{5\sqrt{3}}{3}\)
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The positive root is probably the answer.
I suspect the function came from a "rowing-walking" problem.
A man is on an island 5 miles offshore.
He wants to reach a location 11 miles down the straight shoreline.
He can walk twice as fast as he can row.
He wil row to a point \(\displaystyle P\) on the shore and walk the rest of the way.
Locate point \(\displaystyle P\) to minimize the time for the journey.
