Prove Max/Min of f(x) -> Sqrt of f(x) at x0

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SUMMARY

The discussion focuses on proving that if a function f(x) is non-negative on an interval I and attains a maximum at x0, then the square root of f(x), denoted as sqrt(f(x)), also reaches its maximum at x0. The proof leverages the property that the square root function is increasing for non-negative values. Consequently, if f(y) ≤ f(x0) for all y in I, it follows that sqrt(f(y)) ≤ sqrt(f(x0)), confirming that sqrt(f(x)) achieves its maximum at the same point x0.

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Homework Statement



prove: if f(x) bigger or the same as 0 on an interval I and if f(x) has a maximum value on I at x0(0 is written small beside the x), then sqrt of f(x) also has a maxsimum value at x0. Similarily for minimum values. Hint: Use the fact that sqrt of x is an increasing function on the interval zero to plus infinity.

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The Attempt at a Solution

 
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Looks pretty direct to me. If [itex]f(x_0)[/itex] is a maximum for f, then [itex]f(y)\le f(x_0)[/itex] for all y.Since square root is an increasing function, [itex]\sqrt{f(y)}\le \sqrt{f(x_0)}[/itex].
 

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