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

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If f(x) is non-negative on an interval and has a maximum at x0, then the square root of f(x) also achieves a maximum at x0. This is because if f(x0) is the maximum, then for any y in the interval, f(y) is less than or equal to f(x0). Since the square root function is increasing for non-negative values, it follows that sqrt(f(y)) is less than or equal to sqrt(f(x0)). The same reasoning applies to minimum values, confirming that sqrt(f(x)) will also have a minimum at x0. Thus, the proof hinges on the properties of the square root function and the behavior of f(x) at its extremum.
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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 f(x_0) is a maximum for f, then f(y)\le f(x_0) for all y.Since square root is an increasing function, \sqrt{f(y)}\le \sqrt{f(x_0)}.
 
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