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

[minia2353]

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.

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

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)}.