
#1
May2410, 12:40 AM

P: 13

1. The problem statement, all variables and given/known data
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. 2. Relevant equations 3. The attempt at a solution 



#2
May2410, 07:53 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

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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