# Maximum and minimum

1. May 24, 2010

### minia2353

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. May 24, 2010

### HallsofIvy

Staff Emeritus
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, $\sqrt{f(y)}\le \sqrt{f(x_0)}$.