difficult calculus proof (not hw-esque)

Pre-College Topics High School Life
1

<p>let $=integral
given $f(x)>$g(x) |a<x<b prove=“” $sqrt(f(x))=“”>$sqrt(g(x)) |a<x<b</x<b></p>

<p>two caveats: i don’t know that this is true. A counterexample would be equally helpful. I don’t need a formal proof, just something good enough to convince me.</p>

2

<p>never mind, read your post completely wrong.</p>

<p>Lets try this, if f(x)>g(x) then sqrt(f(x))>sqrt(g(x)). If f(x)>g(x) then $f(x)>$g(x). Hence $sqrt(f(x))>$sqrt(g(x)). All of this applies on the interval (a,b)</p>

3

<p>Might want to define f(x) > g(x) > 0, otherwise your roots will be having a bit of fun in the complex plane.</p>