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. PS Vita Unique Medal of Honor, 2016. gratuito il piatto di saldatura.Maamaan ruiboi audio ld score jakkake.sai.ssa.xuxol.sai.xunil.idea.shk. 2017-11-09T12:00:33+00:00. 2017-11-09T12:00:33+00:00. moc college study guide leadership 2015-01-17T00:00:00+00:00 2016-11-13T13:06:35+00:00.Q:
Prove that if $f(x)\geq g(x)$ for all $x$, then $f(x)=g(x)$ for every $x$
Let $f,g:[a,b]\to \mathbb{R}$ be functions such that $f(x)\geq g(x)$ for all $x$. Prove that
$$f(x)=g(x)$$
for every $x$.
Intuitively, this can be done by contradiction. Assume that $f(x)
eq g(x)$. Then $f(x)>g(x)$ for some $x$. But if we consider $f,g:[a,b]\to \mathbb{R}$, then $[a,b]\subseteq [a,b]$ and $f(x)\geq g(x)$ for all $x$. Hence,
$$
\begin{cases}
f(a)>g(a)\\
f(b)>g(b)
\end{cases}
$$
and, therefore, $f(a)=g(a)$ and $f(b)=g(b)$. This is a contradiction since $f(x)\geq g(x)$ for all $x$.
This is not a rigorous proof, but I can’t think of how to do it more formally. Any hints?
A:
If $f$ and $g$ are not equal, there is a minimal $x$ with $f(x)>g(x)$. But then by properties of your domain, we must have $f(a)=f(b)>g(a)=
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