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\noindent {\bf Question}
\noindent The {\bf minimum value property} states that, if $f$ is
continuous on $[a,b]$, then $f$ achieves its minimum on $[a,b]$;
that is, there exists some $y_0$ in $[a,b]$ so that $f(y_0)\le
f(x)$ for all $x\in [a,b]$. Prove that a continuous function $f:
[a,b]\rightarrow {\bf R}$ satisfies the minimum value property if
it satisfies the maximum value property.
\medskip
\noindent {\bf Answer}
\noindent Since $f$ is continuous on $[a,b]$, so is $g(x) =-f(x)$.
Since $g$ is continuous on the closed interval $[a,b]$, the
maximum value property applied to $g$ yields that there exists
some $x_0$ in $[a,b]$ so that $g(x_0) \ge g(x)$ for all $x$ in
$[a,b]$. Hence, $-f(x_0) \ge -f(x)$ for all $x$ in $[a,b]$, and so
$f(x_0)\le f(x)$ for all $x$ in $[a,b]$. That is, $f$ satisfies
the minimum value property.
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