proof of least and greatest value of function

$f$ is continuous, so it will transform compact sets into compact sets.Thus since $[a,b]$ is compact, $f([a,b])$ is also compact. $f$ will thus attain on the interval $[a,b]$ a maximum and a minimum value because real compact sets are closed and bounded.

Consider the maximum and later use the same argument for $-f$ to consider the minimum.

By a known theorem (http://planetmath.org/FermatsTheoremStationaryPoints) if the maximum is attained in the interior of the domain, $c\\in]a,b[$ then $f(c)\\text{is a maximum}\\implies f^{\\prime}(c)=0$ , since $f$ is differentiable.

If the maximum isn’t attained in $]a,b[$ and since it must be attained in $[a,b]$ either $f(a)$ or $f(b)$ is a maximum.

For the minimum consider $-f$ and note that $-f$ will verify all conditions of the theorem and that a maximum of $-f$ corresponds to a minimum of $f$ and that $-f^{\\prime}(c)=0\\iff f^{\\prime}(c)=0$ .

单词	ProofOfLeastAndGreatestValueOfFunction
释义	proof of least and greatest value of function $f$ is continuous, so it will transform compact sets into compact sets.Thus since $[a,b]$ is compact, $f([a,b])$ is also compact. $f$ will thus attain on the interval $[a,b]$ a maximum and a minimum value because real compact sets are closed and bounded. Consider the maximum and later use the same argument for $-f$ to consider the minimum. By a known theorem (http://planetmath.org/FermatsTheoremStationaryPoints) if the maximum is attained in the interior of the domain, $c\\in]a,b[$ then $f(c)\\text{is a maximum}\\implies f^{\\prime}(c)=0$ , since $f$ is differentiable. If the maximum isn’t attained in $]a,b[$ and since it must be attained in $[a,b]$ either $f(a)$ or $f(b)$ is a maximum. For the minimum consider $-f$ and note that $-f$ will verify all conditions of the theorem and that a maximum of $-f$ corresponds to a minimum of $f$ and that $-f^{\\prime}(c)=0\\iff f^{\\prime}(c)=0$ .
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