If f and g are continuous functions prove that M(x)=max{f(x),g(x)} and m(x)=min{f(x),g(x)} are also continuous.

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The proofs are almost identical, so I'll just prove the first statement. There are two cases.

First, suppose that `f(x)!=g(x),` and without loss of generality, let `f(x)-g(x)=d>0,` ie suppose `f(x)` is greater than `g(x).` By the continuity of `f` and `g,` there is some `delta_1>0` such that for all `yin(x-delta_1,x+delta_1),` we have the two inequalities

`f(y)>f(x)-d/2`

`-g(y)> -g(x)-d/2.`

Adding gives ` ``f(y)-g(y)>f(x)-g(x)-d=d-d=0.` Beneath the...

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