# 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.

## Expert Answers

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),`...

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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 formalism, we've just shown that if one continuous function is greater than another continuous function at some point `x,` then the function remains greater in some neighborhood of ` ``x,` which is what our intuition tells us.

So if `|x-y|<delta_1,` we have

`|M(x)-M(y)|=|` max`{f(x),g(x)}-` max`{f(y),g(y)}|`

`=|f(x)-f(y)|.`

Now again by the continuity of `f,` for any `epsilon>0,` there is some `delta<delta_1` such that `|x-y|<delta` implies `|f(x)-f(y)|<epsilon.`

Combining everything, we see that given `epsilon>0,` we can find `delta>0` such that `|x-y|<delta` implies `|M(x)-M(y)|<epsilon,` so `M` is continuous.

The second case is easier. Suppose `f(x)=g(x).` Then for any `y,`

`|M(x)-M(y)|=|f(x)-f(y)|` if `f(y)>=g(y),` and

`|M(x)-M(y)|=|g(x)-g(y)|` if `f(y)<=g(y).`

So given any `epsilon>0,` pick `delta>0` such that `|x-y|<delta` implies both

`|f(x)-f(y)|<epsilon` and `|g(x)-g(y)|<epsilon,` and continuity of `M` follows.

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