# suppose that f and g are continuous at a. prove that max{f,g} and min{f,g} are continuous at a.help me

cosinusix | College Teacher | (Level 3) Assistant Educator

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If f and g are continuous at a, then  `AAepsi>0, EEalpha_1 | |x-a|<alpha_1 rArr |f(x)-f(a)|<epsi and EEalpha_2| |x-a|<alpha_2 |g(x)-g(a)|<epsi`

Max(f,g)(a) is either f(a) or g(a). Let's suppose it is f(a) ( rename your function if needed such that f(a)=max(f,g)(a)) in this case Min(f,g)(a)=g(a).

2 cases: either f(a)=g(a) or f(a)>g(a)

If f(a)=g(a).

Remark: For a given x close to a, we can't determine is Max(f,g)(x) is f(x) or g(x)  and Min(f,g)(x) is f(x) or g(x). It could be f(x) for some x close to a and g(x) for other x close to a.

The functions f and g are continuous therefore for x close to a, both g(x) and f(x) will stay close to f(a)=g(a). This will guaranty the continuity of the functions M(f,g) and Max(f,g) at x=a.

It is the idea of the proof in this case.

Proof in the case f(a)=g(a)

Max(f,g)(x)-Max(f,g)(a) is either f(x)-f(a) or g(x)-f(a)=g(x)-g(a)

and Min(f,g)(x)-Min(f,g)(a) is either g(x)-g(a) or f(x)-g(a)=f(x)-f(a)

`AAepsi>0, EEalpha<alpha_1 and alpha<alpha_2 `

`alpha_1 and alpha_2 ` are determined by the first property.

`AAx| |x-a|<alpha,`

`|Max(f,g)(x)-Max(f,g)(a)|= |f(x)-f(a)|ltepsi`

or `|Max(f,g)(x)-Max(f,g)(a)|= |g(x)-g(a)|ltepsi`

If f(a)=g(a), Max(f,g) satisfies the definition of continuity therefore Max is continuous.

`|Min(f,g)(x)-Min(f,g)(a)|= |g(x)-g(a)|ltepsi`

or `|Min(f,g)(x)-Min(f,g)(a)|= |f(x)-f(a)|ltepsi`

If f(a)=g(a), Min(f,g) satisfies the definition of continuity therefore Max is continuous.

If `f(a)gtg(a)`

Then f(a)=g(a)+C with C>0

Remark: I will prove that if x is close enough to a, then Max(f,g) (x)is always  f(x) and Min (f,g) is always g(x). Since f and g are continuous, Max and Min will be continuous at a.

Proof in the case f(a)>g(a)

`EE alpha_c | |x-a|<alpha_c, |f(x)-f(a)|<c/2 and |g(x)-g(a)|<c/2`

It implies that `AA x | |x-a|<alpha_c `

` f(x)gtf(a)-c/2=g(a)+c-c/2=g(a)+c/2`

and `g(x)ltg(a)+c/2ltf(x)`

Therefore `AA x | |x-a|<alpha_c `

Max(f,g)(x)=f(x) and Min(f,g)(x)=g(x)

Let's prove that Max and Min satisfy the definition of Continuity.

`AA epsi >0 EE alpha=min(alpha_c, alpha_1, alpha_2) | |x-a|<alpha,`

`|Max(f,g)(x)-Max(f,g)(a)|=|f(x)-f(a)|<epsi`

and

`|Min(f,g)(x)-Min(f,g)(a)|=|g(x)-g(a)|<epsi`

Min(f,g)(x) and Max(f,g)(x) satisty the definition of continuity therefore Min(f,g)(x) and Max(f,g)(x) in the case f(a)>g(a)

In the 2 cases, Min(f,g) and Max(f,g) are continuous at x=a.