# Show that max(f,g) is continious?is my answer wrong?

1. Jun 25, 2013

### Andrax

1. The problem statement, all variables and given/known data
let f and g are continuous functions show that max(f,g) is continious

3. The attempt at a solution
the answer i gave is this
maxf(g)=h(x)
h(x)=f(x)==>f(x)-g(x)>0
h(x)=g(x)==>f(x)-g(x)≤0
thus h is continious

now in the text book they defined max with the normal definition
max=lf(x)-g(x)l+f(x)+g(x)/2

Last edited: Jun 25, 2013
2. Jun 25, 2013

### micromass

Staff Emeritus
Why does this imply that h is continuous?

3. Jun 25, 2013

### Andrax

I thought it's obvious
if f(x)>g(x) then h(x) = f(x) f continious then h continious
if f(x)<=g(x) then h(x)=g(x) g gotinious thus h is continious
Am i doing something wrong here?(spivak's calculus questions are quiet hard )
Edit : i wasn't supposed to study cases?

4. Jun 25, 2013

### Staff: Mentor

"Continious" is not a word - the one you mean is continuous.
Presumably you mean max(f, g) = h(x)
The above assume that f > g for all x in the domain or that g ≤ f for all x in the domain. What happens if one function is greater than the other in some places, but not in others?
For absolute value, use |, not l (lower-case "ell"). Also, what gets divided by 2? As you wrote it, only g(x) is divided by 2. If that's not what you meant, use parentheses.

5. Jun 25, 2013

### micromass

Staff Emeritus
This looks like the following proof:

Consider $f(x) = 1$ for all $x$, $g(x) = -1$ for all $x$ and take

$$h(x) = \left\{\begin{array}{l} 1~\text{if}~x<0\\ -1~\text{if}~x\geq 0\end{array}\right.$$

If $h(x)<0$, then $h(x) = g(x)$ and $g$ is continuous.
If $h(x)>0$, then $h(x) = f(x)$ and $f$ is continuous.
Thus $h$ is continuous.

This is essentially what you did. But the conclusion that $h$ is continuous is false of course.

6. Jun 25, 2013

### Andrax

i was missing this part thank you
micromass made me laugh a bit at my "proof" :).
thanks guys .