# Math Help - continuity

1. ## continuity

If $f:I\rightarrow R$ and $g: \rightarrow R$ are any functions defined on $I$, then we can form new functions $max\{f,g\}(x)=max\{f(x),g(x)\}$ for all $x\in I$, and similarly for $min\{f,g\}$

I'm asked to sketch a graph of $max\{f,g\}$ and $min\{f,g\}$ if $f(x)=sin x$ and $g(x)=cos x$ on $I=[-2\pi , 2\pi ]$

What do the max/min thing mean? Basically, what is the first paragraph telling me? Thanks for any help.

2. Originally Posted by sfspitfire23
If $f:I\rightarrow R$ and $g: \rightarrow R$ are any functions defined on $I$, then we can form new functions $max\{f,g\}(x)=max\{f(x),g(x)\}$ for all $x\in I$, and similarly for $min\{f,g\}$

I'm asked to sketch a graph of $max\{f,g\}$ and $min\{f,g}$ if $f(x)=sin x$ and $g(x)=cos x$ on $I=[-2\pi , 2\pi ]$

What do the max/min thing mean? Basically, what is the first paragraph telling me? Thanks for any help.
If $f,g:X\to\mathbb{R}$ then, $f\wedge g=\frac{f+g-|f-g|}{2}$ and $f\vee g=\frac{f+g+|f-g|}{2}$

P.S. common notation $f\vee g=\min\{f,g\}$ and simlilarly for the other. What it means is "take the bigger value of $f$ or $g$ at x"

3. Originally Posted by Drexel28
If $f,g:X\to\mathbb{R}$ then, $f\wedge g=\frac{f+g-|f-g|}{2}$ and $f\vee g=\frac{f+g+|f-g|}{2}$

P.S. common notation $f\vee g=\min\{f,g\}$ and simlilarly for the other. What it means is "take the bigger value of $f$ or $g$ at x"
Thank you Drexel. However, I'm having trouble seeing where you got $f\wedge g=\frac{f+g-|f-g|}{2}$ and $f\vee g=\frac{f+g+|f-g|}{2}$. If you could explain to me how these came to be? (something with the definition of continuity? I haven't gotten to any value theorems yet)

4. Originally Posted by sfspitfire23
Thank you Drexel. However, I'm having trouble seeing where you got $f\wedge g=\frac{f+g-|f-g|}{2}$ and $f\vee g=\frac{f+g+|f-g|}{2}$. If you could explain to me how these came to be? (something with the definition of continuity? I haven't gotten to any value theorems yet)
No. That's the definition of min and max. Just try it out. "If $f$ is bigger than $f=\max\{f,g\}=\frac{f+g+|f-g|}{2}=\frac{f+g+f-g}{2}=\frac{2f}{2}=f$..."

5. I guess this is a hole in my Analysis. How would you derive f+g+|f-g|/2 from max{f,g}? Thank you for your patience...I'm trying to pick up on it

6. Originally Posted by sfspitfire23
I guess this is a hole in my Analysis. How would you derive f+g+|f-g|/2 from max{f,g}? Thank you for your patience...I'm trying to pick up on it
I just told you.

$\max\{f,g\}=\begin{cases}f(x) &\mbox{if} \quad f(x)\geqslant g(x) \\ g(x) & \mbox{if} \quad g(x)\geqslant f(x)\end{cases}$.

If $f(x)\geqslant g(x)$ then $f(x)-g(x)\geqslant 0$ and so [tex]|f(x)-g(x)|=f(x)-g(x)[/math\ so $\frac{f+g+|f-g|}{2}=\frac{f+g+f-g}{2}=\frac{2f}{2}=f$. If $g(x)\geqslant f(x)$ then $|f-g|=|g-f|=g-f$ and so $\frac{f+g+|f-g|}{2}=\frac{f+g+g-f}{2}=\frac{2g}{2}=g$