# continuity

• Apr 9th 2010, 12:28 PM
sfspitfire23
continuity
If $\displaystyle f:I\rightarrow R$ and $\displaystyle g: \rightarrow R$ are any functions defined on $\displaystyle I$, then we can form new functions $\displaystyle max\{f,g\}(x)=max\{f(x),g(x)\}$ for all $\displaystyle x\in I$, and similarly for $\displaystyle min\{f,g\}$

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

What do the max/min thing mean? Basically, what is the first paragraph telling me? Thanks for any help.
• Apr 9th 2010, 12:33 PM
Drexel28
Quote:

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

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

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

If $\displaystyle f,g:X\to\mathbb{R}$ then, $\displaystyle f\wedge g=\frac{f+g-|f-g|}{2}$ and $\displaystyle f\vee g=\frac{f+g+|f-g|}{2}$

P.S. common notation $\displaystyle f\vee g=\min\{f,g\}$ and simlilarly for the other. What it means is "take the bigger value of $\displaystyle f$ or $\displaystyle g$ at x"
• Apr 9th 2010, 03:36 PM
sfspitfire23
Quote:

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

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

Thank you Drexel. However, I'm having trouble seeing where you got $\displaystyle f\wedge g=\frac{f+g-|f-g|}{2}$ and $\displaystyle f\vee g=\frac{f+g+|f-g|}{2}$. 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)
• Apr 9th 2010, 03:46 PM
Drexel28
Quote:

Originally Posted by sfspitfire23
Thank you Drexel. However, I'm having trouble seeing where you got $\displaystyle f\wedge g=\frac{f+g-|f-g|}{2}$ and $\displaystyle f\vee g=\frac{f+g+|f-g|}{2}$. 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 $\displaystyle f$ is bigger than $\displaystyle f=\max\{f,g\}=\frac{f+g+|f-g|}{2}=\frac{f+g+f-g}{2}=\frac{2f}{2}=f$..."
• Apr 9th 2010, 04:19 PM
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
• Apr 9th 2010, 04:23 PM
Drexel28
Quote:

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.

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

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