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Math Help - continuity

  1. #1
    Senior Member sfspitfire23's Avatar
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    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.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    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"
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  3. #3
    Senior Member sfspitfire23's Avatar
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    Quote Originally Posted by Drexel28 View Post
    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)
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    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..."
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  5. #5
    Senior Member sfspitfire23's Avatar
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    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
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    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
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