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Math Help - Piecewise-defined Function

  1. #1
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    Piecewise-defined Function

    For the piecewise-defined below, find:

    f(-2), f(0) and f(2)


    .........{x^2..........if x < 0
    f(x) = {2..............if x = 0
    .........{2x + 1.......if x > 0
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by magentarita View Post
    For the piecewise-defined below, find:

    f(-2), f(0) and f(2)


    .........{x^2..........if x < 0
    f(x) = {2..............if x = 0
    .........{2x + 1.......if x > 0
    Which of these piecewise functions is defined for negative values of x (i.e. x<0)? Substitute -2 into that function to get the value of f(-2).

    There is only one function that is defined at x=0. So what do you think f(0) would be?

    Which of these piecewise functions is defined for positive values of x (i.e. x>0)? Substitute 2 into that function to get the value of f(2).

    Does this make sense?

    --Chris
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  3. #3
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    ok...

    Quote Originally Posted by Chris L T521 View Post
    Which of these piecewise functions is defined for negative values of x (i.e. x<0)? Substitute -2 into that function to get the value of f(-2).

    There is only one function that is defined at x=0. So what do you think f(0) would be?

    Which of these piecewise functions is defined for positive values of x (i.e. x>0)? Substitute 2 into that function to get the value of f(2).

    Does this make sense?

    --Chris
    It does make slight sense but can you do at least one of them for me?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by magentarita View Post
    It does make slight sense but can you do at least one of them for me?
    Sure.

    f(x)=\left\{\begin{array}{rl}x^2&x<0\\2&x=0\\2x+1&  x>0\end{array}\right.

    I'll evaluate f(-2)

    Since -2<0, you want to focus on the function that is defined for x<0. Looking at the piecewise function, we see something:

    f(x)=\left\{\begin{array}{rl}{\color{red}x^2}&{\co  lor{red}x<0}\\2&x=0\\2x+1&x>0\end{array}\right.

    We see that the function defined for x<0 is x^2. So we can now say that f(-2)=(-2)^2=\color{red}\boxed{4}

    Does this clarify things? Can you try the others on your own?

    --Chris
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  5. #5
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    ok.....

    Quote Originally Posted by Chris L T521 View Post
    Sure.

    f(x)=\left\{\begin{array}{rl}x^2&x<0\\2&x=0\\2x+1&  x>0\end{array}\right.

    I'll evaluate f(-2)

    Since -2<0, you want to focus on the function that is defined for x<0. Looking at the piecewise function, we see something:

    f(x)=\left\{\begin{array}{rl}{\color{red}x^2}&{\co  lor{red}x<0}\\2&x=0\\2x+1&x>0\end{array}\right.

    We see that the function defined for x<0 is x^2. So we can now say that f(-2)=(-2)^2=\color{red}\boxed{4}

    Does this clarify things? Can you try the others on your own?

    --Chris
    Thank you so much.
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