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Math Help - Find f(-1), f(1), f(3)

  1. #1
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    Find f(-1), f(1), f(3)

    Here is another piecewise defined function.

    Find f(-1), f(1), f(3)


    .........{x + 3.....-2<or=tox<1
    f(x)=..{3...........x = 1
    .........{-x + 3....x > 1


    I need someone to explain this question piece by piece.
    I really never really understood this type of function.

    Thanks
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  2. #2
    Lord of certain Rings
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    To understand a piece wise function is simple. Whenever you are given a value of x, to know f(x), see which of the given range x belongs to and then substitute for x in that equation...

    Here say I want f(1), then observe that the definition says f(x) = 3 when x = 1, which means f(1) = 3.

    To know f(-1), observe that definition says whenever x is between -2 and 1, f(x) is x+3. So f(-1) = -1+3 = 2

    Now can you try f(3)?
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  3. #3
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    but....

    Quote Originally Posted by Isomorphism View Post
    To understand a piece wise function is simple. Whenever you are given a value of x, to know f(x), see which of the given range x belongs to and then substitute for x in that equation...

    Here say I want f(1), then observe that the definition says f(x) = 3 when x = 1, which means f(1) = 3.

    To know f(-1), observe that definition says whenever x is between -2 and 1, f(x) is x+3. So f(-1) = -1+3 = 2

    Now can you try f(3)?
    I know that the only number that pertain f(x) = 3, has to be x = 1. So, when x = 1, f(x) = 3 or y = 3.

    However, you said:

    "To know f(-1), observe that definition says whenever x is between -2 and 1, f(x) is x+3. So f(-1) = -1+3 = 2"

    The retriction given is that x lies between -2 and 1. If I replace x with -1 for the function f(x) = x + 3, then how can x be lies between -2 and 1?

    Look:

    -1 + 3 = 2

    Is 2 less than 1?

    Here is the restriction:

    -2 <or= to x < 1

    If I place 2 for x in the restriction, then the inequality does not make sense.

    I thought that we can replace x with -1 for the function
    f(x) = -x + 3 where the restriction is x > 1.

    f(-1) = -(-1) + 3

    f(-1) = 1 + 3

    f(-1) = 4

    I can see that 4 > 1 is true.

    See my point?

    Please, explain your reply more in detail.


    Thanks
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  4. #4
    Lord of certain Rings
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    Quote Originally Posted by magentarita View Post
    Look:

    -1 + 3 = 2

    Is 2 less than 1?

    Here is the restriction:

    -2 <or= to x < 1

    If I place 2 for x in the restriction, then the inequality does not make sense.
    The question says f(x) = x+3 when x is between -2 and 1.

    Your computation -1 + 3 is f(-1), right? The definition does not say f(-1) is between -2 and 1.

    Understand?
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  5. #5
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    got it now....

    Quote Originally Posted by Isomorphism View Post
    The question says f(x) = x+3 when x is between -2 and 1.

    Your computation -1 + 3 is f(-1), right? The definition does not say f(-1) is between -2 and 1.

    Understand?
    I see what you mean.

    We are applying what lies between the parentheses for f(x) to the proper restriction.

    In that case, it makes sense that f(-1) be evaluated for x + 3 because
    -1 lies between -2 and 1.

    I see.

    Thanks
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