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    • Thread Starter
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    Q. f(x)=1/x+1 , x>3

    find the range of f

    i dont know how to do this please help someone!!
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    \frac {1}{x+1}

    or

    \frac {1}{x} + 1

    ??
    • PS Helper
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    PS Helper
    domain is x>3...therefore the range is f(x)<1/4
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    The smaller x is the bigger f is. x>3, so by putting 3 into the equation you can find f<?

    x can be any number greater than 3. So 1/x can be small, almost 0. So you know f>1

    Oh... thought f(x)=(1/x)+1

    Anyway 0<f(x)<1/4
    • Thread Starter
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    (Original post by kosy91)
    domain is x>3...therefore the range is f(x)<1/4
    how did you work it out??
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    (Original post by Pheylan)
    \frac {1}{x+1}

    or

    \frac {1}{x} + 1

    ??
    the first one
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    (Original post by klgyal)
    how did you work it out??
    do you just stick in x=3 to get 1/4
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    (Original post by klgyal)
    do you just stick in x=3 to get 1/4
    In this case yes
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    the range is the range of outputs that f(x) can be. In this case the bigger x gets, the smaller F(x) gets. You know the smallest it can be is 3 ( x>3) hence the range is f(x) > 1/4
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    Wouldnt it be 0<f(x)<1/4 ? as f(x) will never reach 0 but will always be less than or equal to 1/4, As x gets large f(x) gets small.
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    What if x = 999, then f(x) = 1/1000.
    The greatest value f(x) can take is 1/4, as x>3.
    x can be infinitely large, so x+3 tends to infinity, therefore f(x) tends to 0.
    So 0 < f(x) =< 1/4, in my opinion.

    Edit: Answered in above post.
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    ah i only learnt today, should've been quicker in spotting that lower value as well. Thanks guys!
    • PS Helper
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    PS Helper
    you draw the graph of f(x) where the asymptote is x=-1.

    you can now easily get range when domain is x>3
 
 
 
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