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Math Help - determining the largest set D of real numbers

  1. #1
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    determining the largest set D of real numbers

    determine the largest set D of real numbers for which f : D => R is a function:

     h(x) = \sqrt{\frac{1}{x-6}+1}

    any ideas on how to do this practice question? cheers.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by jvignacio View Post
    determine the largest set D of real numbers for which f : D => R is a function:

     h(x) = \sqrt{\frac{1}{x-6}+1}
    any ideas on how to do this practice question? cheers.
    X should be in the set D (called Domain)
    Thus all x which give you a real y are in this domain as co-domain is R
    Solve this inequality

     \frac{1}{x-6}+1 \ge 0......so that we don't have negative inside root

    And obviously x \ne 6

    ...since 0 in denominator is not allowed
    Thus every x which satisfies the above two conditions is your answer
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  3. #3
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    Quote Originally Posted by ADARSH View Post
    X should be in the set D (called Domain)
    Thus all x which give you a real y are in this domain as co-domain is R
    Solve this inequality

     \frac{1}{x-6}+1 \ge 0......so that we don't have negative inside root

    And obviously x \ne 6

    ...since 0 in denominator is not allowed
    Thus every x which satisfies the above two conditions is your answer
    thanks man! appreciated
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  4. #4
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    is this true for

    \frac{1}{x-6}+1 \ge 0 =  x \ge 5

    ??
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  5. #5
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by jvignacio View Post
    is this true for

    \frac{1}{x-6}+1 \ge 0 =  x \ge 5

    ??
    DO you think its correct for 5.5

    Do you think its incorrect for 4


    Answer is (-infinity , 5) And (6, infinity)
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  6. #6
    Like a stone-audioslave ADARSH's Avatar
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    I think you are still in doubt , isn't it , I saw you peeking at this again and again

    \frac{1}{x-6} +1 \ge 0

    \frac{x-5}{x-6} \ge 0

    ---Now if x> 6 the denominator on LHS is positive

    Hence x\ge 5 but in the initial condition we have taken it greater than 6

    Hence this is correct for x>6

    ---Now when 5< x<6 the denominator on LHS is negative

    So there will be a change in inequality sign when we will multiply it by
    (x-6) on both sides

    Hence x-5 < 0

    <br />
\implies x < 5 but this is against initial condtion hence no answer in this range of x


    ----When x \le 5 the denominator on LHS is negative
    So on multiplying the sign of inequality changes

    Hence x-5 \le 0

    thus x \le 5 is an answer

    Now for the complete answer we have to take a combination of all these solution sets(answer as in my last post)

    ------------------------------------

    I am not sure if you were unclear about it, but I think if you were this will be helpful ,ask incase if you still find any trouble (remember the change of inequality sign on multiplying with a negative number on both sides)

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  7. #7
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    Quote Originally Posted by ADARSH View Post
    I think you are still in doubt , isn't it , I saw you peeking at this again and again

    \frac{1}{x-6} +1 \ge 0

    \frac{x-5}{x-6} \ge 0

    ---Now if x> 6 the denominator on LHS is positive

    Hence x\ge 5 but in the initial condition we have taken it greater than 6

    Hence this is correct for x>6

    ---Now when 5< x<6 the denominator on LHS is negative

    So there will be a change in inequality sign when we will multiply it by
    (x-6) on both sides

    Hence x-5 < 0

    <br />
\implies x < 5 but this is against initial condtion hence no answer in this range of x


    ----When x \le 5 the denominator on LHS is negative
    So on multiplying the sign of inequality changes

    Hence x-5 \le 0

    thus x \le 5 is an answer

    Now for the complete answer we have to take a combination of all these solution sets(answer as in my last post)

    ------------------------------------

    I am not sure if you were unclear about it, but I think if you were this will be helpful ,ask incase if you still find any trouble (remember the change of inequality sign on multiplying with a negative number on both sides)

    haha thanks for that, yeah ive been trying to study it and really try understand it. cheers for the last post. made things more simple
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