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Thread: What is the domain of this function??

  1. #1
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    What is the domain of this function??

    The domain of the function f(x)=





    1-(2,+infinity)

    2-all reals

    3-(-infinity,-2)U(2,+infinity)

    4-(-2,2)

    5-(-infinity,-2)U(-2,+infinity)



    ============================


    I try to solve this question :

    4+x2>0

    x2>-4
    x>√-4 The square root is not defined for negative numbers.
    Last edited by Diligent_Learner; Dec 18th 2008 at 12:14 PM.
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  2. #2
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    Is it possible for $\displaystyle 4+x^2<4$?
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  3. #3
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    Quote Originally Posted by Plato View Post
    Is it possible for $\displaystyle 4+x^2<4$?

    I don't understand your question..Can you clarify your idea??
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  4. #4
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    Are the any values of $\displaystyle x$ which gives $\displaystyle 4+x^2 \le 0$?
    If not, then the domain is all real numbers.
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  5. #5
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    Quote Originally Posted by Plato View Post
    Are the any values of $\displaystyle x$ which gives $\displaystyle 4+x^2 \le 0$?
    If not, then the domain is all real numbers.
    we don't have any value of x$\displaystyle 4+x^2 \le 0$

    because the square root for negative number is not defined
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  6. #6
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    Is this true, for all $\displaystyle x$ , $\displaystyle x^2 \ge 0$?
    If that is true add 4 to both sides, $\displaystyle 4+x^2 \ge 4$.
    Therefore, $\displaystyle \sqrt{4+x^2}$ is defined for all $\displaystyle x$.
    Therefore, the domain of the function is all $\displaystyle x$.
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  7. #7
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    Quote Originally Posted by Plato View Post
    Is this true, for all $\displaystyle x$ , $\displaystyle x^2 \ge 0$?
    If that is true add 4 to both sides, $\displaystyle 4+x^2 \ge 4$.
    Therefore, $\displaystyle \sqrt{4+x^2}$ is defined for all $\displaystyle x$.
    Therefore, the domain of the function is all $\displaystyle x$.


    Thank you very much I understand your idea..


    If I want to obtain the range of the following function

    What I must do??

    are there any steps to find range of any function ?and also the domain
    f(x)=$\displaystyle \sqrt{1-x}+x^2$
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  8. #8
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    Quote Originally Posted by Diligent_Learner View Post
    Thank you very much I understand your idea..


    If I want to obtain the range of the following function

    What I must do??

    are there any steps to find range of any function ?and also the domain
    f(x)=$\displaystyle \sqrt{1-x}+x^2$
    To find the domain,

    $\displaystyle 1-x\ge 0$

    $\displaystyle -x \ge -1$

    $\displaystyle \boxed{x \le 1}$

    To find the range,

    $\displaystyle f(1)=\sqrt{1-1}+1^2=1$

    $\displaystyle f(0)=\sqrt{1-0}+0^2=1$

    $\displaystyle f(-1)=\sqrt{1-^-1}+(-1)^2=\sqrt{2}+1$

    Therefore, $\displaystyle \boxed{f(x) \ge 1}$
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  9. #9
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    Quote Originally Posted by masters View Post
    To find the domain,

    $\displaystyle 1-x\ge 0$

    $\displaystyle -x \ge -1$

    $\displaystyle \boxed{x \le 1}$

    To find the range,

    $\displaystyle f(1)=\sqrt{1-1}+1^2=1$

    $\displaystyle f(0)=\sqrt{1-0}+0^2=1$

    $\displaystyle f(-1)=\sqrt{1-^-1}+(-1)^2=\sqrt{2}+1$

    Therefore, $\displaystyle \boxed{f(x) \ge 1}$
    Thank you very much ..


    I understand from your solution that:

    If i want to find the range :

    I should substitute the values of x that included into the domain into function

    and then I notice the the y values that the function will take..
    Last edited by Diligent_Learner; Dec 19th 2008 at 01:42 PM.
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