Combining two sets of domains - how?












1












$begingroup$


I'm doing the exercises related to finding monotonicity/extreme values.



I have given function: $f(x) = -3x + ln x$



Domain of $f$: $D_{f} = (0, infty)$



Derivative: $f'(x) = -3 + frac{1}{x}$



Domain of the derivative (teacher requires this): $D_{f'} = (-infty,0)cup(0,infty)$



Now, how can I combine both domains? I need to write the mutual part of both domains. How to do it in a good math-fashioned style without "syntax mistakes"?



Something like that should do the job?



$begin{cases}
D_{f} = (0, infty) \
D_{f'} = (-infty,0)cup(0,infty) \
end{cases}$



$Rightarrow D_{f} cap D_{f'} = (0, infty)$



Is this correct?



I am asking this because if the solution(s) of $f'(x) = 0$ don't belong to the domain of $D_{f'}$, then I do not take them into account.



Thanks.






real-analysis derivatives







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  • $begingroup$
    Yes, what you wrote looks good to me.
    $endgroup$
    – zipirovich
    Jan 2 at 5:28










  • $begingroup$
    The teacher is wrong with her naive definition of function. The domain of -3 + 1/x is not the domain of f'. They are two different functions.
    $endgroup$
    – William Elliot
    Jan 2 at 8:42












  • $begingroup$
    Why finding domain of derivative is wrong? If I get solution that belongs to domain of function, but does not belong to domain of derivative, it will be invalid solution. Right?
    $endgroup$
    – weno
    Jan 2 at 9:24
















1












$begingroup$


I'm doing the exercises related to finding monotonicity/extreme values.



I have given function: $f(x) = -3x + ln x$



Domain of $f$: $D_{f} = (0, infty)$



Derivative: $f'(x) = -3 + frac{1}{x}$



Domain of the derivative (teacher requires this): $D_{f'} = (-infty,0)cup(0,infty)$



Now, how can I combine both domains? I need to write the mutual part of both domains. How to do it in a good math-fashioned style without "syntax mistakes"?



Something like that should do the job?



$begin{cases}
D_{f} = (0, infty) \
D_{f'} = (-infty,0)cup(0,infty) \
end{cases}$



$Rightarrow D_{f} cap D_{f'} = (0, infty)$



Is this correct?



I am asking this because if the solution(s) of $f'(x) = 0$ don't belong to the domain of $D_{f'}$, then I do not take them into account.



Thanks.






real-analysis derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, what you wrote looks good to me.
    $endgroup$
    – zipirovich
    Jan 2 at 5:28










  • $begingroup$
    The teacher is wrong with her naive definition of function. The domain of -3 + 1/x is not the domain of f'. They are two different functions.
    $endgroup$
    – William Elliot
    Jan 2 at 8:42












  • $begingroup$
    Why finding domain of derivative is wrong? If I get solution that belongs to domain of function, but does not belong to domain of derivative, it will be invalid solution. Right?
    $endgroup$
    – weno
    Jan 2 at 9:24














1












1








1


0



$begingroup$


I'm doing the exercises related to finding monotonicity/extreme values.



I have given function: $f(x) = -3x + ln x$



Domain of $f$: $D_{f} = (0, infty)$



Derivative: $f'(x) = -3 + frac{1}{x}$



Domain of the derivative (teacher requires this): $D_{f'} = (-infty,0)cup(0,infty)$



Now, how can I combine both domains? I need to write the mutual part of both domains. How to do it in a good math-fashioned style without "syntax mistakes"?



Something like that should do the job?



$begin{cases}
D_{f} = (0, infty) \
D_{f'} = (-infty,0)cup(0,infty) \
end{cases}$



$Rightarrow D_{f} cap D_{f'} = (0, infty)$



Is this correct?



I am asking this because if the solution(s) of $f'(x) = 0$ don't belong to the domain of $D_{f'}$, then I do not take them into account.



Thanks.






real-analysis derivatives







share|cite|improve this question











$endgroup$




I'm doing the exercises related to finding monotonicity/extreme values.



I have given function: $f(x) = -3x + ln x$



Domain of $f$: $D_{f} = (0, infty)$



Derivative: $f'(x) = -3 + frac{1}{x}$



Domain of the derivative (teacher requires this): $D_{f'} = (-infty,0)cup(0,infty)$



Now, how can I combine both domains? I need to write the mutual part of both domains. How to do it in a good math-fashioned style without "syntax mistakes"?



Something like that should do the job?



$begin{cases}
D_{f} = (0, infty) \
D_{f'} = (-infty,0)cup(0,infty) \
end{cases}$



$Rightarrow D_{f} cap D_{f'} = (0, infty)$



Is this correct?



I am asking this because if the solution(s) of $f'(x) = 0$ don't belong to the domain of $D_{f'}$, then I do not take them into account.



Thanks.






real-analysis derivatives




real-analysis derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 3:59









twnly

1,2011214




1,2011214










asked Jan 2 at 3:05









wenoweno

35111




35111












  • $begingroup$
    Yes, what you wrote looks good to me.
    $endgroup$
    – zipirovich
    Jan 2 at 5:28










  • $begingroup$
    The teacher is wrong with her naive definition of function. The domain of -3 + 1/x is not the domain of f'. They are two different functions.
    $endgroup$
    – William Elliot
    Jan 2 at 8:42












  • $begingroup$
    Why finding domain of derivative is wrong? If I get solution that belongs to domain of function, but does not belong to domain of derivative, it will be invalid solution. Right?
    $endgroup$
    – weno
    Jan 2 at 9:24


















  • $begingroup$
    Yes, what you wrote looks good to me.
    $endgroup$
    – zipirovich
    Jan 2 at 5:28










  • $begingroup$
    The teacher is wrong with her naive definition of function. The domain of -3 + 1/x is not the domain of f'. They are two different functions.
    $endgroup$
    – William Elliot
    Jan 2 at 8:42












  • $begingroup$
    Why finding domain of derivative is wrong? If I get solution that belongs to domain of function, but does not belong to domain of derivative, it will be invalid solution. Right?
    $endgroup$
    – weno
    Jan 2 at 9:24
















$begingroup$
Yes, what you wrote looks good to me.
$endgroup$
– zipirovich
Jan 2 at 5:28




$begingroup$
Yes, what you wrote looks good to me.
$endgroup$
– zipirovich
Jan 2 at 5:28












$begingroup$
The teacher is wrong with her naive definition of function. The domain of -3 + 1/x is not the domain of f'. They are two different functions.
$endgroup$
– William Elliot
Jan 2 at 8:42






$begingroup$
The teacher is wrong with her naive definition of function. The domain of -3 + 1/x is not the domain of f'. They are two different functions.
$endgroup$
– William Elliot
Jan 2 at 8:42














$begingroup$
Why finding domain of derivative is wrong? If I get solution that belongs to domain of function, but does not belong to domain of derivative, it will be invalid solution. Right?
$endgroup$
– weno
Jan 2 at 9:24




$begingroup$
Why finding domain of derivative is wrong? If I get solution that belongs to domain of function, but does not belong to domain of derivative, it will be invalid solution. Right?
$endgroup$
– weno
Jan 2 at 9:24










2 Answers
2






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I used to use the example of f(x) = ln(x), x > 0, as one in which the derivative, f '(x)=1/x SEEMS to have a "larger" domain, namely the set of all nonzero reals. What has to be emphasized is the so-called "natural domain convention" (domain is implied by formula) does NOT apply for the derivative.



In fact, the domain of f ' is ALWAYS a subset of the domain of f, from the very definition of f '(x) being the limit of the difference quotient, in which appears f(x) itself. So, to say f '(x) makes sense requires that the f(x) that is part of its definition must make sense first.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    f:(0,oo) -> R, x -> -3x + ln x.

    f':(0,oo) -> R, x -> -1 + 1/x.



    g:R{0} -> R, x -> -1 + 1/x.

    g and f' are different because they are defined on different domains.

    The domain of f' cannot have points outside the domain of f

    because f'(x) cannot be calculated when f(x) is not defined.

    However, as you were considering, g restriced to (0,oo) = f'.



    The teacher is wrong.

    As f'(-1) is not defined, -1 cannot be in domain f'.

    g is not f'.






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      1












      $begingroup$

      I used to use the example of f(x) = ln(x), x > 0, as one in which the derivative, f '(x)=1/x SEEMS to have a "larger" domain, namely the set of all nonzero reals. What has to be emphasized is the so-called "natural domain convention" (domain is implied by formula) does NOT apply for the derivative.



      In fact, the domain of f ' is ALWAYS a subset of the domain of f, from the very definition of f '(x) being the limit of the difference quotient, in which appears f(x) itself. So, to say f '(x) makes sense requires that the f(x) that is part of its definition must make sense first.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        I used to use the example of f(x) = ln(x), x > 0, as one in which the derivative, f '(x)=1/x SEEMS to have a "larger" domain, namely the set of all nonzero reals. What has to be emphasized is the so-called "natural domain convention" (domain is implied by formula) does NOT apply for the derivative.



        In fact, the domain of f ' is ALWAYS a subset of the domain of f, from the very definition of f '(x) being the limit of the difference quotient, in which appears f(x) itself. So, to say f '(x) makes sense requires that the f(x) that is part of its definition must make sense first.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          I used to use the example of f(x) = ln(x), x > 0, as one in which the derivative, f '(x)=1/x SEEMS to have a "larger" domain, namely the set of all nonzero reals. What has to be emphasized is the so-called "natural domain convention" (domain is implied by formula) does NOT apply for the derivative.



          In fact, the domain of f ' is ALWAYS a subset of the domain of f, from the very definition of f '(x) being the limit of the difference quotient, in which appears f(x) itself. So, to say f '(x) makes sense requires that the f(x) that is part of its definition must make sense first.






          share|cite|improve this answer









          $endgroup$



          I used to use the example of f(x) = ln(x), x > 0, as one in which the derivative, f '(x)=1/x SEEMS to have a "larger" domain, namely the set of all nonzero reals. What has to be emphasized is the so-called "natural domain convention" (domain is implied by formula) does NOT apply for the derivative.



          In fact, the domain of f ' is ALWAYS a subset of the domain of f, from the very definition of f '(x) being the limit of the difference quotient, in which appears f(x) itself. So, to say f '(x) makes sense requires that the f(x) that is part of its definition must make sense first.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 0:32









          Dr Mike EckerDr Mike Ecker

          511




          511























              0












              $begingroup$

              f:(0,oo) -> R, x -> -3x + ln x.

              f':(0,oo) -> R, x -> -1 + 1/x.



              g:R{0} -> R, x -> -1 + 1/x.

              g and f' are different because they are defined on different domains.

              The domain of f' cannot have points outside the domain of f

              because f'(x) cannot be calculated when f(x) is not defined.

              However, as you were considering, g restriced to (0,oo) = f'.



              The teacher is wrong.

              As f'(-1) is not defined, -1 cannot be in domain f'.

              g is not f'.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                f:(0,oo) -> R, x -> -3x + ln x.

                f':(0,oo) -> R, x -> -1 + 1/x.



                g:R{0} -> R, x -> -1 + 1/x.

                g and f' are different because they are defined on different domains.

                The domain of f' cannot have points outside the domain of f

                because f'(x) cannot be calculated when f(x) is not defined.

                However, as you were considering, g restriced to (0,oo) = f'.



                The teacher is wrong.

                As f'(-1) is not defined, -1 cannot be in domain f'.

                g is not f'.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  f:(0,oo) -> R, x -> -3x + ln x.

                  f':(0,oo) -> R, x -> -1 + 1/x.



                  g:R{0} -> R, x -> -1 + 1/x.

                  g and f' are different because they are defined on different domains.

                  The domain of f' cannot have points outside the domain of f

                  because f'(x) cannot be calculated when f(x) is not defined.

                  However, as you were considering, g restriced to (0,oo) = f'.



                  The teacher is wrong.

                  As f'(-1) is not defined, -1 cannot be in domain f'.

                  g is not f'.






                  share|cite|improve this answer









                  $endgroup$



                  f:(0,oo) -> R, x -> -3x + ln x.

                  f':(0,oo) -> R, x -> -1 + 1/x.



                  g:R{0} -> R, x -> -1 + 1/x.

                  g and f' are different because they are defined on different domains.

                  The domain of f' cannot have points outside the domain of f

                  because f'(x) cannot be calculated when f(x) is not defined.

                  However, as you were considering, g restriced to (0,oo) = f'.



                  The teacher is wrong.

                  As f'(-1) is not defined, -1 cannot be in domain f'.

                  g is not f'.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 2 at 11:16









                  William ElliotWilliam Elliot

                  8,8882820




                  8,8882820






























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