Continuity on Interval












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$begingroup$


I'm not understanding why $g(x)=x^2-3x+2$ is continuous over the interval (-4,4), and why $f(x)=frac{1}{x}+3$ is NOT continuous over the interval (-7,7).



Any help would be appreciated. Thank you!






calculus continuity







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  • 1




    $begingroup$
    Is $f(0)$ even defined?
    $endgroup$
    – Dietrich Burde
    Jan 27 at 20:21










  • $begingroup$
    No, $f(0)$ isn't defined.
    $endgroup$
    – Sailor Moon
    Jan 27 at 20:25










  • $begingroup$
    You should try to plot the two function, you'll understand the issue
    $endgroup$
    – Thomas Lesgourgues
    Jan 27 at 20:28
















0












$begingroup$


I'm not understanding why $g(x)=x^2-3x+2$ is continuous over the interval (-4,4), and why $f(x)=frac{1}{x}+3$ is NOT continuous over the interval (-7,7).



Any help would be appreciated. Thank you!






calculus continuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Is $f(0)$ even defined?
    $endgroup$
    – Dietrich Burde
    Jan 27 at 20:21










  • $begingroup$
    No, $f(0)$ isn't defined.
    $endgroup$
    – Sailor Moon
    Jan 27 at 20:25










  • $begingroup$
    You should try to plot the two function, you'll understand the issue
    $endgroup$
    – Thomas Lesgourgues
    Jan 27 at 20:28














0












0








0





$begingroup$


I'm not understanding why $g(x)=x^2-3x+2$ is continuous over the interval (-4,4), and why $f(x)=frac{1}{x}+3$ is NOT continuous over the interval (-7,7).



Any help would be appreciated. Thank you!






calculus continuity







share|cite|improve this question











$endgroup$




I'm not understanding why $g(x)=x^2-3x+2$ is continuous over the interval (-4,4), and why $f(x)=frac{1}{x}+3$ is NOT continuous over the interval (-7,7).



Any help would be appreciated. Thank you!






calculus continuity




calculus continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 27 at 21:27









Michael Rybkin

3,969421




3,969421










asked Jan 27 at 20:20









Sailor MoonSailor Moon

61




61








  • 1




    $begingroup$
    Is $f(0)$ even defined?
    $endgroup$
    – Dietrich Burde
    Jan 27 at 20:21










  • $begingroup$
    No, $f(0)$ isn't defined.
    $endgroup$
    – Sailor Moon
    Jan 27 at 20:25










  • $begingroup$
    You should try to plot the two function, you'll understand the issue
    $endgroup$
    – Thomas Lesgourgues
    Jan 27 at 20:28














  • 1




    $begingroup$
    Is $f(0)$ even defined?
    $endgroup$
    – Dietrich Burde
    Jan 27 at 20:21










  • $begingroup$
    No, $f(0)$ isn't defined.
    $endgroup$
    – Sailor Moon
    Jan 27 at 20:25










  • $begingroup$
    You should try to plot the two function, you'll understand the issue
    $endgroup$
    – Thomas Lesgourgues
    Jan 27 at 20:28








1




1




$begingroup$
Is $f(0)$ even defined?
$endgroup$
– Dietrich Burde
Jan 27 at 20:21




$begingroup$
Is $f(0)$ even defined?
$endgroup$
– Dietrich Burde
Jan 27 at 20:21












$begingroup$
No, $f(0)$ isn't defined.
$endgroup$
– Sailor Moon
Jan 27 at 20:25




$begingroup$
No, $f(0)$ isn't defined.
$endgroup$
– Sailor Moon
Jan 27 at 20:25












$begingroup$
You should try to plot the two function, you'll understand the issue
$endgroup$
– Thomas Lesgourgues
Jan 27 at 20:28




$begingroup$
You should try to plot the two function, you'll understand the issue
$endgroup$
– Thomas Lesgourgues
Jan 27 at 20:28










2 Answers
2






active

oldest

votes


















1












$begingroup$

$$ f(x) text{ continuous at $0$ } iff lim_{xto 0^-}f(x)=lim_{xto 0^+}f(x)=f(0)$$



But $f(0)$ doesn't exist, so this doesn't hold.



What's more, if it isn't continuous at $0$, it isn't continuous across a range which contains $0$, like $(-7,7)$ does.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $g(x)=x^2-3x+2$ is continuous everywhere and by extension over the interval $(-4, 4)$. It's continuous everywhere because it's a polynomial function. Polynomial functions are contentious over the entire real line.



    $f(x)=frac{1}{x}+3$ is not continuous over the interval $(-7,7)$ because $x=0$ (it lies right in the middle of the interval) is a point where this function is simply not defined. When $x=0$, you are dividing by zero which is not a legitimate operation in mathematics. Therefore, the function $f(x)=frac{1}{x}+3$ can't have a value at that point. In other words, the value of the function $f(x)=frac{1}{x}+3$ at the point $x=0$ does not exit. Graphically, this function looks discontinuous at that point (do you see how the graph of the function is not one single continuous line?):



    graph of the function






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you so much for explaining it to me so clearly! And thank you for graphing it, even though you didn't have to! I can see that x=0 doesn't exist on the function in the graph.
      $endgroup$
      – Sailor Moon
      Jan 27 at 20:58











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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    $$ f(x) text{ continuous at $0$ } iff lim_{xto 0^-}f(x)=lim_{xto 0^+}f(x)=f(0)$$



    But $f(0)$ doesn't exist, so this doesn't hold.



    What's more, if it isn't continuous at $0$, it isn't continuous across a range which contains $0$, like $(-7,7)$ does.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $$ f(x) text{ continuous at $0$ } iff lim_{xto 0^-}f(x)=lim_{xto 0^+}f(x)=f(0)$$



      But $f(0)$ doesn't exist, so this doesn't hold.



      What's more, if it isn't continuous at $0$, it isn't continuous across a range which contains $0$, like $(-7,7)$ does.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $$ f(x) text{ continuous at $0$ } iff lim_{xto 0^-}f(x)=lim_{xto 0^+}f(x)=f(0)$$



        But $f(0)$ doesn't exist, so this doesn't hold.



        What's more, if it isn't continuous at $0$, it isn't continuous across a range which contains $0$, like $(-7,7)$ does.






        share|cite|improve this answer









        $endgroup$



        $$ f(x) text{ continuous at $0$ } iff lim_{xto 0^-}f(x)=lim_{xto 0^+}f(x)=f(0)$$



        But $f(0)$ doesn't exist, so this doesn't hold.



        What's more, if it isn't continuous at $0$, it isn't continuous across a range which contains $0$, like $(-7,7)$ does.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 27 at 20:28









        Rhys HughesRhys Hughes

        7,0801630




        7,0801630























            0












            $begingroup$

            $g(x)=x^2-3x+2$ is continuous everywhere and by extension over the interval $(-4, 4)$. It's continuous everywhere because it's a polynomial function. Polynomial functions are contentious over the entire real line.



            $f(x)=frac{1}{x}+3$ is not continuous over the interval $(-7,7)$ because $x=0$ (it lies right in the middle of the interval) is a point where this function is simply not defined. When $x=0$, you are dividing by zero which is not a legitimate operation in mathematics. Therefore, the function $f(x)=frac{1}{x}+3$ can't have a value at that point. In other words, the value of the function $f(x)=frac{1}{x}+3$ at the point $x=0$ does not exit. Graphically, this function looks discontinuous at that point (do you see how the graph of the function is not one single continuous line?):



            graph of the function






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you so much for explaining it to me so clearly! And thank you for graphing it, even though you didn't have to! I can see that x=0 doesn't exist on the function in the graph.
              $endgroup$
              – Sailor Moon
              Jan 27 at 20:58
















            0












            $begingroup$

            $g(x)=x^2-3x+2$ is continuous everywhere and by extension over the interval $(-4, 4)$. It's continuous everywhere because it's a polynomial function. Polynomial functions are contentious over the entire real line.



            $f(x)=frac{1}{x}+3$ is not continuous over the interval $(-7,7)$ because $x=0$ (it lies right in the middle of the interval) is a point where this function is simply not defined. When $x=0$, you are dividing by zero which is not a legitimate operation in mathematics. Therefore, the function $f(x)=frac{1}{x}+3$ can't have a value at that point. In other words, the value of the function $f(x)=frac{1}{x}+3$ at the point $x=0$ does not exit. Graphically, this function looks discontinuous at that point (do you see how the graph of the function is not one single continuous line?):



            graph of the function






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you so much for explaining it to me so clearly! And thank you for graphing it, even though you didn't have to! I can see that x=0 doesn't exist on the function in the graph.
              $endgroup$
              – Sailor Moon
              Jan 27 at 20:58














            0












            0








            0





            $begingroup$

            $g(x)=x^2-3x+2$ is continuous everywhere and by extension over the interval $(-4, 4)$. It's continuous everywhere because it's a polynomial function. Polynomial functions are contentious over the entire real line.



            $f(x)=frac{1}{x}+3$ is not continuous over the interval $(-7,7)$ because $x=0$ (it lies right in the middle of the interval) is a point where this function is simply not defined. When $x=0$, you are dividing by zero which is not a legitimate operation in mathematics. Therefore, the function $f(x)=frac{1}{x}+3$ can't have a value at that point. In other words, the value of the function $f(x)=frac{1}{x}+3$ at the point $x=0$ does not exit. Graphically, this function looks discontinuous at that point (do you see how the graph of the function is not one single continuous line?):



            graph of the function






            share|cite|improve this answer











            $endgroup$



            $g(x)=x^2-3x+2$ is continuous everywhere and by extension over the interval $(-4, 4)$. It's continuous everywhere because it's a polynomial function. Polynomial functions are contentious over the entire real line.



            $f(x)=frac{1}{x}+3$ is not continuous over the interval $(-7,7)$ because $x=0$ (it lies right in the middle of the interval) is a point where this function is simply not defined. When $x=0$, you are dividing by zero which is not a legitimate operation in mathematics. Therefore, the function $f(x)=frac{1}{x}+3$ can't have a value at that point. In other words, the value of the function $f(x)=frac{1}{x}+3$ at the point $x=0$ does not exit. Graphically, this function looks discontinuous at that point (do you see how the graph of the function is not one single continuous line?):



            graph of the function







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 27 at 20:43

























            answered Jan 27 at 20:29









            Michael RybkinMichael Rybkin

            3,969421




            3,969421












            • $begingroup$
              Thank you so much for explaining it to me so clearly! And thank you for graphing it, even though you didn't have to! I can see that x=0 doesn't exist on the function in the graph.
              $endgroup$
              – Sailor Moon
              Jan 27 at 20:58


















            • $begingroup$
              Thank you so much for explaining it to me so clearly! And thank you for graphing it, even though you didn't have to! I can see that x=0 doesn't exist on the function in the graph.
              $endgroup$
              – Sailor Moon
              Jan 27 at 20:58
















            $begingroup$
            Thank you so much for explaining it to me so clearly! And thank you for graphing it, even though you didn't have to! I can see that x=0 doesn't exist on the function in the graph.
            $endgroup$
            – Sailor Moon
            Jan 27 at 20:58




            $begingroup$
            Thank you so much for explaining it to me so clearly! And thank you for graphing it, even though you didn't have to! I can see that x=0 doesn't exist on the function in the graph.
            $endgroup$
            – Sailor Moon
            Jan 27 at 20:58


















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