Interval for the solutions of ${x+1}<x^2-2x$ where ${x}$ is the fractional part of $x$.












1












$begingroup$



Find the interval(s) which contain solutions of $${x+1}<x^2-2x$$ where ${x}$ is the fractional part of $x$.




I was told that one way of solving this would be graphically. However I generally don't use graphical methods as I find them a tad difficult. Could somebody please show me how to solve this using a graph, as well as through a non-graphical method?



Secondly, from plotting the curves for ${x+1}$ and $x^2-2x$, wouldn't I only get points of intersection and thus values of $x$ which satisfy both equations? How then could I use the method to generate an interval for which the given inequality holds?



Many thanks!






calculus algebra-precalculus functions fractional-part







share|cite|improve this question











$endgroup$












  • $begingroup$
    You have to escape the brackets ({}).
    $endgroup$
    – Bernard
    Jun 5 '16 at 17:21










  • $begingroup$
    What exactly do you mean with the fractional part of $x$?
    $endgroup$
    – Newb
    Jun 5 '16 at 19:49










  • $begingroup$
    ${x}=x-lfloor xrfloor$
    $endgroup$
    – user342209
    Jun 5 '16 at 20:24












  • $begingroup$
    Hint: ${x+1}={x}$
    $endgroup$
    – ZFR
    Jun 9 '16 at 6:34
















1












$begingroup$



Find the interval(s) which contain solutions of $${x+1}<x^2-2x$$ where ${x}$ is the fractional part of $x$.




I was told that one way of solving this would be graphically. However I generally don't use graphical methods as I find them a tad difficult. Could somebody please show me how to solve this using a graph, as well as through a non-graphical method?



Secondly, from plotting the curves for ${x+1}$ and $x^2-2x$, wouldn't I only get points of intersection and thus values of $x$ which satisfy both equations? How then could I use the method to generate an interval for which the given inequality holds?



Many thanks!






calculus algebra-precalculus functions fractional-part







share|cite|improve this question











$endgroup$












  • $begingroup$
    You have to escape the brackets ({}).
    $endgroup$
    – Bernard
    Jun 5 '16 at 17:21










  • $begingroup$
    What exactly do you mean with the fractional part of $x$?
    $endgroup$
    – Newb
    Jun 5 '16 at 19:49










  • $begingroup$
    ${x}=x-lfloor xrfloor$
    $endgroup$
    – user342209
    Jun 5 '16 at 20:24












  • $begingroup$
    Hint: ${x+1}={x}$
    $endgroup$
    – ZFR
    Jun 9 '16 at 6:34














1












1








1





$begingroup$



Find the interval(s) which contain solutions of $${x+1}<x^2-2x$$ where ${x}$ is the fractional part of $x$.




I was told that one way of solving this would be graphically. However I generally don't use graphical methods as I find them a tad difficult. Could somebody please show me how to solve this using a graph, as well as through a non-graphical method?



Secondly, from plotting the curves for ${x+1}$ and $x^2-2x$, wouldn't I only get points of intersection and thus values of $x$ which satisfy both equations? How then could I use the method to generate an interval for which the given inequality holds?



Many thanks!






calculus algebra-precalculus functions fractional-part







share|cite|improve this question











$endgroup$





Find the interval(s) which contain solutions of $${x+1}<x^2-2x$$ where ${x}$ is the fractional part of $x$.




I was told that one way of solving this would be graphically. However I generally don't use graphical methods as I find them a tad difficult. Could somebody please show me how to solve this using a graph, as well as through a non-graphical method?



Secondly, from plotting the curves for ${x+1}$ and $x^2-2x$, wouldn't I only get points of intersection and thus values of $x$ which satisfy both equations? How then could I use the method to generate an interval for which the given inequality holds?



Many thanks!






calculus algebra-precalculus functions fractional-part




calculus algebra-precalculus functions fractional-part






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 9 '16 at 6:24









Prasanna

1289




1289










asked Jun 5 '16 at 17:16









user342209user342209

448211




448211












  • $begingroup$
    You have to escape the brackets ({}).
    $endgroup$
    – Bernard
    Jun 5 '16 at 17:21










  • $begingroup$
    What exactly do you mean with the fractional part of $x$?
    $endgroup$
    – Newb
    Jun 5 '16 at 19:49










  • $begingroup$
    ${x}=x-lfloor xrfloor$
    $endgroup$
    – user342209
    Jun 5 '16 at 20:24












  • $begingroup$
    Hint: ${x+1}={x}$
    $endgroup$
    – ZFR
    Jun 9 '16 at 6:34


















  • $begingroup$
    You have to escape the brackets ({}).
    $endgroup$
    – Bernard
    Jun 5 '16 at 17:21










  • $begingroup$
    What exactly do you mean with the fractional part of $x$?
    $endgroup$
    – Newb
    Jun 5 '16 at 19:49










  • $begingroup$
    ${x}=x-lfloor xrfloor$
    $endgroup$
    – user342209
    Jun 5 '16 at 20:24












  • $begingroup$
    Hint: ${x+1}={x}$
    $endgroup$
    – ZFR
    Jun 9 '16 at 6:34
















$begingroup$
You have to escape the brackets ({}).
$endgroup$
– Bernard
Jun 5 '16 at 17:21




$begingroup$
You have to escape the brackets ({}).
$endgroup$
– Bernard
Jun 5 '16 at 17:21












$begingroup$
What exactly do you mean with the fractional part of $x$?
$endgroup$
– Newb
Jun 5 '16 at 19:49




$begingroup$
What exactly do you mean with the fractional part of $x$?
$endgroup$
– Newb
Jun 5 '16 at 19:49












$begingroup$
${x}=x-lfloor xrfloor$
$endgroup$
– user342209
Jun 5 '16 at 20:24






$begingroup$
${x}=x-lfloor xrfloor$
$endgroup$
– user342209
Jun 5 '16 at 20:24














$begingroup$
Hint: ${x+1}={x}$
$endgroup$
– ZFR
Jun 9 '16 at 6:34




$begingroup$
Hint: ${x+1}={x}$
$endgroup$
– ZFR
Jun 9 '16 at 6:34










1 Answer
1






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oldest

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2












$begingroup$

Here's an algebraic way of doing it.



${x+1 } = { x}$ because the fractional part function has period $1$. Also, $0 leq {x } < 1$ for all $x$. Now, $x^2 - 2x$ has roots $0$ and $2$, and so $x^2 - 2x < 0$ when $x in (0,2)$ and non-negative everywhere else. So, ${x} < x^2 - 2x$ can hold only outside the interval $[0,2]$.



Now, observe that the function $f(x)=x^2 - 2x$ is strictly increasing and concave up for all $x geq 2$. Also, $x^2 - 2x$ and ${x}$ are both zero at $x = 2$. This means that if the slope of the curve $y = x^2 - 2x$ is greater than the slope of $y = {x}$ at $x = 2$, then $x^2 - 2x > {x}$ for all $x > 2$. It turns out that this is indeed the case, because $2x-2$ evaluated at $x=2$ equals $2$, which is greater than $1$.



Now, look at the case when $x < 0$. Observe that $f(x) = x^2 - 2x$ is strictly decreasing and concave up for all $x < 0$. Also, $f(-1) = 3 > 1$, so there is only one solution to $x^2 - 2x = {x}$ when $x < 0$, and it lies in the interval $(-1,0)$. In this interval, the function $g(x) = {x}$ coincides with the function $tilde{g}(x) = x + 1$. Thus, we actually have to find the negative root of the equation $x^2 - 2x = x + 1$, i.e. $x^2 - 3x - 1 = 0$. The roots are
$$
x = frac{3 pm sqrt{13}}{2}
$$
so the negative root is the one with the minus sign. Thus, $x^2 - 2x > { x}$ for all $x$ less than this root.



Thus, there are two intervals which do the job: $(-infty,(3 - sqrt{13})/2)$ and $(2,infty)$.




In general, plotting the curves not only tells you where they intersect, but also things like where the curves are increasing, decreasing, constant, and about their shape (i.e. concavity). A graphical analysis is certainly helpful in this case.



Graph



I used Wolfram|Alpha to plot this graph. The input is Plot[x-Floor[x],x^2-2x,[x,-2,4]].






share|cite|improve this answer











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    $begingroup$

    Here's an algebraic way of doing it.



    ${x+1 } = { x}$ because the fractional part function has period $1$. Also, $0 leq {x } < 1$ for all $x$. Now, $x^2 - 2x$ has roots $0$ and $2$, and so $x^2 - 2x < 0$ when $x in (0,2)$ and non-negative everywhere else. So, ${x} < x^2 - 2x$ can hold only outside the interval $[0,2]$.



    Now, observe that the function $f(x)=x^2 - 2x$ is strictly increasing and concave up for all $x geq 2$. Also, $x^2 - 2x$ and ${x}$ are both zero at $x = 2$. This means that if the slope of the curve $y = x^2 - 2x$ is greater than the slope of $y = {x}$ at $x = 2$, then $x^2 - 2x > {x}$ for all $x > 2$. It turns out that this is indeed the case, because $2x-2$ evaluated at $x=2$ equals $2$, which is greater than $1$.



    Now, look at the case when $x < 0$. Observe that $f(x) = x^2 - 2x$ is strictly decreasing and concave up for all $x < 0$. Also, $f(-1) = 3 > 1$, so there is only one solution to $x^2 - 2x = {x}$ when $x < 0$, and it lies in the interval $(-1,0)$. In this interval, the function $g(x) = {x}$ coincides with the function $tilde{g}(x) = x + 1$. Thus, we actually have to find the negative root of the equation $x^2 - 2x = x + 1$, i.e. $x^2 - 3x - 1 = 0$. The roots are
    $$
    x = frac{3 pm sqrt{13}}{2}
    $$
    so the negative root is the one with the minus sign. Thus, $x^2 - 2x > { x}$ for all $x$ less than this root.



    Thus, there are two intervals which do the job: $(-infty,(3 - sqrt{13})/2)$ and $(2,infty)$.




    In general, plotting the curves not only tells you where they intersect, but also things like where the curves are increasing, decreasing, constant, and about their shape (i.e. concavity). A graphical analysis is certainly helpful in this case.



    Graph



    I used Wolfram|Alpha to plot this graph. The input is Plot[x-Floor[x],x^2-2x,[x,-2,4]].






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Here's an algebraic way of doing it.



      ${x+1 } = { x}$ because the fractional part function has period $1$. Also, $0 leq {x } < 1$ for all $x$. Now, $x^2 - 2x$ has roots $0$ and $2$, and so $x^2 - 2x < 0$ when $x in (0,2)$ and non-negative everywhere else. So, ${x} < x^2 - 2x$ can hold only outside the interval $[0,2]$.



      Now, observe that the function $f(x)=x^2 - 2x$ is strictly increasing and concave up for all $x geq 2$. Also, $x^2 - 2x$ and ${x}$ are both zero at $x = 2$. This means that if the slope of the curve $y = x^2 - 2x$ is greater than the slope of $y = {x}$ at $x = 2$, then $x^2 - 2x > {x}$ for all $x > 2$. It turns out that this is indeed the case, because $2x-2$ evaluated at $x=2$ equals $2$, which is greater than $1$.



      Now, look at the case when $x < 0$. Observe that $f(x) = x^2 - 2x$ is strictly decreasing and concave up for all $x < 0$. Also, $f(-1) = 3 > 1$, so there is only one solution to $x^2 - 2x = {x}$ when $x < 0$, and it lies in the interval $(-1,0)$. In this interval, the function $g(x) = {x}$ coincides with the function $tilde{g}(x) = x + 1$. Thus, we actually have to find the negative root of the equation $x^2 - 2x = x + 1$, i.e. $x^2 - 3x - 1 = 0$. The roots are
      $$
      x = frac{3 pm sqrt{13}}{2}
      $$
      so the negative root is the one with the minus sign. Thus, $x^2 - 2x > { x}$ for all $x$ less than this root.



      Thus, there are two intervals which do the job: $(-infty,(3 - sqrt{13})/2)$ and $(2,infty)$.




      In general, plotting the curves not only tells you where they intersect, but also things like where the curves are increasing, decreasing, constant, and about their shape (i.e. concavity). A graphical analysis is certainly helpful in this case.



      Graph



      I used Wolfram|Alpha to plot this graph. The input is Plot[x-Floor[x],x^2-2x,[x,-2,4]].






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Here's an algebraic way of doing it.



        ${x+1 } = { x}$ because the fractional part function has period $1$. Also, $0 leq {x } < 1$ for all $x$. Now, $x^2 - 2x$ has roots $0$ and $2$, and so $x^2 - 2x < 0$ when $x in (0,2)$ and non-negative everywhere else. So, ${x} < x^2 - 2x$ can hold only outside the interval $[0,2]$.



        Now, observe that the function $f(x)=x^2 - 2x$ is strictly increasing and concave up for all $x geq 2$. Also, $x^2 - 2x$ and ${x}$ are both zero at $x = 2$. This means that if the slope of the curve $y = x^2 - 2x$ is greater than the slope of $y = {x}$ at $x = 2$, then $x^2 - 2x > {x}$ for all $x > 2$. It turns out that this is indeed the case, because $2x-2$ evaluated at $x=2$ equals $2$, which is greater than $1$.



        Now, look at the case when $x < 0$. Observe that $f(x) = x^2 - 2x$ is strictly decreasing and concave up for all $x < 0$. Also, $f(-1) = 3 > 1$, so there is only one solution to $x^2 - 2x = {x}$ when $x < 0$, and it lies in the interval $(-1,0)$. In this interval, the function $g(x) = {x}$ coincides with the function $tilde{g}(x) = x + 1$. Thus, we actually have to find the negative root of the equation $x^2 - 2x = x + 1$, i.e. $x^2 - 3x - 1 = 0$. The roots are
        $$
        x = frac{3 pm sqrt{13}}{2}
        $$
        so the negative root is the one with the minus sign. Thus, $x^2 - 2x > { x}$ for all $x$ less than this root.



        Thus, there are two intervals which do the job: $(-infty,(3 - sqrt{13})/2)$ and $(2,infty)$.




        In general, plotting the curves not only tells you where they intersect, but also things like where the curves are increasing, decreasing, constant, and about their shape (i.e. concavity). A graphical analysis is certainly helpful in this case.



        Graph



        I used Wolfram|Alpha to plot this graph. The input is Plot[x-Floor[x],x^2-2x,[x,-2,4]].






        share|cite|improve this answer











        $endgroup$



        Here's an algebraic way of doing it.



        ${x+1 } = { x}$ because the fractional part function has period $1$. Also, $0 leq {x } < 1$ for all $x$. Now, $x^2 - 2x$ has roots $0$ and $2$, and so $x^2 - 2x < 0$ when $x in (0,2)$ and non-negative everywhere else. So, ${x} < x^2 - 2x$ can hold only outside the interval $[0,2]$.



        Now, observe that the function $f(x)=x^2 - 2x$ is strictly increasing and concave up for all $x geq 2$. Also, $x^2 - 2x$ and ${x}$ are both zero at $x = 2$. This means that if the slope of the curve $y = x^2 - 2x$ is greater than the slope of $y = {x}$ at $x = 2$, then $x^2 - 2x > {x}$ for all $x > 2$. It turns out that this is indeed the case, because $2x-2$ evaluated at $x=2$ equals $2$, which is greater than $1$.



        Now, look at the case when $x < 0$. Observe that $f(x) = x^2 - 2x$ is strictly decreasing and concave up for all $x < 0$. Also, $f(-1) = 3 > 1$, so there is only one solution to $x^2 - 2x = {x}$ when $x < 0$, and it lies in the interval $(-1,0)$. In this interval, the function $g(x) = {x}$ coincides with the function $tilde{g}(x) = x + 1$. Thus, we actually have to find the negative root of the equation $x^2 - 2x = x + 1$, i.e. $x^2 - 3x - 1 = 0$. The roots are
        $$
        x = frac{3 pm sqrt{13}}{2}
        $$
        so the negative root is the one with the minus sign. Thus, $x^2 - 2x > { x}$ for all $x$ less than this root.



        Thus, there are two intervals which do the job: $(-infty,(3 - sqrt{13})/2)$ and $(2,infty)$.




        In general, plotting the curves not only tells you where they intersect, but also things like where the curves are increasing, decreasing, constant, and about their shape (i.e. concavity). A graphical analysis is certainly helpful in this case.



        Graph



        I used Wolfram|Alpha to plot this graph. The input is Plot[x-Floor[x],x^2-2x,[x,-2,4]].







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 21 '18 at 15:45

























        answered Jun 9 '16 at 6:46









        BrahadeeshBrahadeesh

        6,24242361




        6,24242361






























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