Show the following function is increasing












0















If $x<y$ is it true that



$x^{frac{1}{2n+1}}<y^{frac{1}{2n+1}}$?




I considered the following function $f(x)=x^{frac{1}{2n+1}}$



I computed $f^{'}(x)={frac{1}{2n+1}}x^{frac{-2n}{2n+1}}$



Now $x^{frac{-2n}{2n+1}}=(frac{1}{x^2})^{frac{n}{2n+1}}$



Note that $frac{1}{x^2} $ is always positive.



But how to show that $f^{'}(x)$ is positive because we are taking root?



Please help.






real-analysis calculus







share|cite|improve this question






















  • If $frac{1}{2n + 1}$ is a constant you might find it easier just to call that $t$ or something.
    – Trevor Gunn
    Dec 9 at 7:18










  • @TrevorGunn;How would that help?
    – Join_PhD
    Dec 9 at 7:19










  • Then you're less focused on the having a fraction in your exponent and you can think about the general properties of $x^t$ where $0 < t < 1$.
    – Trevor Gunn
    Dec 9 at 7:22










  • Is $x$ assumed to be positive? And what variable denotes $n$?
    – Dr. Sonnhard Graubner
    Dec 9 at 9:37










  • X is a real number and n is constant natural number
    – Join_PhD
    Dec 9 at 9:43
















0















If $x<y$ is it true that



$x^{frac{1}{2n+1}}<y^{frac{1}{2n+1}}$?




I considered the following function $f(x)=x^{frac{1}{2n+1}}$



I computed $f^{'}(x)={frac{1}{2n+1}}x^{frac{-2n}{2n+1}}$



Now $x^{frac{-2n}{2n+1}}=(frac{1}{x^2})^{frac{n}{2n+1}}$



Note that $frac{1}{x^2} $ is always positive.



But how to show that $f^{'}(x)$ is positive because we are taking root?



Please help.






real-analysis calculus







share|cite|improve this question






















  • If $frac{1}{2n + 1}$ is a constant you might find it easier just to call that $t$ or something.
    – Trevor Gunn
    Dec 9 at 7:18










  • @TrevorGunn;How would that help?
    – Join_PhD
    Dec 9 at 7:19










  • Then you're less focused on the having a fraction in your exponent and you can think about the general properties of $x^t$ where $0 < t < 1$.
    – Trevor Gunn
    Dec 9 at 7:22










  • Is $x$ assumed to be positive? And what variable denotes $n$?
    – Dr. Sonnhard Graubner
    Dec 9 at 9:37










  • X is a real number and n is constant natural number
    – Join_PhD
    Dec 9 at 9:43














0












0








0








If $x<y$ is it true that



$x^{frac{1}{2n+1}}<y^{frac{1}{2n+1}}$?




I considered the following function $f(x)=x^{frac{1}{2n+1}}$



I computed $f^{'}(x)={frac{1}{2n+1}}x^{frac{-2n}{2n+1}}$



Now $x^{frac{-2n}{2n+1}}=(frac{1}{x^2})^{frac{n}{2n+1}}$



Note that $frac{1}{x^2} $ is always positive.



But how to show that $f^{'}(x)$ is positive because we are taking root?



Please help.






real-analysis calculus







share|cite|improve this question














If $x<y$ is it true that



$x^{frac{1}{2n+1}}<y^{frac{1}{2n+1}}$?




I considered the following function $f(x)=x^{frac{1}{2n+1}}$



I computed $f^{'}(x)={frac{1}{2n+1}}x^{frac{-2n}{2n+1}}$



Now $x^{frac{-2n}{2n+1}}=(frac{1}{x^2})^{frac{n}{2n+1}}$



Note that $frac{1}{x^2} $ is always positive.



But how to show that $f^{'}(x)$ is positive because we are taking root?



Please help.






real-analysis calculus




real-analysis calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 9 at 7:15









Join_PhD

1968




1968












  • If $frac{1}{2n + 1}$ is a constant you might find it easier just to call that $t$ or something.
    – Trevor Gunn
    Dec 9 at 7:18










  • @TrevorGunn;How would that help?
    – Join_PhD
    Dec 9 at 7:19










  • Then you're less focused on the having a fraction in your exponent and you can think about the general properties of $x^t$ where $0 < t < 1$.
    – Trevor Gunn
    Dec 9 at 7:22










  • Is $x$ assumed to be positive? And what variable denotes $n$?
    – Dr. Sonnhard Graubner
    Dec 9 at 9:37










  • X is a real number and n is constant natural number
    – Join_PhD
    Dec 9 at 9:43


















  • If $frac{1}{2n + 1}$ is a constant you might find it easier just to call that $t$ or something.
    – Trevor Gunn
    Dec 9 at 7:18










  • @TrevorGunn;How would that help?
    – Join_PhD
    Dec 9 at 7:19










  • Then you're less focused on the having a fraction in your exponent and you can think about the general properties of $x^t$ where $0 < t < 1$.
    – Trevor Gunn
    Dec 9 at 7:22










  • Is $x$ assumed to be positive? And what variable denotes $n$?
    – Dr. Sonnhard Graubner
    Dec 9 at 9:37










  • X is a real number and n is constant natural number
    – Join_PhD
    Dec 9 at 9:43
















If $frac{1}{2n + 1}$ is a constant you might find it easier just to call that $t$ or something.
– Trevor Gunn
Dec 9 at 7:18




If $frac{1}{2n + 1}$ is a constant you might find it easier just to call that $t$ or something.
– Trevor Gunn
Dec 9 at 7:18












@TrevorGunn;How would that help?
– Join_PhD
Dec 9 at 7:19




@TrevorGunn;How would that help?
– Join_PhD
Dec 9 at 7:19












Then you're less focused on the having a fraction in your exponent and you can think about the general properties of $x^t$ where $0 < t < 1$.
– Trevor Gunn
Dec 9 at 7:22




Then you're less focused on the having a fraction in your exponent and you can think about the general properties of $x^t$ where $0 < t < 1$.
– Trevor Gunn
Dec 9 at 7:22












Is $x$ assumed to be positive? And what variable denotes $n$?
– Dr. Sonnhard Graubner
Dec 9 at 9:37




Is $x$ assumed to be positive? And what variable denotes $n$?
– Dr. Sonnhard Graubner
Dec 9 at 9:37












X is a real number and n is constant natural number
– Join_PhD
Dec 9 at 9:43




X is a real number and n is constant natural number
– Join_PhD
Dec 9 at 9:43










2 Answers
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0














One way of doing this is to use the elementary fact that inverse of any bijective (strictly) increasing function is (strictly) increasing. Note that $x^{2n+1}$ has positive derivative and it is bijective. Its inverse is your function $f$.



[ If $g$ is strictly incerasing, $x<y$ and $g^{-1}(x) geq g^{-1}(y)$ then $g(g^{-1}(x)) geq (g^{-1}(y))$ which is a contradiction. Hence $g^{-1}(x) < g^{-1}(y)$].






share|cite|improve this answer





























    0














    As you recognised, the claim is true if and only if $f_n(x):=x^{1/(2n+1)}$ is monotonously increasing. However this is not true. For the sake of convenience set $m=1/(2n+1)$, then
    $$frac{d}{dx}f_n(x)=mx^{m-1}$$
    which needs to be always positive for $f_n$ to be monotonously increasing. Let's restrict the domain to $(0,infty)$ to avoid complications with negative powers. Note that anywhere in this domain, $x$ raised to anything is always positive, hence $f_n'$ is positive iff $m$ is. So let's pick $m=-1$ (corresponding to $n=-1$) or any negative number to achieve $f_n'<0$ for all $x$, at which point we get $f_{-1}(x)=1/x$, a decreasing function on $(0,infty)$. Thus the problem condition is true exactly when $m>0$; or in other words, $n>0$.



    Edit: in the comments you said that $ninmathbb N$, which of course implies $n>0$. Hence, the claim is true.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
      2






      active

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      active

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      0














      One way of doing this is to use the elementary fact that inverse of any bijective (strictly) increasing function is (strictly) increasing. Note that $x^{2n+1}$ has positive derivative and it is bijective. Its inverse is your function $f$.



      [ If $g$ is strictly incerasing, $x<y$ and $g^{-1}(x) geq g^{-1}(y)$ then $g(g^{-1}(x)) geq (g^{-1}(y))$ which is a contradiction. Hence $g^{-1}(x) < g^{-1}(y)$].






      share|cite|improve this answer


























        0














        One way of doing this is to use the elementary fact that inverse of any bijective (strictly) increasing function is (strictly) increasing. Note that $x^{2n+1}$ has positive derivative and it is bijective. Its inverse is your function $f$.



        [ If $g$ is strictly incerasing, $x<y$ and $g^{-1}(x) geq g^{-1}(y)$ then $g(g^{-1}(x)) geq (g^{-1}(y))$ which is a contradiction. Hence $g^{-1}(x) < g^{-1}(y)$].






        share|cite|improve this answer
























          0












          0








          0






          One way of doing this is to use the elementary fact that inverse of any bijective (strictly) increasing function is (strictly) increasing. Note that $x^{2n+1}$ has positive derivative and it is bijective. Its inverse is your function $f$.



          [ If $g$ is strictly incerasing, $x<y$ and $g^{-1}(x) geq g^{-1}(y)$ then $g(g^{-1}(x)) geq (g^{-1}(y))$ which is a contradiction. Hence $g^{-1}(x) < g^{-1}(y)$].






          share|cite|improve this answer












          One way of doing this is to use the elementary fact that inverse of any bijective (strictly) increasing function is (strictly) increasing. Note that $x^{2n+1}$ has positive derivative and it is bijective. Its inverse is your function $f$.



          [ If $g$ is strictly incerasing, $x<y$ and $g^{-1}(x) geq g^{-1}(y)$ then $g(g^{-1}(x)) geq (g^{-1}(y))$ which is a contradiction. Hence $g^{-1}(x) < g^{-1}(y)$].







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 at 12:11









          Kavi Rama Murthy

          49.8k31854




          49.8k31854























              0














              As you recognised, the claim is true if and only if $f_n(x):=x^{1/(2n+1)}$ is monotonously increasing. However this is not true. For the sake of convenience set $m=1/(2n+1)$, then
              $$frac{d}{dx}f_n(x)=mx^{m-1}$$
              which needs to be always positive for $f_n$ to be monotonously increasing. Let's restrict the domain to $(0,infty)$ to avoid complications with negative powers. Note that anywhere in this domain, $x$ raised to anything is always positive, hence $f_n'$ is positive iff $m$ is. So let's pick $m=-1$ (corresponding to $n=-1$) or any negative number to achieve $f_n'<0$ for all $x$, at which point we get $f_{-1}(x)=1/x$, a decreasing function on $(0,infty)$. Thus the problem condition is true exactly when $m>0$; or in other words, $n>0$.



              Edit: in the comments you said that $ninmathbb N$, which of course implies $n>0$. Hence, the claim is true.






              share|cite|improve this answer


























                0














                As you recognised, the claim is true if and only if $f_n(x):=x^{1/(2n+1)}$ is monotonously increasing. However this is not true. For the sake of convenience set $m=1/(2n+1)$, then
                $$frac{d}{dx}f_n(x)=mx^{m-1}$$
                which needs to be always positive for $f_n$ to be monotonously increasing. Let's restrict the domain to $(0,infty)$ to avoid complications with negative powers. Note that anywhere in this domain, $x$ raised to anything is always positive, hence $f_n'$ is positive iff $m$ is. So let's pick $m=-1$ (corresponding to $n=-1$) or any negative number to achieve $f_n'<0$ for all $x$, at which point we get $f_{-1}(x)=1/x$, a decreasing function on $(0,infty)$. Thus the problem condition is true exactly when $m>0$; or in other words, $n>0$.



                Edit: in the comments you said that $ninmathbb N$, which of course implies $n>0$. Hence, the claim is true.






                share|cite|improve this answer
























                  0












                  0








                  0






                  As you recognised, the claim is true if and only if $f_n(x):=x^{1/(2n+1)}$ is monotonously increasing. However this is not true. For the sake of convenience set $m=1/(2n+1)$, then
                  $$frac{d}{dx}f_n(x)=mx^{m-1}$$
                  which needs to be always positive for $f_n$ to be monotonously increasing. Let's restrict the domain to $(0,infty)$ to avoid complications with negative powers. Note that anywhere in this domain, $x$ raised to anything is always positive, hence $f_n'$ is positive iff $m$ is. So let's pick $m=-1$ (corresponding to $n=-1$) or any negative number to achieve $f_n'<0$ for all $x$, at which point we get $f_{-1}(x)=1/x$, a decreasing function on $(0,infty)$. Thus the problem condition is true exactly when $m>0$; or in other words, $n>0$.



                  Edit: in the comments you said that $ninmathbb N$, which of course implies $n>0$. Hence, the claim is true.






                  share|cite|improve this answer












                  As you recognised, the claim is true if and only if $f_n(x):=x^{1/(2n+1)}$ is monotonously increasing. However this is not true. For the sake of convenience set $m=1/(2n+1)$, then
                  $$frac{d}{dx}f_n(x)=mx^{m-1}$$
                  which needs to be always positive for $f_n$ to be monotonously increasing. Let's restrict the domain to $(0,infty)$ to avoid complications with negative powers. Note that anywhere in this domain, $x$ raised to anything is always positive, hence $f_n'$ is positive iff $m$ is. So let's pick $m=-1$ (corresponding to $n=-1$) or any negative number to achieve $f_n'<0$ for all $x$, at which point we get $f_{-1}(x)=1/x$, a decreasing function on $(0,infty)$. Thus the problem condition is true exactly when $m>0$; or in other words, $n>0$.



                  Edit: in the comments you said that $ninmathbb N$, which of course implies $n>0$. Hence, the claim is true.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 at 12:18









                  YiFan

                  2,4391421




                  2,4391421






























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