Find values of $a$ if $f$ is a one-one function












2












$begingroup$



If $$f(x) = x^3 + (a+2)x^2 + 5ax + 5$$ is a one-one function then find the set of values of $a$.




I know that I need to find values of $a$ for which function is strictly monotonic increasing or strictly monotonic decreasing and we check monotonicity of function by its derivative $f'(x)$.



here $$f'(x) = 3x^2 + 2x(a+2) + 5a$$



so, what condition should I impose on $f'(x)$ to make the given funtion one one or strictly monotonic?






calculus







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You need only pick $a $ so that $f'(x)$ does not have two zeros. Use its discriminant.
    $endgroup$
    – coffeemath
    Dec 3 '18 at 5:04










  • $begingroup$
    Note: It's OK if $f'$ has only one zero. Then $f$ will still be strictly increasing or strictly decreaing, even though it is flat at a single point. [By strictly increasing is meant if $a<b$ then $f(a)<f(b).$ Similar for strictly decreasing.]
    $endgroup$
    – coffeemath
    Dec 3 '18 at 5:53










  • $begingroup$
    @coffeemath this is what i thought. Now, f is parabola, then it can be one one only if x takes positive real values. So, is it okay to take domain (0,infinity)?
    $endgroup$
    – Mathsaddict
    Dec 3 '18 at 6:07






  • 1




    $begingroup$
    No, look at formula for $f$ -- it's cubic so $f$ will not be a parabola. Its derivative will be. And if that derivative doesn't cross the axis twice, then the original $f$ will be 1-1 for all values of $x$ [i.e.domain all reals]
    $endgroup$
    – coffeemath
    Dec 3 '18 at 6:20


















2












$begingroup$



If $$f(x) = x^3 + (a+2)x^2 + 5ax + 5$$ is a one-one function then find the set of values of $a$.




I know that I need to find values of $a$ for which function is strictly monotonic increasing or strictly monotonic decreasing and we check monotonicity of function by its derivative $f'(x)$.



here $$f'(x) = 3x^2 + 2x(a+2) + 5a$$



so, what condition should I impose on $f'(x)$ to make the given funtion one one or strictly monotonic?






calculus







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You need only pick $a $ so that $f'(x)$ does not have two zeros. Use its discriminant.
    $endgroup$
    – coffeemath
    Dec 3 '18 at 5:04










  • $begingroup$
    Note: It's OK if $f'$ has only one zero. Then $f$ will still be strictly increasing or strictly decreaing, even though it is flat at a single point. [By strictly increasing is meant if $a<b$ then $f(a)<f(b).$ Similar for strictly decreasing.]
    $endgroup$
    – coffeemath
    Dec 3 '18 at 5:53










  • $begingroup$
    @coffeemath this is what i thought. Now, f is parabola, then it can be one one only if x takes positive real values. So, is it okay to take domain (0,infinity)?
    $endgroup$
    – Mathsaddict
    Dec 3 '18 at 6:07






  • 1




    $begingroup$
    No, look at formula for $f$ -- it's cubic so $f$ will not be a parabola. Its derivative will be. And if that derivative doesn't cross the axis twice, then the original $f$ will be 1-1 for all values of $x$ [i.e.domain all reals]
    $endgroup$
    – coffeemath
    Dec 3 '18 at 6:20
















2












2








2


0



$begingroup$



If $$f(x) = x^3 + (a+2)x^2 + 5ax + 5$$ is a one-one function then find the set of values of $a$.




I know that I need to find values of $a$ for which function is strictly monotonic increasing or strictly monotonic decreasing and we check monotonicity of function by its derivative $f'(x)$.



here $$f'(x) = 3x^2 + 2x(a+2) + 5a$$



so, what condition should I impose on $f'(x)$ to make the given funtion one one or strictly monotonic?






calculus







share|cite|improve this question











$endgroup$





If $$f(x) = x^3 + (a+2)x^2 + 5ax + 5$$ is a one-one function then find the set of values of $a$.




I know that I need to find values of $a$ for which function is strictly monotonic increasing or strictly monotonic decreasing and we check monotonicity of function by its derivative $f'(x)$.



here $$f'(x) = 3x^2 + 2x(a+2) + 5a$$



so, what condition should I impose on $f'(x)$ to make the given funtion one one or strictly monotonic?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 5:05









Tianlalu

3,08621038




3,08621038










asked Dec 3 '18 at 4:56









MathsaddictMathsaddict

3049




3049








  • 3




    $begingroup$
    You need only pick $a $ so that $f'(x)$ does not have two zeros. Use its discriminant.
    $endgroup$
    – coffeemath
    Dec 3 '18 at 5:04










  • $begingroup$
    Note: It's OK if $f'$ has only one zero. Then $f$ will still be strictly increasing or strictly decreaing, even though it is flat at a single point. [By strictly increasing is meant if $a<b$ then $f(a)<f(b).$ Similar for strictly decreasing.]
    $endgroup$
    – coffeemath
    Dec 3 '18 at 5:53










  • $begingroup$
    @coffeemath this is what i thought. Now, f is parabola, then it can be one one only if x takes positive real values. So, is it okay to take domain (0,infinity)?
    $endgroup$
    – Mathsaddict
    Dec 3 '18 at 6:07






  • 1




    $begingroup$
    No, look at formula for $f$ -- it's cubic so $f$ will not be a parabola. Its derivative will be. And if that derivative doesn't cross the axis twice, then the original $f$ will be 1-1 for all values of $x$ [i.e.domain all reals]
    $endgroup$
    – coffeemath
    Dec 3 '18 at 6:20
















  • 3




    $begingroup$
    You need only pick $a $ so that $f'(x)$ does not have two zeros. Use its discriminant.
    $endgroup$
    – coffeemath
    Dec 3 '18 at 5:04










  • $begingroup$
    Note: It's OK if $f'$ has only one zero. Then $f$ will still be strictly increasing or strictly decreaing, even though it is flat at a single point. [By strictly increasing is meant if $a<b$ then $f(a)<f(b).$ Similar for strictly decreasing.]
    $endgroup$
    – coffeemath
    Dec 3 '18 at 5:53










  • $begingroup$
    @coffeemath this is what i thought. Now, f is parabola, then it can be one one only if x takes positive real values. So, is it okay to take domain (0,infinity)?
    $endgroup$
    – Mathsaddict
    Dec 3 '18 at 6:07






  • 1




    $begingroup$
    No, look at formula for $f$ -- it's cubic so $f$ will not be a parabola. Its derivative will be. And if that derivative doesn't cross the axis twice, then the original $f$ will be 1-1 for all values of $x$ [i.e.domain all reals]
    $endgroup$
    – coffeemath
    Dec 3 '18 at 6:20










3




3




$begingroup$
You need only pick $a $ so that $f'(x)$ does not have two zeros. Use its discriminant.
$endgroup$
– coffeemath
Dec 3 '18 at 5:04




$begingroup$
You need only pick $a $ so that $f'(x)$ does not have two zeros. Use its discriminant.
$endgroup$
– coffeemath
Dec 3 '18 at 5:04












$begingroup$
Note: It's OK if $f'$ has only one zero. Then $f$ will still be strictly increasing or strictly decreaing, even though it is flat at a single point. [By strictly increasing is meant if $a<b$ then $f(a)<f(b).$ Similar for strictly decreasing.]
$endgroup$
– coffeemath
Dec 3 '18 at 5:53




$begingroup$
Note: It's OK if $f'$ has only one zero. Then $f$ will still be strictly increasing or strictly decreaing, even though it is flat at a single point. [By strictly increasing is meant if $a<b$ then $f(a)<f(b).$ Similar for strictly decreasing.]
$endgroup$
– coffeemath
Dec 3 '18 at 5:53












$begingroup$
@coffeemath this is what i thought. Now, f is parabola, then it can be one one only if x takes positive real values. So, is it okay to take domain (0,infinity)?
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:07




$begingroup$
@coffeemath this is what i thought. Now, f is parabola, then it can be one one only if x takes positive real values. So, is it okay to take domain (0,infinity)?
$endgroup$
– Mathsaddict
Dec 3 '18 at 6:07




1




1




$begingroup$
No, look at formula for $f$ -- it's cubic so $f$ will not be a parabola. Its derivative will be. And if that derivative doesn't cross the axis twice, then the original $f$ will be 1-1 for all values of $x$ [i.e.domain all reals]
$endgroup$
– coffeemath
Dec 3 '18 at 6:20






$begingroup$
No, look at formula for $f$ -- it's cubic so $f$ will not be a parabola. Its derivative will be. And if that derivative doesn't cross the axis twice, then the original $f$ will be 1-1 for all values of $x$ [i.e.domain all reals]
$endgroup$
– coffeemath
Dec 3 '18 at 6:20












4 Answers
4






active

oldest

votes


















6












$begingroup$

HINT. You know that $f'(x)$ is a parabola. To make $f(x)$ always increasing, you want $f'(x)>0$; that is, you want the parabola to always be above the $x$-axis. It touches the $x$-axis when it only has one root. What formula can you use to see what makes this only have one root in terms of $a$? Would this then give you a condition on $a$ to force the parabola to be always above the $x$-axis?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
    $endgroup$
    – Mathsaddict
    Dec 3 '18 at 6:00












  • $begingroup$
    @Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
    $endgroup$
    – Vim
    Dec 3 '18 at 6:15



















1












$begingroup$

A differentiable function is strictly monotonic when its derivative is not $0$ on any segment, meaning if $f'(x)=0$ then there is a small $z>0$ with the property that for any $0<y<z :f'(x+y)neq 0,f'(x-y)neq 0$, and if $forall x:f'(x)geq 0$ or $forall x:f'(x)leq 0$.



Try and use that. You are on the right track.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Consider $y=ax^2+bx+c,$ where $a>0.$ Now $$y=a(x^2+frac{b}{a}x+frac{c}{a})=abig((x+frac{b}{2a})^2+frac{c}{a}-frac{b^2}{4a^2}big)=abig((x+frac{b}{2a})^2+frac{4ac-b^2}{4a^2}big).$$ So if you have $4ac-b^2geq0$ then $y$ will be non-negative for all $x.$



    Try to connect these ideas to your problem.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
      $endgroup$
      – Mathsaddict
      Dec 3 '18 at 5:58










    • $begingroup$
      @Mathsaddict I agree with your comment here. [see my second comment on your question]t
      $endgroup$
      – coffeemath
      Dec 3 '18 at 6:03










    • $begingroup$
      @coffeemath I agree with yall. Edited.
      $endgroup$
      – Janitha357
      Dec 3 '18 at 15:48



















    1












    $begingroup$

    Note that $lim_limits{xto -infty} f(x)=-infty$ and $lim_limits{xto +infty} f(x)=+infty$. That is why you need to make sure the function $f(x)$ is strictly increasing $left(x_1<x_2 Rightarrow f(x_1)<f(x_2)right)$. It implies:
    $$f'(x)ge 0 Rightarrow 3x^2 + 2x(a+2) + 5age 0 Rightarrow Dle 0 Rightarrow (a+2)^2-15ale0 Rightarrow \
    frac12 left(11 - sqrt{105}right)le alefrac12 left(11 + sqrt{105}right).$$
    See WA animation for $f'(x)$.



    Also note that $f'(x)=0$ when $a=frac12 left(11 - sqrt{105}right) left(text{or} a=frac12 left(11 + sqrt{105}right) right)$ and $x = frac{sqrt{35/3}}{2} - frac52 left(text{or} x = -frac52 - frac{sqrt{35/3}}{2}right)$, which implies it is an inflection point (which is still one-to-one).

    See WA animation for $f(x)$.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Why D cannot be equal to 0?
      $endgroup$
      – Mathsaddict
      Dec 3 '18 at 6:50










    • $begingroup$
      Thank you for asking critical question. Please, see the updated answer.
      $endgroup$
      – farruhota
      Dec 3 '18 at 15:55











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    HINT. You know that $f'(x)$ is a parabola. To make $f(x)$ always increasing, you want $f'(x)>0$; that is, you want the parabola to always be above the $x$-axis. It touches the $x$-axis when it only has one root. What formula can you use to see what makes this only have one root in terms of $a$? Would this then give you a condition on $a$ to force the parabola to be always above the $x$-axis?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
      $endgroup$
      – Mathsaddict
      Dec 3 '18 at 6:00












    • $begingroup$
      @Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
      $endgroup$
      – Vim
      Dec 3 '18 at 6:15
















    6












    $begingroup$

    HINT. You know that $f'(x)$ is a parabola. To make $f(x)$ always increasing, you want $f'(x)>0$; that is, you want the parabola to always be above the $x$-axis. It touches the $x$-axis when it only has one root. What formula can you use to see what makes this only have one root in terms of $a$? Would this then give you a condition on $a$ to force the parabola to be always above the $x$-axis?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
      $endgroup$
      – Mathsaddict
      Dec 3 '18 at 6:00












    • $begingroup$
      @Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
      $endgroup$
      – Vim
      Dec 3 '18 at 6:15














    6












    6








    6





    $begingroup$

    HINT. You know that $f'(x)$ is a parabola. To make $f(x)$ always increasing, you want $f'(x)>0$; that is, you want the parabola to always be above the $x$-axis. It touches the $x$-axis when it only has one root. What formula can you use to see what makes this only have one root in terms of $a$? Would this then give you a condition on $a$ to force the parabola to be always above the $x$-axis?






    share|cite|improve this answer









    $endgroup$



    HINT. You know that $f'(x)$ is a parabola. To make $f(x)$ always increasing, you want $f'(x)>0$; that is, you want the parabola to always be above the $x$-axis. It touches the $x$-axis when it only has one root. What formula can you use to see what makes this only have one root in terms of $a$? Would this then give you a condition on $a$ to force the parabola to be always above the $x$-axis?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 3 '18 at 5:05









    mathematics2x2lifemathematics2x2life

    8,06121738




    8,06121738












    • $begingroup$
      Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
      $endgroup$
      – Mathsaddict
      Dec 3 '18 at 6:00












    • $begingroup$
      @Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
      $endgroup$
      – Vim
      Dec 3 '18 at 6:15


















    • $begingroup$
      Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
      $endgroup$
      – Mathsaddict
      Dec 3 '18 at 6:00












    • $begingroup$
      @Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
      $endgroup$
      – Vim
      Dec 3 '18 at 6:15
















    $begingroup$
    Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
    $endgroup$
    – Mathsaddict
    Dec 3 '18 at 6:00






    $begingroup$
    Thanks for this simple explanation. Since, f is a parabola, so is there also any condition on x? I think domain of f should be positive real numbers, since f is one one.
    $endgroup$
    – Mathsaddict
    Dec 3 '18 at 6:00














    $begingroup$
    @Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
    $endgroup$
    – Vim
    Dec 3 '18 at 6:15




    $begingroup$
    @Mathsaddict there's no restriction on f taking any real number. The way I interpreted f being "one-to-one" is f is a bijection from R to itself.
    $endgroup$
    – Vim
    Dec 3 '18 at 6:15











    1












    $begingroup$

    A differentiable function is strictly monotonic when its derivative is not $0$ on any segment, meaning if $f'(x)=0$ then there is a small $z>0$ with the property that for any $0<y<z :f'(x+y)neq 0,f'(x-y)neq 0$, and if $forall x:f'(x)geq 0$ or $forall x:f'(x)leq 0$.



    Try and use that. You are on the right track.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      A differentiable function is strictly monotonic when its derivative is not $0$ on any segment, meaning if $f'(x)=0$ then there is a small $z>0$ with the property that for any $0<y<z :f'(x+y)neq 0,f'(x-y)neq 0$, and if $forall x:f'(x)geq 0$ or $forall x:f'(x)leq 0$.



      Try and use that. You are on the right track.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        A differentiable function is strictly monotonic when its derivative is not $0$ on any segment, meaning if $f'(x)=0$ then there is a small $z>0$ with the property that for any $0<y<z :f'(x+y)neq 0,f'(x-y)neq 0$, and if $forall x:f'(x)geq 0$ or $forall x:f'(x)leq 0$.



        Try and use that. You are on the right track.






        share|cite|improve this answer









        $endgroup$



        A differentiable function is strictly monotonic when its derivative is not $0$ on any segment, meaning if $f'(x)=0$ then there is a small $z>0$ with the property that for any $0<y<z :f'(x+y)neq 0,f'(x-y)neq 0$, and if $forall x:f'(x)geq 0$ or $forall x:f'(x)leq 0$.



        Try and use that. You are on the right track.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 5:05









        NL1992NL1992

        7311




        7311























            1












            $begingroup$

            Consider $y=ax^2+bx+c,$ where $a>0.$ Now $$y=a(x^2+frac{b}{a}x+frac{c}{a})=abig((x+frac{b}{2a})^2+frac{c}{a}-frac{b^2}{4a^2}big)=abig((x+frac{b}{2a})^2+frac{4ac-b^2}{4a^2}big).$$ So if you have $4ac-b^2geq0$ then $y$ will be non-negative for all $x.$



            Try to connect these ideas to your problem.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
              $endgroup$
              – Mathsaddict
              Dec 3 '18 at 5:58










            • $begingroup$
              @Mathsaddict I agree with your comment here. [see my second comment on your question]t
              $endgroup$
              – coffeemath
              Dec 3 '18 at 6:03










            • $begingroup$
              @coffeemath I agree with yall. Edited.
              $endgroup$
              – Janitha357
              Dec 3 '18 at 15:48
















            1












            $begingroup$

            Consider $y=ax^2+bx+c,$ where $a>0.$ Now $$y=a(x^2+frac{b}{a}x+frac{c}{a})=abig((x+frac{b}{2a})^2+frac{c}{a}-frac{b^2}{4a^2}big)=abig((x+frac{b}{2a})^2+frac{4ac-b^2}{4a^2}big).$$ So if you have $4ac-b^2geq0$ then $y$ will be non-negative for all $x.$



            Try to connect these ideas to your problem.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
              $endgroup$
              – Mathsaddict
              Dec 3 '18 at 5:58










            • $begingroup$
              @Mathsaddict I agree with your comment here. [see my second comment on your question]t
              $endgroup$
              – coffeemath
              Dec 3 '18 at 6:03










            • $begingroup$
              @coffeemath I agree with yall. Edited.
              $endgroup$
              – Janitha357
              Dec 3 '18 at 15:48














            1












            1








            1





            $begingroup$

            Consider $y=ax^2+bx+c,$ where $a>0.$ Now $$y=a(x^2+frac{b}{a}x+frac{c}{a})=abig((x+frac{b}{2a})^2+frac{c}{a}-frac{b^2}{4a^2}big)=abig((x+frac{b}{2a})^2+frac{4ac-b^2}{4a^2}big).$$ So if you have $4ac-b^2geq0$ then $y$ will be non-negative for all $x.$



            Try to connect these ideas to your problem.






            share|cite|improve this answer











            $endgroup$



            Consider $y=ax^2+bx+c,$ where $a>0.$ Now $$y=a(x^2+frac{b}{a}x+frac{c}{a})=abig((x+frac{b}{2a})^2+frac{c}{a}-frac{b^2}{4a^2}big)=abig((x+frac{b}{2a})^2+frac{4ac-b^2}{4a^2}big).$$ So if you have $4ac-b^2geq0$ then $y$ will be non-negative for all $x.$



            Try to connect these ideas to your problem.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 3 '18 at 15:48

























            answered Dec 3 '18 at 5:23









            Janitha357Janitha357

            1,756516




            1,756516












            • $begingroup$
              if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
              $endgroup$
              – Mathsaddict
              Dec 3 '18 at 5:58










            • $begingroup$
              @Mathsaddict I agree with your comment here. [see my second comment on your question]t
              $endgroup$
              – coffeemath
              Dec 3 '18 at 6:03










            • $begingroup$
              @coffeemath I agree with yall. Edited.
              $endgroup$
              – Janitha357
              Dec 3 '18 at 15:48


















            • $begingroup$
              if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
              $endgroup$
              – Mathsaddict
              Dec 3 '18 at 5:58










            • $begingroup$
              @Mathsaddict I agree with your comment here. [see my second comment on your question]t
              $endgroup$
              – coffeemath
              Dec 3 '18 at 6:03










            • $begingroup$
              @coffeemath I agree with yall. Edited.
              $endgroup$
              – Janitha357
              Dec 3 '18 at 15:48
















            $begingroup$
            if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
            $endgroup$
            – Mathsaddict
            Dec 3 '18 at 5:58




            $begingroup$
            if 4ac - b^2 = 0, then y will be zero at -b/2a. There should be no problem if f' has only one root. Why do i need 4ac - b^2 to be greater then 0? I think it can also be 0 at one point.
            $endgroup$
            – Mathsaddict
            Dec 3 '18 at 5:58












            $begingroup$
            @Mathsaddict I agree with your comment here. [see my second comment on your question]t
            $endgroup$
            – coffeemath
            Dec 3 '18 at 6:03




            $begingroup$
            @Mathsaddict I agree with your comment here. [see my second comment on your question]t
            $endgroup$
            – coffeemath
            Dec 3 '18 at 6:03












            $begingroup$
            @coffeemath I agree with yall. Edited.
            $endgroup$
            – Janitha357
            Dec 3 '18 at 15:48




            $begingroup$
            @coffeemath I agree with yall. Edited.
            $endgroup$
            – Janitha357
            Dec 3 '18 at 15:48











            1












            $begingroup$

            Note that $lim_limits{xto -infty} f(x)=-infty$ and $lim_limits{xto +infty} f(x)=+infty$. That is why you need to make sure the function $f(x)$ is strictly increasing $left(x_1<x_2 Rightarrow f(x_1)<f(x_2)right)$. It implies:
            $$f'(x)ge 0 Rightarrow 3x^2 + 2x(a+2) + 5age 0 Rightarrow Dle 0 Rightarrow (a+2)^2-15ale0 Rightarrow \
            frac12 left(11 - sqrt{105}right)le alefrac12 left(11 + sqrt{105}right).$$
            See WA animation for $f'(x)$.



            Also note that $f'(x)=0$ when $a=frac12 left(11 - sqrt{105}right) left(text{or} a=frac12 left(11 + sqrt{105}right) right)$ and $x = frac{sqrt{35/3}}{2} - frac52 left(text{or} x = -frac52 - frac{sqrt{35/3}}{2}right)$, which implies it is an inflection point (which is still one-to-one).

            See WA animation for $f(x)$.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Why D cannot be equal to 0?
              $endgroup$
              – Mathsaddict
              Dec 3 '18 at 6:50










            • $begingroup$
              Thank you for asking critical question. Please, see the updated answer.
              $endgroup$
              – farruhota
              Dec 3 '18 at 15:55
















            1












            $begingroup$

            Note that $lim_limits{xto -infty} f(x)=-infty$ and $lim_limits{xto +infty} f(x)=+infty$. That is why you need to make sure the function $f(x)$ is strictly increasing $left(x_1<x_2 Rightarrow f(x_1)<f(x_2)right)$. It implies:
            $$f'(x)ge 0 Rightarrow 3x^2 + 2x(a+2) + 5age 0 Rightarrow Dle 0 Rightarrow (a+2)^2-15ale0 Rightarrow \
            frac12 left(11 - sqrt{105}right)le alefrac12 left(11 + sqrt{105}right).$$
            See WA animation for $f'(x)$.



            Also note that $f'(x)=0$ when $a=frac12 left(11 - sqrt{105}right) left(text{or} a=frac12 left(11 + sqrt{105}right) right)$ and $x = frac{sqrt{35/3}}{2} - frac52 left(text{or} x = -frac52 - frac{sqrt{35/3}}{2}right)$, which implies it is an inflection point (which is still one-to-one).

            See WA animation for $f(x)$.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Why D cannot be equal to 0?
              $endgroup$
              – Mathsaddict
              Dec 3 '18 at 6:50










            • $begingroup$
              Thank you for asking critical question. Please, see the updated answer.
              $endgroup$
              – farruhota
              Dec 3 '18 at 15:55














            1












            1








            1





            $begingroup$

            Note that $lim_limits{xto -infty} f(x)=-infty$ and $lim_limits{xto +infty} f(x)=+infty$. That is why you need to make sure the function $f(x)$ is strictly increasing $left(x_1<x_2 Rightarrow f(x_1)<f(x_2)right)$. It implies:
            $$f'(x)ge 0 Rightarrow 3x^2 + 2x(a+2) + 5age 0 Rightarrow Dle 0 Rightarrow (a+2)^2-15ale0 Rightarrow \
            frac12 left(11 - sqrt{105}right)le alefrac12 left(11 + sqrt{105}right).$$
            See WA animation for $f'(x)$.



            Also note that $f'(x)=0$ when $a=frac12 left(11 - sqrt{105}right) left(text{or} a=frac12 left(11 + sqrt{105}right) right)$ and $x = frac{sqrt{35/3}}{2} - frac52 left(text{or} x = -frac52 - frac{sqrt{35/3}}{2}right)$, which implies it is an inflection point (which is still one-to-one).

            See WA animation for $f(x)$.






            share|cite|improve this answer











            $endgroup$



            Note that $lim_limits{xto -infty} f(x)=-infty$ and $lim_limits{xto +infty} f(x)=+infty$. That is why you need to make sure the function $f(x)$ is strictly increasing $left(x_1<x_2 Rightarrow f(x_1)<f(x_2)right)$. It implies:
            $$f'(x)ge 0 Rightarrow 3x^2 + 2x(a+2) + 5age 0 Rightarrow Dle 0 Rightarrow (a+2)^2-15ale0 Rightarrow \
            frac12 left(11 - sqrt{105}right)le alefrac12 left(11 + sqrt{105}right).$$
            See WA animation for $f'(x)$.



            Also note that $f'(x)=0$ when $a=frac12 left(11 - sqrt{105}right) left(text{or} a=frac12 left(11 + sqrt{105}right) right)$ and $x = frac{sqrt{35/3}}{2} - frac52 left(text{or} x = -frac52 - frac{sqrt{35/3}}{2}right)$, which implies it is an inflection point (which is still one-to-one).

            See WA animation for $f(x)$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 3 '18 at 15:54

























            answered Dec 3 '18 at 6:33









            farruhotafarruhota

            19.8k2738




            19.8k2738








            • 1




              $begingroup$
              Why D cannot be equal to 0?
              $endgroup$
              – Mathsaddict
              Dec 3 '18 at 6:50










            • $begingroup$
              Thank you for asking critical question. Please, see the updated answer.
              $endgroup$
              – farruhota
              Dec 3 '18 at 15:55














            • 1




              $begingroup$
              Why D cannot be equal to 0?
              $endgroup$
              – Mathsaddict
              Dec 3 '18 at 6:50










            • $begingroup$
              Thank you for asking critical question. Please, see the updated answer.
              $endgroup$
              – farruhota
              Dec 3 '18 at 15:55








            1




            1




            $begingroup$
            Why D cannot be equal to 0?
            $endgroup$
            – Mathsaddict
            Dec 3 '18 at 6:50




            $begingroup$
            Why D cannot be equal to 0?
            $endgroup$
            – Mathsaddict
            Dec 3 '18 at 6:50












            $begingroup$
            Thank you for asking critical question. Please, see the updated answer.
            $endgroup$
            – farruhota
            Dec 3 '18 at 15:55




            $begingroup$
            Thank you for asking critical question. Please, see the updated answer.
            $endgroup$
            – farruhota
            Dec 3 '18 at 15:55


















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