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?










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?










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?










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






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











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023653%2ffind-values-of-a-if-f-is-a-one-one-function%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023653%2ffind-values-of-a-if-f-is-a-one-one-function%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Bundesstraße 106

            Verónica Boquete

            Ida-Boy-Ed-Garten