(Calculus) Derivative Thinking Question












4












$begingroup$


Recently, my Calculus and Vectors (Grade 12) teacher gave our class a thinking question/assignment to work on over the march break, and after working on for some time, I've become stuck on it.



The Question:



Consider f(x), a general quadratic function in standard form, and g(x) its reciprocal. For which values of x are the slopes of their respective tangent lines equal?



Consider two cases: one where it is true for exactly one value of x, and the other where it is true for exactly two values of x. In the latter case, you can assume that the steepness a does not equal 0 of f(x) is equal to both it’s y-intercept and also to the slope of its tangent at x = 1.



Find the required conditions on the parameters a,b,c in terms of a.



My Progress So Far:



So I know that $f(x) = ax^2 +bx + c$ and $g(x) = 1/f(x)$. After I solved for the derivative of each function, I set them both equal to each other and started solving for it. But then I came to a equation of $(ax^2 + bx + c)^2 = -1$ and that doesn't work.



Next I tried something else. Since I know from the 2nd case that $f(x) = c$ and $f(x) = f'(1)$ when a cannot equal 0, I set $c = f'(1)$ and got $c = 2a + b$. But after that, I don't know where to go.



I'm not expecting a full solution, but if anyone could give me a hint for solving this question, I would appreciate it. I know that it says to write everything in terms of a, but I'm not sure how to approach that method.










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  • $begingroup$
    I made a typo in solving and so my prior comment did not make sense. To get your equation $f(x)^2=-1$ you likely made a cancellation on both sides, but what happens if the term you canceled is zero?
    $endgroup$
    – Michael
    2 hours ago












  • $begingroup$
    The term that I cancelled was 2ax+b (which is f'(x)), I don't know if the term cancelled can be zero since I do not know the unknowns. (I apologize if I didn't answer your question).
    $endgroup$
    – Rasheed Amanzai
    1 hour ago






  • 1




    $begingroup$
    So you got $-f'(x)=f'(x)f(x)^2$ and want to find an $x$ where this is possible. It is not possible if $f'(x) neq 0$ as you already showed and so the only way possible is if $f'(x)=0$. So then for what values of $a,b$ is it possible to have $2ax+b=0$?
    $endgroup$
    – Michael
    1 hour ago












  • $begingroup$
    Now that made a lot more sense to me. So then the values of a and b would both have to equal 0 in order to get f'(x) = 0.
    $endgroup$
    – Rasheed Amanzai
    1 hour ago
















4












$begingroup$


Recently, my Calculus and Vectors (Grade 12) teacher gave our class a thinking question/assignment to work on over the march break, and after working on for some time, I've become stuck on it.



The Question:



Consider f(x), a general quadratic function in standard form, and g(x) its reciprocal. For which values of x are the slopes of their respective tangent lines equal?



Consider two cases: one where it is true for exactly one value of x, and the other where it is true for exactly two values of x. In the latter case, you can assume that the steepness a does not equal 0 of f(x) is equal to both it’s y-intercept and also to the slope of its tangent at x = 1.



Find the required conditions on the parameters a,b,c in terms of a.



My Progress So Far:



So I know that $f(x) = ax^2 +bx + c$ and $g(x) = 1/f(x)$. After I solved for the derivative of each function, I set them both equal to each other and started solving for it. But then I came to a equation of $(ax^2 + bx + c)^2 = -1$ and that doesn't work.



Next I tried something else. Since I know from the 2nd case that $f(x) = c$ and $f(x) = f'(1)$ when a cannot equal 0, I set $c = f'(1)$ and got $c = 2a + b$. But after that, I don't know where to go.



I'm not expecting a full solution, but if anyone could give me a hint for solving this question, I would appreciate it. I know that it says to write everything in terms of a, but I'm not sure how to approach that method.










share|cite|improve this question









New contributor




Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    I made a typo in solving and so my prior comment did not make sense. To get your equation $f(x)^2=-1$ you likely made a cancellation on both sides, but what happens if the term you canceled is zero?
    $endgroup$
    – Michael
    2 hours ago












  • $begingroup$
    The term that I cancelled was 2ax+b (which is f'(x)), I don't know if the term cancelled can be zero since I do not know the unknowns. (I apologize if I didn't answer your question).
    $endgroup$
    – Rasheed Amanzai
    1 hour ago






  • 1




    $begingroup$
    So you got $-f'(x)=f'(x)f(x)^2$ and want to find an $x$ where this is possible. It is not possible if $f'(x) neq 0$ as you already showed and so the only way possible is if $f'(x)=0$. So then for what values of $a,b$ is it possible to have $2ax+b=0$?
    $endgroup$
    – Michael
    1 hour ago












  • $begingroup$
    Now that made a lot more sense to me. So then the values of a and b would both have to equal 0 in order to get f'(x) = 0.
    $endgroup$
    – Rasheed Amanzai
    1 hour ago














4












4








4





$begingroup$


Recently, my Calculus and Vectors (Grade 12) teacher gave our class a thinking question/assignment to work on over the march break, and after working on for some time, I've become stuck on it.



The Question:



Consider f(x), a general quadratic function in standard form, and g(x) its reciprocal. For which values of x are the slopes of their respective tangent lines equal?



Consider two cases: one where it is true for exactly one value of x, and the other where it is true for exactly two values of x. In the latter case, you can assume that the steepness a does not equal 0 of f(x) is equal to both it’s y-intercept and also to the slope of its tangent at x = 1.



Find the required conditions on the parameters a,b,c in terms of a.



My Progress So Far:



So I know that $f(x) = ax^2 +bx + c$ and $g(x) = 1/f(x)$. After I solved for the derivative of each function, I set them both equal to each other and started solving for it. But then I came to a equation of $(ax^2 + bx + c)^2 = -1$ and that doesn't work.



Next I tried something else. Since I know from the 2nd case that $f(x) = c$ and $f(x) = f'(1)$ when a cannot equal 0, I set $c = f'(1)$ and got $c = 2a + b$. But after that, I don't know where to go.



I'm not expecting a full solution, but if anyone could give me a hint for solving this question, I would appreciate it. I know that it says to write everything in terms of a, but I'm not sure how to approach that method.










share|cite|improve this question









New contributor




Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Recently, my Calculus and Vectors (Grade 12) teacher gave our class a thinking question/assignment to work on over the march break, and after working on for some time, I've become stuck on it.



The Question:



Consider f(x), a general quadratic function in standard form, and g(x) its reciprocal. For which values of x are the slopes of their respective tangent lines equal?



Consider two cases: one where it is true for exactly one value of x, and the other where it is true for exactly two values of x. In the latter case, you can assume that the steepness a does not equal 0 of f(x) is equal to both it’s y-intercept and also to the slope of its tangent at x = 1.



Find the required conditions on the parameters a,b,c in terms of a.



My Progress So Far:



So I know that $f(x) = ax^2 +bx + c$ and $g(x) = 1/f(x)$. After I solved for the derivative of each function, I set them both equal to each other and started solving for it. But then I came to a equation of $(ax^2 + bx + c)^2 = -1$ and that doesn't work.



Next I tried something else. Since I know from the 2nd case that $f(x) = c$ and $f(x) = f'(1)$ when a cannot equal 0, I set $c = f'(1)$ and got $c = 2a + b$. But after that, I don't know where to go.



I'm not expecting a full solution, but if anyone could give me a hint for solving this question, I would appreciate it. I know that it says to write everything in terms of a, but I'm not sure how to approach that method.







calculus derivatives quadratics tangent-line slope






share|cite|improve this question









New contributor




Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Jon due

948




948






New contributor




Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 hours ago









Rasheed AmanzaiRasheed Amanzai

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213




New contributor




Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Rasheed Amanzai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    I made a typo in solving and so my prior comment did not make sense. To get your equation $f(x)^2=-1$ you likely made a cancellation on both sides, but what happens if the term you canceled is zero?
    $endgroup$
    – Michael
    2 hours ago












  • $begingroup$
    The term that I cancelled was 2ax+b (which is f'(x)), I don't know if the term cancelled can be zero since I do not know the unknowns. (I apologize if I didn't answer your question).
    $endgroup$
    – Rasheed Amanzai
    1 hour ago






  • 1




    $begingroup$
    So you got $-f'(x)=f'(x)f(x)^2$ and want to find an $x$ where this is possible. It is not possible if $f'(x) neq 0$ as you already showed and so the only way possible is if $f'(x)=0$. So then for what values of $a,b$ is it possible to have $2ax+b=0$?
    $endgroup$
    – Michael
    1 hour ago












  • $begingroup$
    Now that made a lot more sense to me. So then the values of a and b would both have to equal 0 in order to get f'(x) = 0.
    $endgroup$
    – Rasheed Amanzai
    1 hour ago


















  • $begingroup$
    I made a typo in solving and so my prior comment did not make sense. To get your equation $f(x)^2=-1$ you likely made a cancellation on both sides, but what happens if the term you canceled is zero?
    $endgroup$
    – Michael
    2 hours ago












  • $begingroup$
    The term that I cancelled was 2ax+b (which is f'(x)), I don't know if the term cancelled can be zero since I do not know the unknowns. (I apologize if I didn't answer your question).
    $endgroup$
    – Rasheed Amanzai
    1 hour ago






  • 1




    $begingroup$
    So you got $-f'(x)=f'(x)f(x)^2$ and want to find an $x$ where this is possible. It is not possible if $f'(x) neq 0$ as you already showed and so the only way possible is if $f'(x)=0$. So then for what values of $a,b$ is it possible to have $2ax+b=0$?
    $endgroup$
    – Michael
    1 hour ago












  • $begingroup$
    Now that made a lot more sense to me. So then the values of a and b would both have to equal 0 in order to get f'(x) = 0.
    $endgroup$
    – Rasheed Amanzai
    1 hour ago
















$begingroup$
I made a typo in solving and so my prior comment did not make sense. To get your equation $f(x)^2=-1$ you likely made a cancellation on both sides, but what happens if the term you canceled is zero?
$endgroup$
– Michael
2 hours ago






$begingroup$
I made a typo in solving and so my prior comment did not make sense. To get your equation $f(x)^2=-1$ you likely made a cancellation on both sides, but what happens if the term you canceled is zero?
$endgroup$
– Michael
2 hours ago














$begingroup$
The term that I cancelled was 2ax+b (which is f'(x)), I don't know if the term cancelled can be zero since I do not know the unknowns. (I apologize if I didn't answer your question).
$endgroup$
– Rasheed Amanzai
1 hour ago




$begingroup$
The term that I cancelled was 2ax+b (which is f'(x)), I don't know if the term cancelled can be zero since I do not know the unknowns. (I apologize if I didn't answer your question).
$endgroup$
– Rasheed Amanzai
1 hour ago




1




1




$begingroup$
So you got $-f'(x)=f'(x)f(x)^2$ and want to find an $x$ where this is possible. It is not possible if $f'(x) neq 0$ as you already showed and so the only way possible is if $f'(x)=0$. So then for what values of $a,b$ is it possible to have $2ax+b=0$?
$endgroup$
– Michael
1 hour ago






$begingroup$
So you got $-f'(x)=f'(x)f(x)^2$ and want to find an $x$ where this is possible. It is not possible if $f'(x) neq 0$ as you already showed and so the only way possible is if $f'(x)=0$. So then for what values of $a,b$ is it possible to have $2ax+b=0$?
$endgroup$
– Michael
1 hour ago














$begingroup$
Now that made a lot more sense to me. So then the values of a and b would both have to equal 0 in order to get f'(x) = 0.
$endgroup$
– Rasheed Amanzai
1 hour ago




$begingroup$
Now that made a lot more sense to me. So then the values of a and b would both have to equal 0 in order to get f'(x) = 0.
$endgroup$
– Rasheed Amanzai
1 hour ago










1 Answer
1






active

oldest

votes


















6












$begingroup$

If $g(x)=1/f(x)$, then, by the chain rule,
$$
g'(x)=-frac{f'(x)}{f(x)^2}
$$

Thus, assuming of course that $f(x)ne0$, we have $g'(x)=f'(x)$ if and only if
$$
-frac{f'(x)}{f(x)^2}=f'(x)
$$

which can only happen if $f'(x)=0$. Indeed, if $f'(x)ne0$, the equation becomes $f(x)^2=-1$, which is exactly the condition you find.



You can observe that this is independent of $f$ being a quadratic polynomial.



In your particular case, the condition is $x=-b/(2a)$, provided that $f(-b/(2a))ne0$ (or $g(-b/(2a))$ would be undefined.



enter image description here






share|cite|improve this answer









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    1 Answer
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    active

    oldest

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    active

    oldest

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    6












    $begingroup$

    If $g(x)=1/f(x)$, then, by the chain rule,
    $$
    g'(x)=-frac{f'(x)}{f(x)^2}
    $$

    Thus, assuming of course that $f(x)ne0$, we have $g'(x)=f'(x)$ if and only if
    $$
    -frac{f'(x)}{f(x)^2}=f'(x)
    $$

    which can only happen if $f'(x)=0$. Indeed, if $f'(x)ne0$, the equation becomes $f(x)^2=-1$, which is exactly the condition you find.



    You can observe that this is independent of $f$ being a quadratic polynomial.



    In your particular case, the condition is $x=-b/(2a)$, provided that $f(-b/(2a))ne0$ (or $g(-b/(2a))$ would be undefined.



    enter image description here






    share|cite|improve this answer









    $endgroup$


















      6












      $begingroup$

      If $g(x)=1/f(x)$, then, by the chain rule,
      $$
      g'(x)=-frac{f'(x)}{f(x)^2}
      $$

      Thus, assuming of course that $f(x)ne0$, we have $g'(x)=f'(x)$ if and only if
      $$
      -frac{f'(x)}{f(x)^2}=f'(x)
      $$

      which can only happen if $f'(x)=0$. Indeed, if $f'(x)ne0$, the equation becomes $f(x)^2=-1$, which is exactly the condition you find.



      You can observe that this is independent of $f$ being a quadratic polynomial.



      In your particular case, the condition is $x=-b/(2a)$, provided that $f(-b/(2a))ne0$ (or $g(-b/(2a))$ would be undefined.



      enter image description here






      share|cite|improve this answer









      $endgroup$
















        6












        6








        6





        $begingroup$

        If $g(x)=1/f(x)$, then, by the chain rule,
        $$
        g'(x)=-frac{f'(x)}{f(x)^2}
        $$

        Thus, assuming of course that $f(x)ne0$, we have $g'(x)=f'(x)$ if and only if
        $$
        -frac{f'(x)}{f(x)^2}=f'(x)
        $$

        which can only happen if $f'(x)=0$. Indeed, if $f'(x)ne0$, the equation becomes $f(x)^2=-1$, which is exactly the condition you find.



        You can observe that this is independent of $f$ being a quadratic polynomial.



        In your particular case, the condition is $x=-b/(2a)$, provided that $f(-b/(2a))ne0$ (or $g(-b/(2a))$ would be undefined.



        enter image description here






        share|cite|improve this answer









        $endgroup$



        If $g(x)=1/f(x)$, then, by the chain rule,
        $$
        g'(x)=-frac{f'(x)}{f(x)^2}
        $$

        Thus, assuming of course that $f(x)ne0$, we have $g'(x)=f'(x)$ if and only if
        $$
        -frac{f'(x)}{f(x)^2}=f'(x)
        $$

        which can only happen if $f'(x)=0$. Indeed, if $f'(x)ne0$, the equation becomes $f(x)^2=-1$, which is exactly the condition you find.



        You can observe that this is independent of $f$ being a quadratic polynomial.



        In your particular case, the condition is $x=-b/(2a)$, provided that $f(-b/(2a))ne0$ (or $g(-b/(2a))$ would be undefined.



        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        egregegreg

        184k1486205




        184k1486205






















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