Critical points in a function

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Consider the function $f(x)=(a^2 -3a+2)cos(x/2) + (a-1)x$. We have to find set of values of $a$ for which $f(x)$ possess critical points.
When we put $a=1$, we get both $f(x) = 0$ and $f'(x) =0$. So we can say $f(x)$ possess critical points for $a=1$. But in the answer, $a=1$ is not included. Why?




calculus graphing-functions




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    You said it yourself: If $a=1$, then $f(x) = 0$ for all $x$. Not a very interesting function ... Technically, indeed also $f'(x) = 0$ but I would not say that a constant function has any critical points.
    – Matti P.
    Aug 14 at 6:58






  • 1




    You can play around with this: desmos.com/calculator/nkyzggmvbc
    – Matti P.
    Aug 14 at 6:59










  • What is the definition of a critical point that your textbook uses?
    – dxiv
    Aug 14 at 7:01






  • 1




    Critical points of a function are points where either f'(x)=0 or f'(x) doesn't exist
    – Lakshay
    Aug 14 at 7:05










  • @Lakshay Then you are correct in that $,f,$ has critical points for $,a=1,$, in fact the entire domain.
    – dxiv
    Aug 14 at 7:07














up vote
2
down vote

favorite












Consider the function $f(x)=(a^2 -3a+2)cos(x/2) + (a-1)x$. We have to find set of values of $a$ for which $f(x)$ possess critical points.
When we put $a=1$, we get both $f(x) = 0$ and $f'(x) =0$. So we can say $f(x)$ possess critical points for $a=1$. But in the answer, $a=1$ is not included. Why?




calculus graphing-functions




share|cite|improve this question


















  • 1




    You said it yourself: If $a=1$, then $f(x) = 0$ for all $x$. Not a very interesting function ... Technically, indeed also $f'(x) = 0$ but I would not say that a constant function has any critical points.
    – Matti P.
    Aug 14 at 6:58






  • 1




    You can play around with this: desmos.com/calculator/nkyzggmvbc
    – Matti P.
    Aug 14 at 6:59










  • What is the definition of a critical point that your textbook uses?
    – dxiv
    Aug 14 at 7:01






  • 1




    Critical points of a function are points where either f'(x)=0 or f'(x) doesn't exist
    – Lakshay
    Aug 14 at 7:05










  • @Lakshay Then you are correct in that $,f,$ has critical points for $,a=1,$, in fact the entire domain.
    – dxiv
    Aug 14 at 7:07












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Consider the function $f(x)=(a^2 -3a+2)cos(x/2) + (a-1)x$. We have to find set of values of $a$ for which $f(x)$ possess critical points.
When we put $a=1$, we get both $f(x) = 0$ and $f'(x) =0$. So we can say $f(x)$ possess critical points for $a=1$. But in the answer, $a=1$ is not included. Why?




calculus graphing-functions




share|cite|improve this question














Consider the function $f(x)=(a^2 -3a+2)cos(x/2) + (a-1)x$. We have to find set of values of $a$ for which $f(x)$ possess critical points.
When we put $a=1$, we get both $f(x) = 0$ and $f'(x) =0$. So we can say $f(x)$ possess critical points for $a=1$. But in the answer, $a=1$ is not included. Why?





calculus graphing-functions





share|cite|improve this question













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edited Aug 14 at 7:00









Taroccoesbrocco

3,74451431




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asked Aug 14 at 6:52









Lakshay

114




114







  • 1




    You said it yourself: If $a=1$, then $f(x) = 0$ for all $x$. Not a very interesting function ... Technically, indeed also $f'(x) = 0$ but I would not say that a constant function has any critical points.
    – Matti P.
    Aug 14 at 6:58






  • 1




    You can play around with this: desmos.com/calculator/nkyzggmvbc
    – Matti P.
    Aug 14 at 6:59










  • What is the definition of a critical point that your textbook uses?
    – dxiv
    Aug 14 at 7:01






  • 1




    Critical points of a function are points where either f'(x)=0 or f'(x) doesn't exist
    – Lakshay
    Aug 14 at 7:05










  • @Lakshay Then you are correct in that $,f,$ has critical points for $,a=1,$, in fact the entire domain.
    – dxiv
    Aug 14 at 7:07












  • 1




    You said it yourself: If $a=1$, then $f(x) = 0$ for all $x$. Not a very interesting function ... Technically, indeed also $f'(x) = 0$ but I would not say that a constant function has any critical points.
    – Matti P.
    Aug 14 at 6:58






  • 1




    You can play around with this: desmos.com/calculator/nkyzggmvbc
    – Matti P.
    Aug 14 at 6:59










  • What is the definition of a critical point that your textbook uses?
    – dxiv
    Aug 14 at 7:01






  • 1




    Critical points of a function are points where either f'(x)=0 or f'(x) doesn't exist
    – Lakshay
    Aug 14 at 7:05










  • @Lakshay Then you are correct in that $,f,$ has critical points for $,a=1,$, in fact the entire domain.
    – dxiv
    Aug 14 at 7:07







1




1




You said it yourself: If $a=1$, then $f(x) = 0$ for all $x$. Not a very interesting function ... Technically, indeed also $f'(x) = 0$ but I would not say that a constant function has any critical points.
– Matti P.
Aug 14 at 6:58




You said it yourself: If $a=1$, then $f(x) = 0$ for all $x$. Not a very interesting function ... Technically, indeed also $f'(x) = 0$ but I would not say that a constant function has any critical points.
– Matti P.
Aug 14 at 6:58




1




1




You can play around with this: desmos.com/calculator/nkyzggmvbc
– Matti P.
Aug 14 at 6:59




You can play around with this: desmos.com/calculator/nkyzggmvbc
– Matti P.
Aug 14 at 6:59












What is the definition of a critical point that your textbook uses?
– dxiv
Aug 14 at 7:01




What is the definition of a critical point that your textbook uses?
– dxiv
Aug 14 at 7:01




1




1




Critical points of a function are points where either f'(x)=0 or f'(x) doesn't exist
– Lakshay
Aug 14 at 7:05




Critical points of a function are points where either f'(x)=0 or f'(x) doesn't exist
– Lakshay
Aug 14 at 7:05












@Lakshay Then you are correct in that $,f,$ has critical points for $,a=1,$, in fact the entire domain.
– dxiv
Aug 14 at 7:07




@Lakshay Then you are correct in that $,f,$ has critical points for $,a=1,$, in fact the entire domain.
– dxiv
Aug 14 at 7:07










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Hints:



  1. Compute $f'(x)$ and distinguish three cases: $a=1,a=2$ and $ 1 ne a ne 2$


  2. Compute all $x$ with $f'(x)=0$






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    Hints:



    1. Compute $f'(x)$ and distinguish three cases: $a=1,a=2$ and $ 1 ne a ne 2$


    2. Compute all $x$ with $f'(x)=0$






    share|cite|improve this answer
























      up vote
      0
      down vote













      Hints:



      1. Compute $f'(x)$ and distinguish three cases: $a=1,a=2$ and $ 1 ne a ne 2$


      2. Compute all $x$ with $f'(x)=0$






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        Hints:



        1. Compute $f'(x)$ and distinguish three cases: $a=1,a=2$ and $ 1 ne a ne 2$


        2. Compute all $x$ with $f'(x)=0$






        share|cite|improve this answer












        Hints:



        1. Compute $f'(x)$ and distinguish three cases: $a=1,a=2$ and $ 1 ne a ne 2$


        2. Compute all $x$ with $f'(x)=0$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 14 at 7:16









        Fred

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