Verify whether there exists a polynomial [tex]f(x)[/tex] such that [tex]f(f(x))=x^2-x+1[/tex], and find [tex]f(0)[/tex].

Question

Verify whether there exists a polynomial  f(x)  such that  f(f(x))=x^2-x+1 , and find  f(0) .

Sophia · Answer

Answer:    f(0) =  {0}^{2}  - 0 + 1     = 0 - 0 + 1        f(0) = 1

Isabelle · Answer

large underline{ sf{Given- }}      :   :   :   :   :   bull  :   sf  :   :  f(f(x)) =  {x}^{2}  - x + 1       large underline{ sf{To :Find - }}      :   :   :   :   :   bull  :   sf  :   : f(0)       large underline{ sf{Solution-}}     Given that       rm : longmapsto : sf  :   :  f(f(x)) =  {x}^{2}  - x + 1 -  -  - (1)       Put x = 0, in equation (1), we get        rm : longmapsto : sf  :   :  f(f(0)) =  {0}^{2}  - 0 + 1     rm : longmapsto : sf  :   :  f(f(0)) =1 -  -  - (2)      Put x = 1, in equation (1) we get        rm : longmapsto : sf  :   :  f(f(1)) =  {1}^{2}  - 1 + 1     rm : longmapsto : sf  :   :  f(f(1)) = 1 -  -  - (3)     Hᴇɴᴄᴇ,     ➢ Pair of points of the given equation are shown in the below table.      begin{gathered} boxed{ begin{array}{c|c}  bf x &  bf f(f(x))     frac{ qquad  qquad}{} &  frac{ qquad  qquad}{}     sf 0 &  sf 1        sf 1 &  sf 1  end{array}}     end{gathered}     Now,      Again we have given that        rm : longmapsto : sf  :   :  f(f(x)) =  {x}^{2}  - x + 1     rm : longmapsto :  red{  bf  : Put  : x  :  =   : f(x),  : we  : get}     rm : longmapsto : sf  :   :  f(f(f(x))) =  {f(x)}^{2}  - f(x) + 1 -  - (4)     rm : longmapsto :  blue{  bf  : Put  : x = 0,  : we  : get}        rm : longmapsto : sf  :   :  f(f(f(0))) =  {f(0)}^{2}  - f(0) + 1     rm : longmapsto : sf  :   :  f(1) =  {f(0)}^{2}  - f(0) + 1 -  -  - (5)       :   :   :   :   :   :   :   :   :   :   :   :   :   :   :   :   :   :   :   :  boxed{  because  bf  : f(f(0)) = 1}     Now,      if we get the value of f(1), we are able to find f(0)     So,     Put x = 1, in equation (4), we get        rm : longmapsto : sf  :   :  f(f(f(1))) =  {f(1)}^{2}  - f(1) + 1      rm : longmapsto : sf  :   :  f(1) =  {f(1)}^{2}  - f(1) + 1   :   :    { because  : f(f(1)) = 1  }     rm : longmapsto : sf  :   : {f(1)}^{2}  -2 f(1) + 1= 0     rm : longmapsto : {(f(1) - 1)}^{2}  = 0     rm : longmapsto :f(1) - 1 = 0     bf implies  :f(1) = 1 -  -  - (6)     On substituting f(1) = 1 in equation (5), we get       rm : longmapsto : sf  :   :    cancel1 =  {f(0)}^{2}  - f(0) +   cancel1     rm : longmapsto : {f(0)}^{2}  - f(0) = 0     rm : longmapsto :f(0) bigg(f(0) - 1  bigg)  = 0     bf implies  :f(0)  :  =   : 0  :   : or  :   : f(0)  :  =   : 1      Verification  :-     Case :- 1     When f(0) = 0     Then,      Value of f(f(x)) when x = 0, is        rm : longmapsto : f(f(0)) = f(0) = 0     rm : longmapsto :which  : is  : contradiction  : as  : f(f(0)) = 1   So,      bf implies  :f(0)  :   ne  : 0     Case :- 2     When f(0) = 1     Then    Value of f(f(x)) when x = 0,is       rm : longmapsto : f(f(0)) = f(1) = 1  : satisfies.     bf implies  :f(0) = 1      :   :   :   :   :   :  overbrace{  underline {  boxed {  bf  therefore  : The  : value  : of  : f(0) = 1}}}

Ava

17 th June is Father’s Day. Make it special for your father and surprise him in a thoughtful manner. Write how you made it special

Write the coefficient of x2 in each of the following:
(a) 3x-5

please help me this question please .

2 thoughts on “Verify whether there exists a polynomial [tex]f(x)[/tex] such that [tex]f(f(x))=x^2-x+1[/tex], and find [tex]f(0)[/tex].”

Verification :-

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