
#1
Oct3108, 02:32 PM

P: 365

1. The problem statement, all variables and given/known data
Find the function g(x), if there are [itex]h=g \circ f[/itex] f(x)=x+1, [itex]x \in \mathbb{R}[/itex] and [itex]h(x)=x^3+3x+1[/itex] 2. Relevant equations [tex](g \circ f)(x)=g(f(x))=h(x)[/tex] [tex]f \circ g \neq g \circ f [/tex] 3. The attempt at a solution [tex]h=g \circ f[/tex] [tex]h(x)=g(f(x))[/tex] [tex]x^3+3x+1=g(x+1)[/tex] Is there any way that I will directly find the result of g(x), or I should guess and try some things? I tried something at home, but useless. Please help me! 



#2
Oct3108, 03:12 PM

Mentor
P: 4,499

My guess is g is going to be a polynomial. You can use the degree of f and h to find what the degree of g is going to be, then plug f into a generic polynomial of that degree and see what conditions the coefficients have to satisfy




#3
Oct3108, 03:15 PM

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Well, you can figure g ought to be a polynomial function of degree 3, right? So put g(x+1)=A(x+1)^3+B(x+1)^2+C(x+1)+D. If you set that equal to x^3+3x+1, can you find A,B,C and D? You should get four equations in four unknowns if you equate the powers of x. Of course, you can also be clever and pick some special values of x that might make the job easier (like x=1).




#4
Oct3108, 03:29 PM

P: 365

Composition
Thanks for the posts.
I found A=1, B=3, C=6, D=3. [tex]g(x+1)=(x+1)^33(x+1)^2+6(x+1)3[/tex] So [tex]g(x)=x^33x^2+6x3[/tex] Yes it is correct. How will x=1 make the job easier? 



#5
Oct3108, 03:38 PM

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#6
Oct3108, 06:05 PM

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#7
Oct3108, 06:47 PM

P: 365

Thanks for the replys.
gabbagabbahey, do you think like [itex](u1)^3+3(u1)+1=g(u)[/itex] ? I got the same result [itex]u^33u^2+6u3=g(u)[/itex], so if I write [itex]g(x)=u^33u^2+6u3[/itex]. Thanks again. 



#8
Oct3108, 07:21 PM

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Yes, [itex](u1)^3+3(u1)+1=g(u)[/itex] so automatically, [itex]g(x)=(x1)^3+3(x1)+1[/itex] and you could leave the answer just like that. (or you could expand it in powers of x and get [itex]g(x)=x^33x^2+6x3[/itex])




#9
Oct3108, 09:54 PM

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