## Given f find g such that g(x)=f(x-1)

1. The problem statement, all variables and given/known data
Find two complex polynomials f and g such that f(0)=f(-1)=1, f(1)=3 and g(x)=f(x-1)

2. The attempt at a solution
Using Lagrange polynomial I got f such that f(0)=f(-1)=1, f(1)=3
Such f is defined by

f(x)=x2+x+1

Now that I've found f I need to find g such that g(x)=f(x-1), but I don't have any idea of how to do that. Any hint would be appreciated.
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Mentor
 Quote by krozer 1. The problem statement, all variables and given/known data Find two complex polynomials f and g such that f(0)=f(-1)=1, f(1)=3 and g(x)=f(x-1) 2. The attempt at a solution Using Lagrange polynomial I got f such that f(0)=f(-1)=1, f(1)=3 Such f is defined by f(x)=x2+x+1 Now that I've found f I need to find g such that g(x)=f(x-1), but I don't have any idea of how to do that. Any hint would be appreciated.
Given that f(x) = x2+x+1

Find f(x-1) .
 Recognitions: Gold Member Science Advisor Staff Emeritus You clearly know how to do some fairly complicated Calculus so surely you know how to evaluate a function! Just replace each "x" in $f(x)= x^2+ x+ 12$ with x- 1.

## Given f find g such that g(x)=f(x-1)

Notice that if you define:
$$h(x) \equiv f(x) - 1$$
then x = 0, and x = -1 are zeros of h. The simplest polynomial that can be written is:
$$h(x) = A x (x + 1)$$
Then, use the fact that $h(1) = f(1) - 1 = 3 - 1 = 2$ to determine A. You can go back to f trivially then.

Once you have chosen a particular choice for f, finding g, as explained in the above posts, is fairly trivial (just substitute $x \rightarrow x - 1$, expand and simplify).

 Tags linear algebra, polynomial

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