Let a polynomail function f satisfies the relation f(x)+f((1)/(x))=f(x) ·f((1)/(x)) for all x ∈R

Let a polynomail function f satisfies the relation...

Key Concepts

Functional Equations

This concept involves equations where the unknowns are functions and the equations involve the functions evaluated at different points. Solving these often requires finding patterns or substitutions that reduce the functional relation to a simpler form, allowing the function's behavior to be deduced across its domain.

Polynomial Functions

A polynomial function is an expression consisting of variables raised to nonnegative integer powers with coefficients, which are closed under operations like addition and multiplication. Recognizing that the function in question is a polynomial restricts the possible forms it may take, particularly when additional conditions are imposed by a functional equation.

Reciprocal Invariance

This concept focuses on the behavior of functions under the transformation x ? 1/x. Functional equations that relate f(x) to f(1/x) exhibit symmetry properties that can be exploited to simplify the problem. Recognizing this inversion symmetry is crucial for determining the form of the function.

Algebraic Transformation

Algebraic transformations, such as substituting g(x) = f(x) - 1, help in rewriting the functional equation into a factorized form. This new form can reveal multiplicative identities that the transformed function must satisfy, leading to restrictions on its structure and ultimately to the general solution.

Transcript

00:01 Hello everyone from the question we are given that polynomial function f f -satisfied the relation fx plus f1 upon x is equals to fx into f1 upon x for all x belongs to r minus 0 then we have to find the value of fx so for this first we let fx is equals a 0 x to the power n plus a 1 x to the power n minus 1 plus a 2 x to the power n minus 1 plus a 2 plus a 2 plus a 2 up to a n n so f x plus f1 upon x is equal to f x into f1 upon x this implies so put the value of f x then we get a 0 x to the power n plus a 1 x to the power n minus 1 plus a 2 x to the power n minus 2 plus up to a n plus f 1 upon x is equal to a 0 upon x to the power n plus a 1 upon x to the power n minus 1 plus a 2 upon x to the power n minus 1 plus a 2 upon x to the power n minus 2 plus a 2 plus a 2 upon x to the power n minus 2 plus to an.

01:02 This is equals to a0 x to the power n plus a1 x to the power n minus 1 plus a2 x to the power n minus 2 plus up to a n multiply by a0 divided by x to the power n plus a1 divided by x to the power n minus 1 plus a2 divided by x to the power n minus 2 plus up to a n.

01:23 By simplifying this we get a 0 multiply by x to the power 2 n plus a 1 x to the power 2 n minus 1 plus a2x to the power 2 n minus 2 plus plus a n x to the power n plus a0 plus a1 x square plus up to an x to the power n is equal to a0 x to the power n minus 1 plus a2x to the power n minus 1 plus a2 plus a0 by a1 x plus a2 x square plus up to an x to the power n by comparing both sides then we get an is equals to 1, an minus 1 is equal to 0, an minus 2 is equal to 0, an minus 2 is equal to 0, an minus 3 is equal to 0, to a2 equals to 0, a1 is equal to 0.

02:13 Now comparing the coefficients of x to the power n, then we get 2 an is equal to a0 square plus a n square.

02:21 Then we get 2 is equals to a0 square plus 1...

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