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Finding the Polynomial Function f(x) Given Conditions

April 03, 2025Film4840
Exploring the Polynomial Function f(x) Given Conditions In this articl

Exploring the Polynomial Function f(x) Given Conditions

In this article, we delve into the problem of finding a polynomial function f(x) such that when composed with itself twice, the result is a function of the form 3x2. Specifically, we will determine the values of a and b in the general polynomial fx axb that satisfy this condition.

Rational Behind the Solution

We start by assuming that fx is a polynomial function in the form:

fx axb, where a and b are constants.

When we compose fx with itself, we obtain:

ffx a(axb)b a2xab

We are given that:

ffx 3x2

From this, we can derive the following equations by comparing the coefficients:

1. a2 3 ? a ±√3

2. ab 2

Deriving the Values of a and b

From the first equation, we have two possible values for a:

a √3 or a -√3

Substituting these values into the second equation, we get:

Case 1: a √3

ab 2 ? b 2 / (1 √3) (2(1 - √3)) / (1 - 3) (2 - 2√3) / -2 1 - √3

Case 2: a -√3

ab 2 ? b 2 / (1 - √3) (2(1 √3)) / -2 -1 - √3

Thus, the polynomial functions that satisfy the given condition are:

fx √3x^(1 - √3)

or

fx -√3x^(-1 - √3)

Verification

To verify, let's compute the composition of each function with itself:

1. For fx √3x^(1 - √3):

ffx (√3x^(1 - √3))(√3x^(1 - √3)) 3x^(2 - 2√3)

Since 2 - 2√3 ≠ 2, this function does not satisfy the given condition.

2. For fx -√3x^(-1 - √3):

ffx (-√3x^(-1 - √3))(-√3x^(-1 - √3)) 3x^(2 2√3)

Since 2 2√3 ≠ 2, this function also does not satisfy the given condition.

The correct approach is to consider the simpler form of the function, where the polynomial is affine (i.e., linear):

Case 1: fx √3x - √3

ffx (√3x - √3)(√3x - √3) 3x2 - 2√3x 3

Case 2: fx -√3x √3

ffx (-√3x √3)(-√3x √3) 3x2 - 2√3x 3

Both cases yield the correct result, confirming that the function fx -√3x √3 is a valid solution.

Conclusion

The polynomial function fx -√3x √3 satisfies the given condition. This solution demonstrates the process of solving polynomial function compositions and highlights the importance of careful algebraic manipulation in finding the correct solution.