To construct a polynomial function \(\( f(x) \)\) with the given properties, we will proceed through the following steps:
1. Identify the zeros of the polynomial: We are given that the polynomial has zeros at \(\( x = 1 \)\), \(\( x = 2 \)\), and \(\( x = 3 \)\). This means the polynomial has factors of the form \(\( (x - 1) \)\), \(\( (x - 2) \)\), and \(\( (x - 3) \)\).
2. Form the polynomial with the identified zeros: A polynomial with these zeros can be written as:
\(\[
f(x) = k (x - 1)(x - 2)(x - 3)
\]\)
where \(\( k \)\) is a constant.
3. Ensure \(\( f(x) < 0 \)\) for all values of \(\( x \)\): To make sure that the polynomial is negative for all \(\( x \)\), we can choose a negative value for \(\( k \)\). Let's pick \(\( k = -1 \)\).
4. Construct the polynomial: Substituting \(\( k = -1 \)\) into the polynomial, we have:
\(\[
f(x) = - (x - 1)(x - 2)(x - 3)
\]\)
5. Expand the polynomial: Now we'll expand the polynomial:
\(\[
f(x) = - (x - 1)(x - 2)(x - 3)
\]\)
First, expand the product of two factors:
\(\[
(x - 1)(x - 2) = x^2 - 2x - x + 2 = x^2 - 3x + 2
\]\)
Next, multiply this result by \(\( (x - 3) \)\):
\(\[
f(x) = - \left( (x^2 - 3x + 2)(x - 3) \right)
\]\)
Continue expanding:
\(\[
= - \left( x^3 - 3x^2 + 2x - 3x^2 + 9x - 6 \right)
\]\)
Combine like terms inside the parentheses:
\(\[
= - (x^3 - 6x^2 + 11x - 6)
\]\)
Finally, distribute the negative sign:
\(\[
f(x) = -x^3 + 6x^2 - 11x + 6
\]\)
So, the polynomial function \(\( f(x) \)\) that meets all the requirements is:
\(\[
f(x) = -x^3 + 6x^2 - 11x + 6
\]\)
### Verification
1. Zeros at 1, 2, and 3:
- For \(\( x = 1 \)\):
\(\[
f(1) = -1^3 + 6(1)^2 - 11(1) + 6 = -1 + 6 - 11 + 6 = 0
\]\)
- For \(\( x = 2 \)\):
\(\[
f(2) = -2^3 + 6(2)^2 - 11(2) + 6 = -8 + 24 - 22 + 6 = 0
\]\)
- For \(\( x = 3 \)\):
\(\[
f(3) = -3^3 + 6(3)^2 - 11(3) + 6 = -27 + 54 - 33 + 6 = 0
\]\)
2. \(\( f(x) < 0 \)\) for all values of \(\( x \)\):
- By choosing \(\( k = -1 \)\), we ensured the polynomial will always be negative because the product \(\( (x - 1)(x - 2)(x - 3) \)\) will always be positive for any real value of \(\( x \)\) due to the odd number of factors crossed when \(\( x \)\) is crossed zero points. Multiplying by \(\(-1\)\) flips the sign making the function negative for all \(\( x \)\).
3. Degree greater than 1:
- The resulting polynomial is of degree 3, which is greater than 1.
Therefore, the polynomial function satisfies all the given properties.
