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Let f (x) be a polynomial function with a zero of multiplicity of 1 at 2 and a zero of multiplicity of 2 at 1. Let g(x) be the radical function g of x equals the cube root of x minus 2 Part A: Using the Factor Theorem, determine the polynomial function f (x) in expanded form. Show all necessary calculations. (5 points) Part B: Let h (x) be the piecewise defined function of h of x is the piecewise function of f of x if x is less than 0 and g of x if x is greater than or equal to 0 Are there any breaks in the domain of h (x)? Explain why or why not. (5 points)

Answer :

Azikennamdigo

The polynomial functions in their expanded form is given as follows. It is right to state that there are no breaks in the domain of h(x).



What is a polynomial function?

In an equation such as the quadratic equation, cubic equation, etc., a polynomial function is a function that only uses non-negative integer powers or only positive integer exponents of a variable.

For instance, the polynomial 3x+4 has an exponent of 1.

Part A: F(x) has zero at 2 and multiplicity of 1; and

1 at the multiplicity of 2

f(x) = x-2) (x-1)²

= (x-2) (x² - 2x + 1)

= x³ - 4x² + 5x -2

Part B: h (x) = [tex]\left \{ {{x^3 -4x^2 + 5x -2; X < 0} \atop {\sqrt[3]{x-2} ; X\geq 0 }} \right.[/tex]

The domain of X is X ∈ R

Hence it is correct to state that there are no breaks in the domain of h(x).

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