Question

Let f (x) be a polynomial function with a zero of multiplicity of 1 at 4 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 4 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 1

Using the Factor Theorem, it is found that:

a) $f(x) = x^3 - 6x^2 + 9x - 4$

b) Due to equal lateral limits, there are no breaks in the domain of h(x).

What is the Factor Theorem?

• The Factor Theorem states that a polynomial function with roots $x_1, x_2, \codts, x_n$ is given by:

$f(x) = a(x - x_1)(x - x_2) \cdots (x - x_n)$

• In which a is the leading coefficient.

Item a:

The zeroes are:

• Zero of multiplicity of 1 at 4, hence $x_1 = 4$.

• Zero of multiplicity of 2 at 1, hence $x_2 = x_3 = 1$.

• Leading coefficient of $a = 1$.

Then:

$f(x) = (x - 4)(x - 1)(x - 1)$

$f(x) = (x - 4)(x^2 - 2x + 1)$

$f(x) = x^3 - 6x^2 + 9x - 4$

Item b:

The piece-wise function is defined by:

$h(x) = x^3 - 6x^2 + 9x - 4, x \leq 0$

$h(x) = \sqrt[3]{x} - 4, x > 0$

Using lateral limits at the point in which the definition is changed:

$\lim_{x \rightarrow 0^-} h(x) = \lim_{x \rightarrow 0} x^3 - 6x^2 + 9x - 4 = -4$

$\lim_{x \rightarrow 0^+} h(x) = \lim_{x \rightarrow 0} \sqrt[3]{x} - 4 = -4$

Due to equal lateral limits, there are no breaks in the domain of h(x).

To learn more about the Factor Theorem, you can take a look at https://brainly.com/question/24380382