Sure, let's tackle each part of the problem step-by-step.
### Part A: Determining the polynomial function [tex]\( f(x) \)[/tex]
The polynomial function [tex]\( f(x) \)[/tex] has zeros at [tex]\( x = 4 \)[/tex] with multiplicity 1 and at [tex]\( x = 1 \)[/tex] with multiplicity 2. Using the Factor Theorem, we can express [tex]\( f(x) \)[/tex] in its factored form:
[tex]\[ f(x) = (x - 4)(x - 1)^2 \][/tex]
Now, we will expand this expression to obtain [tex]\( f(x) \)[/tex] in its expanded form.
First, let's expand [tex]\( (x - 1)^2 \)[/tex]:
[tex]\[ (x - 1)^2 = (x - 1)(x - 1) = x^2 - 2x + 1 \][/tex]
Next, we need to expand:
[tex]\[ f(x) = (x - 4)(x^2 - 2x + 1) \][/tex]
To expand this, we use the distributive property:
[tex]\[ f(x) = x(x^2 - 2x + 1) - 4(x^2 - 2x + 1) \][/tex]
[tex]\[ f(x) = (x^3 - 2x^2 + x) - (4x^2 - 8x + 4) \][/tex]
[tex]\[ f(x) = x^3 - 2x^2 + x - 4x^2 + 8x - 4 \][/tex]
Finally, combine like terms:
[tex]\[ f(x) = x^3 - 6x^2 + 9x - 4 \][/tex]
So, the polynomial function [tex]\( f(x) \)[/tex] in expanded form is:
[tex]\[ f(x) = x^3 - 6x^2 + 9x - 4 \][/tex]
### Part B: Examining continuity of the piecewise function [tex]\( h(x) \)[/tex]
The piecewise function [tex]\( h(x) \)[/tex] is defined as:
[tex]\[ h(x) = \begin{cases}
f(x) & \text{if } x < 0 \\
g(x) & \text{if } x \geq 0
\end{cases} \][/tex]
To determine if [tex]\( h(x) \)[/tex] has any breaks or discontinuities, we need to check if [tex]\( h(x) \)[/tex] is continuous at [tex]\( x = 0 \)[/tex]. In particular, we need to compare the values of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] at [tex]\( x = 0 \)[/tex].
Let's evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^3 - 6(0)^2 + 9(0) - 4 = -4 \][/tex]
Next, let's evaluate [tex]\( g(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ g(0) = \sqrt[3]{0} - 4 = -4 \][/tex]
We notice that [tex]\( f(0) = g(0) = -4 \)[/tex]. Since the value of [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex] is equal to the value of [tex]\( g(x) \)[/tex] at [tex]\( x = 0 \)[/tex], there is no break or discontinuity at [tex]\( x = 0 \)[/tex].
Therefore, the piecewise function [tex]\( h(x) \)[/tex] is continuous, and there are no breaks or discontinuities in the domain of [tex]\( h(x) \)[/tex].
Hence, the function [tex]\( h(x) \)[/tex] is continuous over its entire domain.
