Answer:
A) See attached for graph.
B) (-3, 0) (0, 0) (18, 0)
C) (-3, 0) ∪ [3, 18)
Step-by-step explanation:
Piecewise functions have <u>multiple pieces</u> of curves/lines where each piece corresponds to its definition over an <u>interval</u>.
Given piecewise function:
![g(x)=\begin{cases}x^3-9x \quad \quad \quad \quad \quad \textsf{if }x < 3\\-\log_4(x-2)+2 \quad \textsf{if }x\geq 3\end{cases}](https://tex.z-dn.net/?f=g%28x%29%3D%5Cbegin%7Bcases%7Dx%5E3-9x%20%5Cquad%20%5Cquad%20%5Cquad%20%5Cquad%20%5Cquad%20%5Ctextsf%7Bif%20%7Dx%20%3C%203%5C%5C-%5Clog_4%28x-2%29%2B2%20%5Cquad%20%20%5Ctextsf%7Bif%20%7Dx%5Cgeq%203%5Cend%7Bcases%7D)
Therefore, the function has two definitions:
<h3><u>Part A</u></h3>
When <u>graphing</u> piecewise functions:
- Use an open circle where the value of x is <u>not included</u> in the interval.
- Use a closed circle where the value of x is <u>included</u> in the interval.
- Use an arrow to show that the function <u>continues indefinitely</u>.
<u>First piece of function</u>
Substitute the endpoint of the interval into the corresponding function:
![\implies g(3)=(3)^3-9(3)=0 \implies (3,0)](https://tex.z-dn.net/?f=%5Cimplies%20g%283%29%3D%283%29%5E3-9%283%29%3D0%20%5Cimplies%20%283%2C0%29)
Place an open circle at point (3, 0).
Graph the cubic curve, adding an arrow at the other endpoint to show it continues indefinitely as x → -∞.
<u>Second piece of function</u>
Substitute the endpoint of the interval into the corresponding function:
![\implies g(3)=-\log_4(3-2)+2=2 \implies (3,2)](https://tex.z-dn.net/?f=%5Cimplies%20g%283%29%3D-%5Clog_4%283-2%29%2B2%3D2%20%5Cimplies%20%283%2C2%29)
Place an closed circle at point (3, 2).
Graph the curve, adding an arrow at the other endpoint to show it continues indefinitely as x → ∞.
See attached for graph.
<h3><u>Part B</u></h3>
The x-intercepts are where the curve crosses the x-axis, so when y = 0.
Set the <u>first piece</u> of the function to zero and solve for x:
![\begin{aligned}g(x) & = 0\\\implies x^3-9x & = 0\\x(x^2-9) & = 0\\\\\implies x^2-9 & = 0 \quad \quad \quad \implies x=0\\x^2 & = 9\\\ x & = \pm 3\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dg%28x%29%20%26%20%3D%200%5C%5C%5Cimplies%20x%5E3-9x%20%26%20%3D%200%5C%5Cx%28x%5E2-9%29%20%26%20%3D%200%5C%5C%5C%5C%5Cimplies%20x%5E2-9%20%26%20%3D%200%20%5Cquad%20%5Cquad%20%5Cquad%20%5Cimplies%20x%3D0%5C%5Cx%5E2%20%26%20%3D%209%5C%5C%5C%20x%20%26%20%3D%20%5Cpm%203%5Cend%7Baligned%7D)
Therefore, as x < 3, the x-intercepts are (-3, 0) and (0, 0) for the first piece.
Set the <u>second piece</u> to zero and solve for x:
![\begin{aligned}\implies g(x) & =0\\-\log_4(x-2)+2 & =0\\\log_4(x-2) & =2\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cimplies%20g%28x%29%20%26%20%3D0%5C%5C-%5Clog_4%28x-2%29%2B2%20%26%20%3D0%5C%5C%5Clog_4%28x-2%29%20%26%20%3D2%5Cend%7Baligned%7D)
![\textsf{Apply log law}: \quad \log_ab=c \iff a^c=b](https://tex.z-dn.net/?f=%5Ctextsf%7BApply%20log%20law%7D%3A%20%5Cquad%20%5Clog_ab%3Dc%20%5Ciff%20a%5Ec%3Db)
![\begin{aligned}\implies 4^2&=x-2\\x & = 16+2\\x & = 18 \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cimplies%204%5E2%26%3Dx-2%5C%5Cx%20%26%20%3D%2016%2B2%5C%5Cx%20%26%20%3D%2018%20%5Cend%7Baligned%7D)
Therefore, the x-intercept for the second piece is (18, 0).
So the x-intercepts for the piecewise function are (-3, 0), (0, 0) and (18, 0).
<h3><u>Part C</u></h3>
From the graph from part A, and the calculated x-intercepts from part B, the function g(x) is positive between the intervals -3 < x < 0 and 3 ≤ x < 18.
Interval notation: (-3, 0) ∪ [3, 18)
Learn more about piecewise functions here:
brainly.com/question/11562909