Question

PLEASE HELP 100 POINTS!!!!!!
A piecewise function g(x) is defined by g of x is equal to the piecewise function of x cubed minus 9 times x for x is less than 3 and negative log base 4 of the quantity x minus 2 end quantity plus 2 for x is greater than or equal to 3

Part A: Graph the piecewise function g(x) and determine the domain. (5 points)

Part B: Determine the x-intercepts of g(x). Show all necessary calculations. (5 points)

Part C: Describe the interval(s) in which the graph of g(x) is positive. (5 points)

Answer 1

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 multiple pieces of curves/lines where each piece corresponds to its definition over an interval.

Given piecewise function:

$g(x)=\begin{cases}x^3-9x \quad \quad \quad \quad \quad \textsf{if }x < 3\\-\log_4(x-2)+2 \quad \textsf{if }x\geq 3\end{cases}$

Therefore, the function has two definitions:

• $g(x)=x^3-9x \quad \textsf{when x is less than 3}$

• $g(x)=-\log_4(x-2)+2 \quad \textsf{when x is more than or equal to 3}$

Part A

When graphing piecewise functions:

• Use an open circle where the value of x is not included in the interval.
• Use a closed circle where the value of x is included in the interval.
• Use an arrow to show that the function continues indefinitely.

First piece of function

Substitute the endpoint of the interval into the corresponding function:

$\implies g(3)=(3)^3-9(3)=0 \implies (3,0)$

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 → -∞.

Second piece of function

Substitute the endpoint of the interval into the corresponding function:

$\implies g(3)=-\log_4(3-2)+2=2 \implies (3,2)$

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.

Part B

The x-intercepts are where the curve crosses the x-axis, so when y = 0.

Set the first piece of the function to zero and solve for x:

$\begin{aligned}g(x) & = 0\\\implies x^3-9x & = 0\\x(x^2-9) & = 0\\\\\implies x^2-9 & = 0 \quad \quad \quad \implies x=0\\x^2 & = 9\\\ x & = \pm 3\end{aligned}$

Therefore, as x < 3, the x-intercepts are (-3, 0) and (0, 0) for the first piece.

Set the second piece to zero and solve for x:

$\begin{aligned}\implies g(x) & =0\\-\log_4(x-2)+2 & =0\\\log_4(x-2) & =2\end{aligned}$

$\textsf{Apply log law}: \quad \log_ab=c \iff a^c=b$

$\begin{aligned}\implies 4^2&=x-2\\x & = 16+2\\x & = 18 \end{aligned}$

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).

Part C

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:

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