Question

A piecewise function g(x) is defined by 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

ANSWERS

PART A

• Domain: (-∞, ∞)

,

• Graph:

PART B

• x-intercepts: -2, 0, 17

PART C

• Intervals in which g(x) is positive: (-2, 0) U [2, 17)

Step-by-step explanation

PART A

The first part of this function is a cubic function. If factor x² out, we can see clearly what are the roots:

`g_1(x)=x^2(x-4)`

So the roots are x = 0, x = -2 and x = 2.

The second part of the function is a logarithmic function. It starts at x = 2, where the logarithmic part is 0, so its value is 2 at that point and then decreases and intercepts the x-axis at x = 17.

So, the graph of this function is:

As we can see, the graph is continuous for all x, except for x = 2, where it has a jump.

The only restriction is that the argument of the logarithm cannot be zero, but since this happens at x = 1 and that value is where the function is defined as cubic, then there is no restriction there.

Hence, the domain of this function is all real values: (-∞, ∞).

PART B

As explained in part A, the roots of the cubic part of the function are x = -2, x = 0 and x = 2. However, at x = 2 the function is already defined as logarithmic, so the cubic part of the function does not intercept the x-axis at this point.

Additionally, there is another x-intercept given when the second part of the function is 0,

`\begin{gathered} -\log_4(x-1)+2=0 \\ \log_4(x-1)=2 \end{gathered}`

To find x raise 4 to each side of the equation,

`\begin{gathered} 4^(\log_4(x-1))=4^2 \\ \\ x-1=16\text{ }\Rightarrow\text{ }x=16+1=17 \end{gathered}`

Hence, this function has three x-intercepts: -2, 0, 17.

PART C

Based on the graph, we can describe the intervals for which g(x) is positive:

As we can observe, the function is positive between x = -2 and x = 0, not including these values since the function is 0 there, and also between x = 2 and x = 17, including 2 because the function's value is 2 there, but not 17 because it is a zero.

Hence, the intervals in which the function is positive are (-2, 0) U [2, 17).