Given The Piecewise Function F(x) For X

Given The Following Piecewise Functionfxpiecewise 4x4for X

1. Given the following piecewise function: f(x) = piecewise { 4x + 4 } for x 4

a. Find the domain.

b. Find the range.

c. Find the intercepts.

d. Is f continuous on its domain? If not, state where f is discontinuous.

2. Graph the following function using transformations. Be sure to graph all of the stages on one graph. State the domain and range.

For example, if you were asked to graph y = x² + 1 using transformation, you would show the graph of y = x² and the graph shifted up 1 unit. Please do not show only the final graph. y = -√x + 3

Paper For Above instruction

The piecewise function provided offers a comprehensive view of how different rules apply to different segments of the domain. To analyze this function, we first examine its domain, which encompasses all real numbers because each segment is defined for a subset of the real number line. Specifically, the function is defined for x 4, thus covering all real values of x, making the domain (-∞, ∞).

The range of the function depends on the outputs produced by each piece. For x 4, the function is f(x) = -2x, a decreasing linear function. At x just greater than 4, f(x) is slightly less than -8, tending toward -∞ as x increases. At x = 4, f(4) = -2(4) = -8. Combining these segments, the range includes all y-values from -∞ up to just above -4, plus the constant value -2 at x between -2 and 4, and continues downward beyond x=4. Overall, the range is (-∞, 4] because at x = -2, y = -2, and the function takes all values less than -4 for x > 4, tending toward -∞.

To find intercepts, the y-intercept occurs where x=0. Since 0 is within -2 ≤ x ≤ 4, f(0) = -2, so the y-intercept is at (0, -2). The x-intercept occurs where f(x) = 0. For the first piece, 4x + 4 = 0; solving gives x = -1. Since -1 is in the interval x -2, so in the second segment where f(x) = -2, the value is -2, which does not equal zero. For the third segment, -2x = 0 when x=0, which is within the domain segment for x > 4? No, x=0 is less than 4, so not in that segment. Therefore, the x-intercept is at (-1, 0).

Regarding continuity, the function is continuous within each segment. At x = -2, the left limit is lim x→-2^- (4x+4) = 4(-2)+4 = -8+4 = -4, and the function value at x=-2 is -2, so there's a discontinuity there because the left limit (-4) does not equal the function value (-2). Similarly, at x=4, the limit from the left is f(4) = -2, but approaching from the right, f(x) = -2x, yields f(4) = -8, which does not match the earlier value, indicating another discontinuity at x=4.

Graphically, to plot this function, start by plotting the linear segment with slope 4 for x

In the second part of the assignment, the function y = -√x + 3 can be analyzed through transformations. The base graph y=√x is a standard square root curve starting at (0,0), opening to the right. The transformation y= -√x + 3 involves reflecting y=√x over the x-axis and shifting the graph upward by 3 units. This results in a graph that opens downward starting at (0, 3), with the maximum point at the origin reflected over the x-axis, creating a downward opening curve starting at (0, 3). The domain of this transformed function is x ≥ 0, as with the basic square root function. The range is y ≤ 3, as the graph peaks at y=3 when x=0 and decreases as x increases.

To graph this step by step, first plot the basic y=√x curve, then reflect it over the x-axis to get y= -√x, and finally shift the entire graph upward by 3 units. This involves moving the starting point from (0, 0) to (0, 3), and the entire curve opening downward from that point. This process highlights the importance of understanding transformations for accurate graphing and interpretation.

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