Question

(a) Find the domain of each function. (b) Locate any intercepts. (c) Graph each function. (d) Based on the graph, find the range. $$ f(x)=\left\\{\begin{array}{ll} 2 x+5 & \text { if }-3 \leq x<0 \\ -3 & \text { if } x=0 \\ -5 x & \text { if } x>0 \end{array}\right. $$

Short answer

Domain: \(-3 \leq x < \infty\). Intercept: (0, -3). Range: \((-\infty, -3] \cup (-1, 5)\).

Step-by-step solution

Find the Domain

Evaluate the given piecewise function to determine the values of x for which each piece is defined.- For the first piece, \(2x + 5\), the domain is \(-3 \leq x < 0\).- For the second piece, \(-3\), the domain is \(x = 0\).- For the third piece, \(-5x\), the domain is \(x > 0\).Combine these to get the overall domain: \(-3 \leq x < \infty\).

Locate the Intercepts

Find the x-intercepts and y-intercepts.- For the first piece \(2x + 5\): - Y-intercept: When \(x = 0\), \( y = 2(0) + 5 = 5\). But since \(x \eq 0\) for this piece, there's no y-intercept here. - X-intercept: Set \(2x + 5 = 0\), solving for x gives \(x = -2.5\).- For the second piece \(-3\): - Y-intercept: When \(x = 0\), \(y = -3\). Hence, there is a y-intercept at (0, -3). - X-intercept: None, since \(-3 \eq 0\).- For the third piece \(-5x\): - Y-intercept: None, since it is defined only for \(x > 0\). - X-intercept: Set \(-5x = 0\), solving for x gives \(x = 0\), but already assigned to \(x = 0\) in a different piece.

Graph the Function

Plot each piece of the function based on its domain and range. Identify key points and behavior:- For \(2x + 5\) in \(-3 \leq x < 0\), plot points: - At \(x = -3\), \(y = 2(-3) + 5 = -6 + 5 = -1\). - At \(x = 0\), \(y = 2(0) + 5 = 5\). - Linearly connect points between \(-3\) and \(0\) (not including 0).- For \(-3\) at \(x = 0\), plot: - Point at (0, -3), since this represents the value exactly at \(x = 0\).- For \(-5x\) in \(x > 0\), plot points: - At \(x = 1\), \(y = -5(1) = -5\). - At \(x = 2\), \(y = -10\). - Linearly connect points for \(x > 0\).

Identify the Range from the Graph

From the graph, observe the y-values:- From \(-3 \leq x < 0\): The function increases from \(-1\) to just less than 5.- At \(x = 0\): The function value is -3.- From \(x > 0\): The function decreases towards negative infinity as x increases.Overall range: \( (-\infty, -3] \cup (-1, 5) \).

Key concepts

Domain

The *domain* of a function includes all the x-values that the function can take. For a piecewise function, you must look at each piece separately and combine their intervals.

- For the first piece, given by the equation 2x + 5 for -3 ≤ x < 0, the domain is all x-values between -3 and 0, including -3 but not including 0.
- For the second piece, given by the constant -3 at x = 0, the domain includes exactly one point: x = 0.
- For the third piece, given by the equation -5x for x > 0, the domain includes all x-values greater than 0.

Now, combine these intervals to get the full domain of the piecewise function: \[-3 \leq x < \infty\]. This means that x can take any value from -3 to infinity, but not including positive infinity.

Intercepts

Understanding *intercepts* is crucial for graphing functions.

- **X-intercepts** occur where the function crosses the x-axis (when y = 0).
- **Y-intercepts** occur where the function crosses the y-axis (when x = 0).

Let's find the intercepts for each piece:
- For the piece 2x + 5 (valid for -3 ≤ x < 0):
- Find the x-intercept by setting 2x + 5 = 0 and solving for x, which gives x = -2.5.
- It doesn't have a y-intercept because x = 0 isn't within its domain.
- For the piece -3 (valid at x = 0):
- The y-intercept is -3, occurring exactly at (0, -3).
- It doesn't have an x-intercept because y = -3 is never zero.
- For the piece -5x (valid for x > 0):
- When x = 0, y = 0, but x > 0 means x = 0 isn't within its domain, so no intercept exists here.

Therefore, we have:
- X-intercept at (-2.5, 0)
- Y-intercept at (0, -3)

Graphing Functions

Graphing a piecewise function involves plotting each piece based on its specified range.
Follow these steps:
- **Plotting 2x + 5** for -3 ≤ x < 0:
- At x = -3, y = 2(-3) + 5 = -1. So the point is (-3, -1).
- At x = 0, y = 2(0) + 5 = 5. But since it doesn't include 0, we only consider values just less than 0.
- **Plotting -3** at x = 0:
- Simply plot the point (0, -3). This is just a single point as -3 is constant only at x = 0.
- **Plotting -5x** for x > 0:
- At x = 1, y = -5(1) = -5. So the point is (1, -5).
- At x = 2, y = -5(2) = -10. So the point is (2, -10).

Once these points are plotted, connect those in each piece's interval. Remember, only join points within the piece's domain. For 2x + 5, draw from (-3, -1) to almost (0, 5). For -5x, extend the line beyond (1, -5) and (2, -10).

Range

Determining the *range* means identifying possible y-values for the function.

From the graph or directly from the function:
- For -3 ≤ x < 0: As x changes from -3 to just under 0 in 2x + 5, y starts at -1 and approaches 5, but not including 5. So, the range covers [-1, 5).
- At x = 0: The value y = -3 makes -3 part of the range.
- For x > 0: As x grows positively, -5x becomes more negative. Hence, the range for this part is (-∞, 0].

Combining these: The total range of the function is \((-\infty, -3] \cup [-1, 5)\). Notice the union of separate segments capturing all y-values.