Question

(a) Find the domain of each function. (b) Locate any intercepts. (c) Graph each function. (d) Based on the graph, find the range. $$f(x)=\left\\{\begin{array}{ll}2 x & \text { if } x \neq 0 \\\1 & \text { if } x=0\end{array}\right.$$

Short answer

The domain is \( (-\infty, \infty) \, the y-intercept is \((0,1)\), there is no x-intercept, and the range is \( (-\infty, 0) \cup (0, \infty)\). Graph the line \(2x\) for \(x eq 0\) and point at \((0,1)\).

Step-by-step solution

- Find the Domain

The domain of a function consists of all the possible input values (x-values). For the function given, \(f(x) = 2x\) is defined for all \(x eq 0\), and \(f(x) = 1\) is specifically defined for \(x = 0\). Therefore, the domain of the function is all real numbers. In interval notation, this is \( (-\infty, \infty) \).

- Locate the Intercepts

To find the intercepts, determine the points where the function crosses the x-axis and the y-axis. \For the y-intercept, evaluate \(f(0)\): \( f(0) = 1 \) \So the y-intercept is at \( (0, 1)\). \For the x-intercept, set \(f(x)=0\) and solve for x. Since \(f(x) = 2x\) when \(x eq 0\), \0 = 2x implies x = 0. But when \(x = 0\), \(f(x) = 1\), so the function never actually crosses the x-axis. Thus, there is no x-intercept.

- Graph the Function

To graph the function, plot the two parts. \- For \(x eq 0\), the function is a straight line \(f(x) = 2x\). This will be a line passing through the origin but excluding the point \((0, 0)\). \- For \(x = 0\), the point will be \((0, 1)\). \Combine these two graphs accordingly.

- Find the Range

The range of a function is the set of all possible output (y-values). Based on the graph: \- For \(x eq 0\), \(f(x) = 2x\) can take any value except 0. \- When \(x = 0\), \(f(x) = 1\). \Therefore, the range is all real numbers except 0, plus 1. The range is \( (-\infty, 0) \cup (0, \infty) \).

Key concepts

Domain of a Function

The domain of a function includes all the possible input values, or x-values, that you can plug into the function and get a valid output. For the given piecewise function:
$$f(x)=\begin{cases} 2x & \text{if } x eq 0 \ 1 & \text{if } x = 0 \end{cases}$$
It's important to look at the different segments of the function to understand its domain. Here, the function \(2x\) works for all x-values except when \(x = 0\). But the function is also defined as \(1\) when \(x = 0\).
Therefore, when you combine these intervals, the domain covers all real numbers.
**In interval notation**, this is $$(-\infty, \infty)$$. This means you can input any real number into the function, and it will give you a defined output.

Intercepts of a Function

Intercepts are points where the graph of the function crosses the x-axis or the y-axis. Finding these helps in understanding the behavior of the function.
**Finding the y-intercept:** This is where the function crosses the y-axis (when x = 0).
For the given function, if \(x = 0\), then $$f(0) = 1$$. So, the y-intercept is at $$(0, 1)$$.
**Finding the x-intercept:** This is where the function crosses the x-axis (when \(f(x) = 0\)).
For the given function, let's set \(f(x) = 0\): $$0 = 2x$$
Solving for x gives $$x = 0$$. But remember, when \(x = 0\), the function's value is actually $$f(0) = 1$$, not zero.
This means the function never actually crosses the x-axis, so there is no x-intercept.

Graphing Piecewise Functions

Graphing piecewise functions involves plotting different parts of the function on the same set of axes.
For the function:
$$f(x)=\begin{cases} 2x & \text{if } x eq 0 \ 1 & \text{if } x = 0 \end{cases}$$
**Step-by-step process:**

• Start with the part of the function defined for \(x eq 0\). This is the line \(f(x) = 2x\).
• Plot the line \(f(x) = 2x\), but leave out the point where \(x = 0\).


• Next, plot the point where \(x = 0\). Here, \(f(x) = 1\), so the point is \((0, 1)\).

Combining these two parts gives you the graph of the piecewise function. You'll have a straight line for \(x eq 0\), and a separate point at \((0, 1)\).

Range of a Function

The range of a function is the set of all possible output values, or y-values, that the function can produce.
For our function:
$$f(x)=\begin{cases} 2x & \text{if } x eq 0 \ 1 & \text{if } x = 0 \end{cases}$$
**Step-by-step process:**

• Consider the part of the function where \(x eq 0\): $$f(x) = 2x$$. This will give you all possible y-values except for 0, because \(2x = 0\) only when \(x = 0\), but at \(x = 0\), the function takes the value of 1.
• Next, consider the value at \(x = 0\): $$f(0) = 1$$. Therefore, 1 is included in the range.

By combining these observations, we get that the range of the function covers all real numbers except 0, plus 1.
**In interval notation,** this is written as: $$(-\infty, 0) \cup (0, \infty)$$.