Question

(a) Find the domain of each function. (b) Locate any intercepts. (c) Graph each function. (d) Based on the graph, find the range. $$f(x)=\left\\{\begin{array}{ll}2-x & \text { if }-3 \leq x<1 \\\\\sqrt{x} & \text { if } x>1\end{array}\right.$$

Short answer

Domain: \(]-3, 1[ \cup ]1, \infty[\). Intercepts: (0, 2) Graph as described. Range: \([1, \infty[\).

Step-by-step solution

- Find the domain of the function

The domain includes all the possible values of the variable x. For the function \(f(x)=\begin{cases}2-x & \text { if } -3 \leq x < 1 \ \sqrt{x} & \text { if } x > 1 \end{cases} \),we split it into two parts. The first part, \(2-x\), is defined for \(-3 \leq x < 1\), which means x can take any value from -3 to just less than 1. The second part, \(bsp;\sqrt{x}\), is defined for \(x > 1\), which means x must be greater than 1. Combining these two intervals, the domain is \(-3 \leq x < 1 \text{{ and }} x > 1\), which in interval notation is \(]-3, 1[ \cup ]1, \infty[\).

- Find the intercepts

To find the intercepts, we see where the function crosses the axes. For the y-intercept (x = 0), we use the piece of the function that covers 0, which is \(2-x\). Plugging x = 0 into \(2-x\) gives: \(bsp;f(0) = 2-0 = 2\). Therefore, the y-intercept is (0, 2). For the x-intercept (f(x) = 0), setting \(2-x = 0\) gives x = 2, but since \(x < 1\) in this part, there is no x-intercept in this portion. For \(bsp;\sqrt{x} = 0\), x must be 0, which is not within the domain of \(\sqrt{x}\). Therefore, there are no x-intercepts.

- Graph the function

Graph the piecewise function in two parts. First part: \(2-x\) from x = -3 to just before x = 1. This is a line with a downward slope. It starts at (2, -3) and ends just before (1, 1). Second part: \(\sqrt{x}\) starting just after x = 1. This is the square root function starting just after x = 1.

- Find the range based on the graph

Inspect the graph to determine all possible y-values the function can take. For the first part, as x goes from -3 to just less than 1, y starts from 5 and goes to 1. For the second part, as x goes from just over 1 to infinity, y starts from 1 and goes to infinity. Therefore, combining these intervals, the range is \([1, \infty[\).

Key concepts

Domain and Range

The domain of a function includes all possible values of the variable (x) that can be input into the function without causing any undefined behavior, such as division by zero or taking the square root of a negative number. For a piecewise function, the domain is determined by the individual domains of each piece.

For the function given:
\[f(x)=\begin{cases}2-x & \text{if } -3 \leq x < 1 \ \sqrt{x} & \text{if } x > 1 \end{cases}\]
The first part \(2-x\) is valid for \(-3 \leq x <1\). The second part \(\sqrt{x}\), is valid when \(x > 1\). Therefore, combining these intervals, the complete domain is \([-3, 1[\text{ and } ]1, \infty[\). In interval notation, this is \(]-3,1[ \cup ]1, \infty [\).

The range of a function is all possible output values (y-values) it can produce. For the graph of the first segment \(2-x\), y-values range from 5 (when \(x = -3\)) to just under 1 (as \(x\) approaches 1). For \(\sqrt{x}\), starting just after \(x >1\), y-values start at just over 1 up to infinity. Therefore, the range of the function is \([1, \infty[\).

Function Intercepts

Function intercepts are points where the graph crosses the axes. These include:

• Y-intercept: Where the function crosses the y-axis (x = 0). For our function, the relevant piece when \(x=0\) is \(2-x\), giving \(f(0)=2-0=2\), resulting in a y-intercept at (0, 2).
• X-intercepts: Where the function crosses the x-axis (f(x)=0). Setting \(2 - x = 0\): solving \(x=2\) but this doesn't fit within the part \( x < 1 \) of the domain. Thus, there are no x-intercepts for \(2 - x\). For \(\sqrt{x} = 0\), x=0 is not in the domain \(x>1\). Therefore, there are no x-intercepts.

Graphing Piecewise Functions

To graph a piecewise function, you need to plot each piece within its domain:

• First part, \(2-x\): Plot this line from \(x = -3\) to just before \(x = 1\). It has a downward slope with points like (0, 2) and ends just before (1, 1).
• Second part, \(\sqrt{x}\): Start graphing this just after \(x = 1\), beginning just above (1, 1) and then the curve moves upwards.

Use different colors or line styles to distinguish different pieces. Expect to make jumps or gaps at domain boundaries to illustrate where each piece applies.