Question

the graph of a piecewise - defined function is given. write a definition for the function that best describes this graph.
$f(x)=\begin{cases}-x\text{ if } - 3leq xleqsquare\\1\text{ if }0lt xleqsquareend{cases}$
(simplify your answers. use integers or fractions for any numbers in the expressions.)

Explanation:

Step1: Analyze the first piece

The first part of the piecewise function is \( f(x)=-x \). Let's find its domain. From the graph, the line with slope -1 (since \( y = -x \)) passes through points like \((-3, 3)\) and \((0, 0)\). Wait, actually, let's check the points. Wait, the first segment: let's see the x - values. The left - most point of the first segment (the line with negative slope) seems to be when \( x=-3 \) (since the point is \((-3, 3)\)) and it goes up to \( x = 0 \) (since at \( x = 0 \), \( y=0 \)). Wait, but the given form has \( -x \) if \( - 3\leq x\leq\square \). Let's check the value at \( x = 0 \), \( f(0)=-0 = 0 \), which matches the point \((0,0)\). Then the next segment: when \( 0\lt x\leq\square \), \( f(x) = 1 \)? Wait, no, wait the graph: after \( x = 0 \), the next part? Wait, no, looking at the graph, there are two segments? Wait, no, the points are \((-3,3)\), \((0,0)\) and \((3, - 3)\)? Wait, no, the user's graph has points \((-3,2)\)? Wait, maybe I misread. Wait, the problem says "the graph of a piecewise - defined function is given". Let's re - examine.

Wait, the first part: the function \( f(x)=-x \). Let's find the domain for the first part. Let's take two points on the first line. Suppose the first line passes through \((-3,3)\) (since \( f(-3)=-(-3)=3 \)) and \((0,0)\) (since \( f(0)=-0 = 0 \)). So the domain for the first part \( f(x)=-x \) is \( - 3\leq x\leq0 \).

Step2: Analyze the second piece

The second part of the piecewise function is \( f(x)=1 \)? Wait, no, wait the graph. Wait, maybe I made a mistake. Wait, the second segment: after \( x = 0 \), let's see the points. Wait, the point \((3,-3)\)? No, wait the user's graph has a point \((3, - 3)\)? Wait, no, the original problem's graph: let's look at the coordinates. The first segment: from \( x=-3 \) to \( x = 0 \), the function is \( y=-x \) (since when \( x=-3 \), \( y = 3 \); \( x = 0 \), \( y = 0 \)). The second segment: from \( x = 0 \) to \( x = 3 \), let's check the slope. Wait, if we take two points, say \((0,0)\) and \((3,-3)\), the slope is \( m=\frac{-3 - 0}{3 - 0}=-1 \), so the equation is \( y=-x \)? No, that can't be. Wait, maybe the second part is a constant? Wait, no, the given form is \( f(x)=\begin{cases}-x&\text{if }-3\leq x\leq\square\\1&\text{if }0\lt x\leq\square\end{cases} \)? Wait, no, maybe I misread the graph. Wait, the user's graph has points \((-3,2)\)? Wait, the original problem's image: the first point is \((-3,2)\), then \((0,0)\), then \((3,-3)\)? No, that doesn't fit. Wait, maybe the first segment: when \( x=-3 \), \( f(-3)=3 \) (since \( -(-3)=3 \)), and when \( x = 0 \), \( f(0)=0 \). Then the second segment: from \( x = 0 \) to \( x = 3 \), let's see, if \( x = 3 \), \( f(3)=-3 \) (since \( -3=-3 \)). Wait, no, maybe the domain for the first part is \( - 3\leq x\leq0 \), and for the second part, when \( 0\lt x\leq3 \), \( f(x)=-x \)? No, that can't be. Wait, the problem's given form is \( f(x)=\begin{cases}-x&\text{if }-3\leq x\leq\square\\1&\text{if }0\lt x\leq\square\end{cases} \)? Wait, maybe there is a mistake in my initial analysis. Wait, let's go back.

Wait, the first part: \( f(x)=-x \). Let's find the upper limit of its domain. At \( x = 0 \), \( f(0)=0 \). Then the second part: when \( 0\lt x\leq3 \), what is \( f(x) \)? Wait, if we look at the graph, maybe the second segment is a line with slope 1? No, the point \((3,-3)\) would have \( y=-x \). Wait, maybe the domain for the first part \( f(x)=-x \) is \( - 3\leq x\leq0 \), and for the second part, when \( 0\lt x\leq3 \), \( f(x)=-x \)? No, that would make it a single function. Wait, no, the problem's given form has two pieces: one is \( -x \) with domain \( -3\leq x\leq\square \), and the other is 1 with domain \( 0\lt x\leq\square \). Wait, maybe the graph has a horizontal line after \( x = 0 \)? But the point \((3,-3)\) doesn't fit. Wait, maybe the user made a typo, but according to the given structure:

For the first piece \( f(x)=-x \), the domain is \( -3\leq x\leq0 \) (because when \( x=-3 \), \( f(-3)=3 \); when \( x = 0 \), \( f(0)=0 \)).

For the second piece, when \( 0\lt x\leq3 \), let's check: if \( x = 3 \), what is \( f(3) \)? If we follow the line \( y=-x \), \( f(3)=-3 \), but the given form has \( f(x)=1 \)? No, that's not right. Wait, maybe the second piece is \( f(x)=-x \) as well? No, the problem's form is different. Wait, maybe the first piece: domain \( -3\leq x\leq0 \), and the second piece: domain \( 0\lt x\leq3 \), and the function for the second piece is also \( -x \)? But that would be a single function. Wait, no, the graph must have two pieces. Wait, maybe the first piece is \( f(x)=-x \) for \( -3\leq x\leq0 \), and the second piece is \( f(x)=x \) for \( 0\lt x\leq3 \)? No, the slope would be positive. Wait, I think I made a mistake. Let's start over.

Let's consider the two - point form of a line. For the first segment: let's take two points \((x_1,y_1)=(-3,3)\) and \((x_2,y_2)=(0,0)\). The slope \( m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{0 - 3}{0+3}=-1 \). The equation of the line is \( y - y_1=m(x - x_1) \), so \( y - 3=-1(x + 3) \), which simplifies to \( y=-x \). So this line is \( y = -x \) with \( x\in[-3,0] \).

For the second segment: let's take two points \((x_1,y_1)=(0,0)\) and \((x_2,y_2)=(3,-3)\). The slope \( m=\frac{-3-0}{3 - 0}=-1 \), and the equation is \( y-0=-1(x - 0) \), so \( y=-x \). Wait, that's the same function. But the problem's piecewise form has two pieces. Maybe the graph is different. Wait, the problem's given form has \( f(x)=\begin{cases}-x&\text{if }-3\leq x\leq\square\\1&\text{if }0\lt x\leq\square\end{cases} \). Maybe the second piece is a horizontal line. Wait, if \( f(x)=1 \) for \( 0\lt x\leq3 \), but the point \((3,-3)\) doesn't fit. This is confusing. Wait, maybe the first piece: domain \( -3\leq x\leq0 \), and the second piece: domain \( 0\lt x\leq3 \), and the function for the second piece is \( -x \). But the problem's form is written as two pieces. Wait, maybe the first piece is \( -x \) for \( -3\leq x\leq0 \), and the second piece is \( -x \) for \( 0\lt x\leq3 \), but that's not piecewise in a non - trivial way. Alternatively, maybe the first piece is \( -x \) for \( -3\leq x\leq0 \), and the second piece is \( 1 \) for \( 0\lt x\leq3 \), but that doesn't match the graph. Wait, maybe the user's graph has a typo, but according to the given structure, let's assume that for the first part \( f(x)=-x \), the domain is \( -3\leq x\leq0 \), and for the second part, when \( 0\lt x\leq3 \), we need to find the upper limit. Since the line \( y=-x \) at \( x = 3 \) gives \( y=-3 \), so if we follow the first function's pattern, the second part's domain upper limit is 3. But the second function is given as 1? No, that's wrong. Wait, maybe I misread the function. Wait, the problem says "Simplify your answers. Use integers or fractions for any numbers in the expressions."

Wait, let's re - express:

The first piece: \( f(x)=-x \), domain \( -3\leq x\leq0 \) (because when \( x=-3 \), \( f(-3)=3 \); \( x = 0 \), \( f(0)=0 \)).

The second piece: when \( 0\lt x\leq3 \), what is \( f(x) \)? If we look at the graph, the line continues as \( y=-x \), but the problem's form has \( f(x)=1 \)? No, that's incorrect. Wait, maybe the second piece is \( f(x)=-x \) as well, but the problem's form is written with two pieces. Maybe the first piece is \( -x \) for \( -3\leq x\leq0 \), and the second piece is \( -x \) for \( 0\lt x\leq3 \), but that's a single function. Alternatively, maybe the graph has a horizontal line at \( y = 1 \) for \( 0\lt x\leq3 \), but that doesn't match the points. I think there is a misinterpretation. Let's go with the given structure:

For the first part, \( f(x)=-x \), the domain is \( -3\leq x\leq0 \) (since at \( x=-3 \), \( f(-3)=3 \); at \( x = 0 \), \( f(0)=0 \)).

For the second part, \( f(x)=1 \) (given in the problem's form) with domain \( 0\lt x\leq3 \) (since the line \( y=-x \) at \( x = 3 \) would be \( y=-3 \), but maybe the graph is different. Wait, the problem's form is:

\( f(x)=\begin{cases}-x&\text{if }-3\leq x\leq\boldsymbol{0}\\1&\text{if }0\lt x\leq\boldsymbol{3}\end{cases} \)

Wait, no, that doesn't make sense. Wait, maybe the first piece: \( -3\leq x\leq0 \), and the second piece: \( 0\lt x\leq3 \), and the function for the second piece is \( -x \). But the problem's form has \( 1 \) as the function. I think there is a mistake in the problem's given function form, but based on the line \( y=-x \) passing through \((-3,3)\), \((0,0)\) and \((3,-3)\), the piecewise function should be \( f(x)=-x \) for \( -3\leq x\leq3 \), but it's given as two pieces. So, assuming the first piece is \( -x \) with domain \( -3\leq x\leq0 \), and the second piece is \( -x \) with domain \( 0\lt x\leq3 \), but the problem's form has \( 1 \) as the second function. I think the intended answer is:

For the first piece, the domain is \( -3\leq x\leq0 \), and for the second piece, the domain is \( 0\lt x\leq3 \). But the function for the second piece should be \( -x \), but the problem's form has \( 1 \). This is confusing. Alternatively, maybe the first piece: \( -3\leq x\leq0 \), and the second piece: \( 0\lt x\leq3 \), and the function for the second piece is \( -x \). So the completed function is:

\( f(x)=\begin{cases}-x&\text{if }-3\leq x\leq0\\-x&\text{if }0\lt x\leq3\end{cases} \)

But that's a single function. I think there is a mistake in the problem's function form. However, based on the given structure, the first blank (upper limit for the first piece) is \( 0 \), and the second blank (upper limit for the second piece) is \( 3 \).

Answer:

The piecewise function is \( f(x)=

$$\begin{cases}-x&\text{if }-3\leq x\leq0\\1&\text{if }0\lt x\leq3\end{cases}$$

\) (Note: There might be a discrepancy between the graph and the given function form, but based on the problem's structure, the domains are \( -3\leq x\leq0 \) for the first piece and \( 0\lt x\leq3 \) for the second piece.)

So the blanks are filled as: for the first piece, the upper limit of \( x \) is \( 0 \); for the second piece, the upper limit of \( x \) is \( 3 \).