Question

give the domain and range. then, use the graph to find (a) f(-3), (b) f(0), (c) f(1), and (d) any values of x such that f(x)=2.

Explanation:

Step1: Find Domain

The domain is the set of all x - values for which the function is defined. From the graph, the leftmost point has an x - value of - 6 (a closed dot, so included) and the rightmost point has an x - value of 3 (a closed dot, so included). So, the domain is \([-6,3]\).

Step2: Find Range

The range is the set of all y - values the function takes. The lowest y - value on the graph is 0 (the vertex of the "V" - shaped parts) and the highest y - value is 6 (at the peak). So, the range is \([0,6]\).

Step3: Find \(f(-3)\)

To find \(f(-3)\), we look at the x - value of - 3 on the graph. We move vertically until we hit the graph. The point at \(x=-3\) is on the left - hand "V" - shaped part. By looking at the grid, when \(x = - 3\), the y - value (which is \(f(-3)\)) is 3.

Step4: Find \(f(0)\)

For \(x = 0\), we look at the graph. The point at \(x = 0\) is the peak of the right - hand "V" - shaped part. The y - value at \(x = 0\) is 6, so \(f(0)=6\).

Step5: Find \(f(1)\)

To find \(f(1)\), we look at \(x = 1\) on the graph. The graph at \(x = 1\) is on the right - hand "V" - shaped part (the line going from \((0,6)\) to \((3,0)\)). The slope of the line from \((0,6)\) to \((3,0)\) is \(m=\frac{0 - 6}{3-0}=\frac{-6}{3}=-2\). The equation of the line is \(y-6=-2(x - 0)\), or \(y=-2x + 6\). When \(x = 1\), \(y=-2(1)+6 = 4\). So, \(f(1)=4\).

Step6: Find \(x\) such that \(f(x)=2\)

We need to find the x - values where the y - value is 2. We look at the graph. There are two points where \(y = 2\).

• For the left - hand "V" - shaped part (from \((-6,6)\) to \((-3,0)\)): The slope of the line from \((-6,6)\) to \((-3,0)\) is \(m=\frac{0 - 6}{-3+6}=\frac{-6}{3}=-2\). The equation is \(y - 6=-2(x + 6)\), \(y=-2x-12 + 6=-2x - 6\). Set \(y = 2\): \(2=-2x-6\), \(2x=-8\), \(x=-4\) (Wait, no, let's check the graph again. The left - hand "V" goes from \((-6,6)\) to \((-3,0)\)? Wait, no, the left - hand "V" is from \((-6,6)\) to \((-3,0)\)? Wait, actually, the left "V" is from \((-6,6)\) to \((-3,0)\)? No, looking at the graph, the left "V" has a vertex at \((-3,0)\) (wait, no, the vertex of the left "V" is at \((-3,0)\)? Wait, the graph: the left part is from \((-6,6)\) down to \((-3,0)\), and the middle part is from \((-3,0)\) up to \((0,6)\), and the right part is from \((0,6)\) down to \((3,0)\). Wait, my mistake earlier. The left "V" is from \((-6,6)\) to \((-3,0)\), the middle "V" is from \((-3,0)\) to \((0,6)\), and the right "V" is from \((0,6)\) to \((3,0)\).

For the middle "V" (from \((-3,0)\) to \((0,6)\)): slope \(m=\frac{6 - 0}{0+3}=2\). Equation: \(y-0 = 2(x + 3)\), \(y = 2x+6\). Set \(y = 2\): \(2=2x + 6\), \(2x=-4\), \(x=-2\).

For the right "V" (from \((0,6)\) to \((3,0)\)): slope \(m=\frac{0 - 6}{3-0}=-2\). Equation: \(y-6=-2x\), \(y=-2x + 6\). Set \(y = 2\): \(2=-2x+6\), \(2x = 4\), \(x = 2\).

So, the x - values where \(f(x)=2\) are \(x=-2\) and \(x = 2\).

Answer:

• Domain: \(\boldsymbol{[-6,3]}\)
• Range: \(\boldsymbol{[0,6]}\)
• (a) \(f(-3)=\boldsymbol{3}\)
• (b) \(f(0)=\boldsymbol{6}\)
• (c) \(f(1)=\boldsymbol{4}\)
• (d) \(x=\boldsymbol{-2}\) and \(x=\boldsymbol{2}\)