Intermediate Algebra: Functions and Graphs

1. The coordinates of point \(P\) tell us that \(f(15) = 2412\text{,}\) so the DJIA was 2412 at noon on October 15.

2. We can say that \(f(20) = 1726\text{,}\) so the point \((20, 1726)\) lies on the graph of \(f\text{.}\) This point is labeled \(Q\) in the figure above.

1. To find \(g(-2)\text{,}\) we look for the point with \(t\)-coordinate \(-2\text{.}\) The point \((-2, 0)\) lies on the graph of \(g\text{,}\) so \(g(-2) = 0\text{.}\) Similarly, the point \((5, 1)\) lies on the graph, so \(g(5) = 1\text{.}\)

2. We look for points on the graph with \(y\)-coordinate \(-2\text{.}\) Because the points \((-5, -2)\text{,}\) \((-3, -2)\text{,}\) and \((3, -2)\) lie on the graph, we know that \(g(-5) = -2\text{,}\) \(g(-3) = -2\text{,}\) and \(g(3) = -2\text{.}\) Thus, the \(t\)-values we want are \(-5\text{,}\) \(-3\text{,}\) and \(3\text{.}\)

3. The highest point on the graph is \((1, 4)\text{,}\) so the largest \(y\)-value is \(4\text{.}\) Thus, the maximum value of \(g(t)\) is \(4\text{,}\) and it occurs when \(t = 1\text{.}\)

4. A graph is increasing if the \(y\)-values get larger as we read from left to right. The graph of \(g\) is increasing for \(t\)-values between \(-4\) and \(1\text{,}\) and between \(3\) and \(5\text{.}\) Thus, \(g\) is increasing on the intervals \((-4, 1)\) and \((3, 5)\text{.}\)

We choose several convenient values for \(x\) and evaluate the function to find the corresponding \(f(x)\)-values. For this function we cannot choose \(x\)-values less than \(-4\text{,}\) because the square root of a negative number is not a real number.

The results are shown in the table.

\(x\)	\(f(x)\)
\(-4\)	\(0\)
\(-3\)	\(1\)
\(0\)	\(2\)
\(2\)	\(\sqrt{6}\)
\(5\)	\(3\)

Points on the graph have coordinates \((x, f(x))\text{,}\) so the vertical coordinate of each point is given by the value of \(f(x)\text{.}\) We plot the points and connect them with a smooth curve, as shown in the figure. Notice that no points on the graph have \(x\)-coordinates less than \(-4\text{.}\)

• Graph (a) represents a function, because it passes the vertical line test.

• Graph (b) is not the graph of a function, because the vertical line at (for example) \(x = 1\) intersects the graph at two points.

• For graph (c), notice the break in the curve at \(x = 2\text{:}\) The solid dot at \((2, 1)\) is the only point on the graph with \(x = 2\text{;}\) the open circle at \((2, 3)\) indicates that \((2, 3)\) is not a point on the graph. Thus, graph (c) is a function, with \(f(2) = 1\text{.}\)

If we sketch in the horizontal line \(y = 15\text{,}\) we can see that there are three points on the graph of \(f\) that have \(y\)-coordinate \(15\text{,}\) as shown below. The \(x\)-coordinates of these points are the solutions of the equation.

From the graph, we see that the solutions are \(x = -3\text{,}\) \(x = 1\text{,}\) and approximately \(x = 2.5\text{.}\) We can verify each solution algebraically.

For example, if \(x = \alert{-3}\text{,}\) we have

so \(-3\) is a solution. Similarly, you can check that \(x = 1\) and \(x = 2.5\) are solutions.

We first locate all points on the graph that have \(y\)-coordinates greater than or equal to \(15\text{.}\) The \(x\)-coordinates of these points are the solutions of the inequality.

The figure below shows the points in red, and their \(x\)-coordinates as intervals on the \(x\)-axis. The solutions are \(x \le -3\) and \(1\le x \le 2.5\text{,}\) or in interval notation, \((-\infty, -3] \cup [1, 2.5]\text{.}\)

To find the value of \(C\) that corresponds to \(x=\alert{500}\text{,}\) we evaluate \(C(\alert{500})\text{:}\)

The cost of producing 500 backpacks is $13,000.