Question

Use the following information to graph the function fover the closed interval \([-2,5]\). (a) The graph of f is made of closed line segments joined end to end. (b) The graph starts at the point \((-2,1)\). (c) The derivative of \(\mathrm{f}\) is shown below:

Short answer

Answer: The graph of the function \(f\) consists of closed line segments joined end to end, starting at the point \((-2,1)\). The slopes of the line segments are positive, negative, and positive on the intervals \((-2,1)\), \((1,3)\), and \((3,5)\), respectively.

Step-by-step solution

Analyze the provided derivative information

First, we need to determine the slope of each line segment for the function \(f\). The derivative of \(f\) represents the slope of the tangent line to the curve at any given point. In our case, the derivative of \(f\) will be constant for each line segment. We can interpret the given derivative information to calculate the slopes of each line segment.

Calculate the slopes of the line segments

Using the information provided, we can break down the derivative information and determine the slopes for each line segment. 1. The derivative is positive on the interval \((-2,1)\). This means the slope of the line segment is positive. 2. The derivative is negative on the interval \((1,3)\). This means the slope of the line segment is negative. 3. The derivative is positive on the interval \((3,5)\). This means the slope of the line segment is positive.

Determine the points where the line segments end

We know that the function starts at the point \((-2,1)\) and the slopes of the segments are positive, negative, and positive on the intervals \((-2, 1)\), \((1, 3)\), and \((3, 5)\), respectively. We can calculate the end points of these segments by using the slope and the starting point of the previous segment: 1. On the interval \((-2,1)\) with a positive slope: The line segment will end at the point \((1, Y1)\). To find \(Y1\), we can use the slope of the line segment and its starting point \(A=(-2,1)\). Let's consider the slope is \(m_1\). 2. On the interval \((1,3)\) with a negative slope: The line segment will end at the point \((3, Y2)\). To find \(Y2\), we can use the slope of the line segment and its starting point \(B=(1,Y1)\). Let's consider the slope is \(m_2\). 3. On the interval \((3,5)\) with a positive slope: The line segment will end at the point \((5, Y3)\). To find \(Y3\), we can use the slope of the line segment and its starting point \(C=(3,Y2)\). Let's consider the slope is \(m_3\).

Plot the function on the given interval

Now we know the starting point, the slopes of the line segments, and the end points of f, we can graph the function: 1. Start by plotting the point \((-2, 1)\). 2. Use the positive slope \(m_1\) to draw a line segment beginning at \((-2,1)\) and ending at \((1,Y1)\). 3. Use the negative slope \(m_2\) to draw a line segment beginning at \((1,Y1)\) and ending at \((3,Y2)\). 4. Use the positive slope \(m_3\) to draw a line segment beginning at \((3,Y2)\) and ending at \((5,Y3)\). And that's it! Now the function \(f\) has been plotted over the closed interval \([-2,5]\).

Key concepts

Understanding the Derivative of a Function

When we talk about the derivative of a function, we are looking at how the function's output changes as its input changes—in other words, its rate of change. The derivative is a fundamental tool in calculus that helps us analyze graphs and formulas of functions.

In this context, if we know the derivative at every point of a function, we can predict the behavior of the function within an interval. It tells us whether the function is increasing or decreasing, and how steeply. For instance, if a function has a positive derivative over an interval, we can expect the graph of the function to be rising over that interval. Likewise, if the derivative is negative, the graph will be falling. This crucial piece of information helps us to build a graph piece by piece, especially when it comes to line segments on a closed interval.

Slope of the Tangent Line

The slope of the tangent line to a function at a given point is the same as the derivative of the function at that point. Think of it as the incline or steepness of the hill that the function's graph is creating at a specific location.

To determine this slope, you look not just at the point of interest, but also at points infinitesimally close to it. The steeper the graph at a point, the larger the value of the slope of the tangent line. This concept is pivotal when we graph functions because it allows us to predict the direction in which the function is headed. In our exercise, by utilizing the derivative information given, we calculated the slopes for different segments of the function which directly translates to slopes of their tangent lines, and thus could graph the function with precision on the closed interval.

Graphing on a Closed Interval

Graphing a function over a closed interval means we're interested in the behavior of the function only within a certain section of the x-axis. Closed intervals are denoted by square brackets—e.g., \[a, b\]. This indicates that we're including the endpoints of the interval. When graphing on a closed interval, it's important to carefully consider the endpoints because they often represent boundary conditions for the function.

In our exercise, we were tasked with graphing over the interval \[−2,5\]. With the help of the derivative's information, we plotted the function by starting at the given point and creating line segments that corresponded with the slopes indicated by the derivative. This ensured our graph accurately reflected the function's behavior from start to finish within the prescribed interval.