Problem 22
Question
In Exercises \(21-36,\) find the absolute maximum and minimum values of each function on the given interval. Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates. $$ f(x)=-x-4, \quad-4 \leq x \leq 1 $$
Step-by-Step Solution
Verified Answer
The absolute maximum is 0 at \((-4, 0)\), and the absolute minimum is -5 at \((1, -5)\).
1Step 1: Identify the Function and Interval
The given function is \( f(x) = -x - 4 \), and the interval is \(-4 \leq x \leq 1\). Our task is to find the absolute maximum and minimum values of this function within the given interval.
2Step 2: Evaluate the Function at End Points
To find absolute extrema, we evaluate the function at the endpoints of the interval. Calculate \( f(x) \) at \( x = -4 \) and \( x = 1 \).- \( f(-4) = -(-4) - 4 = 4 - 4 = 0 \)- \( f(1) = -(1) - 4 = -1 - 4 = -5 \)
3Step 3: Compare and Determine Extrema
Compare the values obtained from evaluating at the endpoints. The largest value is the absolute maximum, and the smallest is the absolute minimum.- Absolute maximum: \( f(-4) = 0 \)- Absolute minimum: \( f(1) = -5 \)
4Step 4: Identify Coordinates of Extrema
Identify the coordinates on the graph where the extrema occur. - The absolute maximum point is \((-4, 0)\)- The absolute minimum point is \((1, -5)\)
5Step 5: Graph the Function with Extrema Points
Draw the graph of the function \( f(x) = -x - 4 \) from \( x = -4 \) to \( x = 1 \). Mark the points \((-4, 0)\) and \((1, -5)\) on the graph as the locations of the absolute maximum and minimum, respectively.
Key Concepts
Function EvaluationGraphing FunctionsEndpointsCoordinates of Extrema
Function Evaluation
When it comes to determining the absolute extrema of a function, the evaluation at specific points is crucial. Function evaluation is the process of obtaining the function's output, or value, given a specific input, which is usually an x-value. In the original exercise, we evaluate the linear function \( f(x) = -x - 4 \) at the endpoints of the interval \([-4, 1]\). This gives us outputs at \( x = -4 \) and \( x = 1 \).
- At \( x = -4 \), the function evaluates as \( f(-4) = 0 \).
- At \( x = 1 \), the function evaluates as \( f(1) = -5 \).
Graphing Functions
Graphing functions provides a visual way to understand how a function behaves over a specific interval. For the function \( f(x) = -x - 4 \), we need to create a graph between \( x = -4 \) and \( x = 1 \). A linear function like this results in a straight line graph. To graph, we plot points using the function evaluation results from the endpoints:
- The point \((-4, 0)\) is plotted for \( x = -4 \).
- The point \((1, -5)\) is plotted for \( x = 1 \).
Endpoints
Endpoints are crucial in analyzing functions over a closed interval. They mark the beginning and end of the interval over which we analyze the function. In this case, the given endpoints are \( x = -4 \) and \( x = 1 \).Evaluating a function at its endpoints allows us to determine potential extrema:
- Endpoints represent the boundary of our interval.
- Extrema, such as maximum and minimum values, are often located at these points.
Coordinates of Extrema
Finding the coordinates of extrema is the process of identifying the specific points where a function reaches its highest or lowest value within an interval. For the function \( f(x) = -x - 4 \), we evaluated the function at both endpoints and found:
- The absolute maximum value occurs at the coordinate \((-4, 0)\).
- The absolute minimum value exists at the coordinate \((1, -5)\).
Other exercises in this chapter
Problem 22
In Exercises \(17-56,\) find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answe
View solution Problem 22
Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. \begin{equation} y=x\left(\frac{x}{2}-5\right)^{4}
View solution Problem 22
a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying wher
View solution Problem 23
The graphs of \(y=x^{2}(x+1)\) and \(y=1 / x(x>0)\) intersect at one point \(x=r .\) Use Newton's method to estimate the value of \(r\) to four decimal places.
View solution