Problem 21
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. $$ f(x)=\frac{2}{3} x-5, \quad-2 \leq x \leq 3 $$
Step-by-Step Solution
Verified Answer
The maximum value is -3 at \(x=3\) and minimum is \(-\frac{19}{3}\) at \(x=-2\).
1Step 1: Understand the Function and Interval
The function given is a linear function, described by the equation \( f(x) = \frac{2}{3}x - 5 \). The interval of interest is \(-2 \leq x \leq 3\). Our task is to find the absolute maximum and minimum values of this function such that \( x \) is within this interval.
2Step 2: Evaluate the Function at Interval Endpoints
To find absolute extrema on a closed interval for a linear function, evaluate the function at each endpoint. For \( x = -2 \), \( f(x) = \frac{2}{3}(-2) - 5 = -\frac{4}{3} - 5 = -\frac{19}{3} \). For \( x = 3 \), \( f(x) = \frac{2}{3}(3) - 5 = 2 - 5 = -3 \).
3Step 3: Identify the Absolute Extremes
Comparing the values at each endpoint, \( f(-2) = -\frac{19}{3} \) and \( f(3) = -3 \), \(-\frac{19}{3} \approx -6.33\) is less than \(-3\). Therefore, the absolute minimum value is \( f(-2) = -\frac{19}{3} \) and the absolute maximum value is \( f(3) = -3 \).
4Step 4: Plot the Function and Identify Extrema Points
On graphing the function, which is a line, we plot the points \((-2, -\frac{19}{3})\) and \((3, -3)\). The line steadily increases from left to right. The point \((-2, -\frac{19}{3})\) is the minimum, and the point \((3, -3)\) is the maximum within the given interval.
Key Concepts
Absolute MaximumAbsolute MinimumClosed IntervalLinear Function Analysis
Absolute Maximum
An absolute maximum is the highest point over the entire range of a function on a specified interval. For a linear function like the one in our exercise, determining the maximum involves checking the function's output at the endpoints of the closed interval. As linear functions have constant slope and no turning points, the absolute maximum will be found at one of these endpoints.
Here, we evaluated the function at the boundary points of the interval \(-2 \leq x \leq 3\). Calculating for these values showed that the endpoint at \(x = 3\) gave \(f(3) = -3\), which is the highest function value on our interval. Thus, this value represents the absolute maximum.
Here, we evaluated the function at the boundary points of the interval \(-2 \leq x \leq 3\). Calculating for these values showed that the endpoint at \(x = 3\) gave \(f(3) = -3\), which is the highest function value on our interval. Thus, this value represents the absolute maximum.
- Occurs at a single boundary of the interval.
- No other point within the interval exceeds this value.
Absolute Minimum
The absolute minimum refers to the lowest point in the value of the function over an entire interval. Similar to finding the maximum in a linear function, this involves checking the endpoints of our closed interval. Since the slope in linear functions does not change, we will also find the minimum at one of these two endpoints.
Our function on \(-2 \leq x \leq 3\) is evaluated at these interval limits. For \(x = -2\), the function value is \(f(-2) = -\frac{19}{3}\), which is approximately \(-6.33\). This is smaller compared to the other endpoint, thus representing the absolute minimum.
Our function on \(-2 \leq x \leq 3\) is evaluated at these interval limits. For \(x = -2\), the function value is \(f(-2) = -\frac{19}{3}\), which is approximately \(-6.33\). This is smaller compared to the other endpoint, thus representing the absolute minimum.
- Lowest function value within the defined interval.
- Found in a linear function at one of the interval's borders.
Closed Interval
A closed interval is a range of \(x\)-values for which the function is defined, including the endpoints. It's expressed as \[a, b\]\, meaning all \(x\) between and including \(a\) and \(b\). This is crucial in finding extrema as endpoints are potential locations for maximum and minimum values.
In the exercise, the closed interval is specified as \(-2 \leq x \leq 3\). By considering only closed intervals, we ensure that the function values at these boundary points are included when evaluating for absolute extrema. This inclusion is essential for complete linear function analysis as endpoints can provide the extremum points.
In the exercise, the closed interval is specified as \(-2 \leq x \leq 3\). By considering only closed intervals, we ensure that the function values at these boundary points are included when evaluating for absolute extrema. This inclusion is essential for complete linear function analysis as endpoints can provide the extremum points.
Linear Function Analysis
Linear function analysis involves understanding characteristics like slope, intercepts, and how they relate to graphing and determining extrema. A linear function such as \(f(x) = \frac{2}{3}x - 5\) draws a straight line when plotted.
- **Slope**: The coefficient of \(x\), here \(\frac{2}{3}\), indicates the rate of increase. A positive slope implies an upward trend as \(x\) increases.
- **Intercept**: The constant, \(-5\), denotes where the line crosses the \(y\)-axis.
- When analyzing linear functions over a closed interval, you verify at the endpoints to determine extreme values. Linear functions do not bend or change direction, making analysis straightforward. By checking the endpoints, one can conclusively determine where the absolute maximum and minimum lie within the interval.
Other exercises in this chapter
Problem 20
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 21
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 21
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 21
Show that the functions in Exercises \(19-26\) have exactly one zero in the given interval. \begin{equation}g(t)=\sqrt{t}+\sqrt{1+t}-4, \quad(0, \infty)\end{equ
View solution