Problem 30
Question
Sketch the graph of \(f(t)=4-t-|4-t|\) for \(2 \leqslant t \leqslant 5\) and find its slope and range.
Step-by-Step Solution
Verified Answer
The graph is 0 on [2,4] then decreases linearly with slope -2 on (4,5]. Range: [-2, 0].
1Step 1: Analyze the absolute value
For \( 2 \leq t \leq 4 \): \( 4-t \geq 0 \), so \( |4-t| = 4-t \). Then \( f(t) = 4-t-(4-t) = 0 \).
For \( 4 < t \leq 5 \): \( 4-t < 0 \), so \( |4-t| = t-4 \). Then \( f(t) = 4-t-(t-4) = 8-2t \).
For \( 4 < t \leq 5 \): \( 4-t < 0 \), so \( |4-t| = t-4 \). Then \( f(t) = 4-t-(t-4) = 8-2t \).
2Step 2: Describe the graph
On \( [2,4] \): \( f(t) = 0 \) (horizontal line).
On \( (4,5] \): \( f(t) = 8-2t \) with slope \( -2 \). At \( t=5 \): \( f(5) = -2 \).
Range: \( [-2, 0] \).
On \( (4,5] \): \( f(t) = 8-2t \) with slope \( -2 \). At \( t=5 \): \( f(5) = -2 \).
Range: \( [-2, 0] \).
Key Concepts
Graphing Piecewise FunctionsSlope AnalysisFinding Range
Graphing Piecewise Functions
Graphing piecewise functions involves plotting different expressions over different intervals. For the given function, \( f(t) = 4 - t - |4-t| \), we need to split it into pieces based on the behavior of the absolute value.
First, consider when \( 4 - t \geq 0 \) (or \( t \leq 4 \)). In this interval, \(|4-t| = 4-t\), which simplifies the function to \( f(t) = 4 - t - (4 - t) = 0 \). Hence, the graph is a horizontal line at \( f(t) = 0 \) for \( 2 \leq t \leq 4 \).
Next, consider when \( 4 - t < 0 \) (or \( t > 4 \)). Here, \(|4-t| = t - 4\), which simplifies the function to \( f(t) = 4 - t - (t - 4) = 8 - 2t \). This part of the graph is a straight line that decreases since its slope is negative. Plot this line from \( t = 4 \) to \( t = 5 \).
By combining these two parts, you've sketched the complete piecewise graph of \( f(t) \). The function changes behavior at \( t = 4 \), remembering that piecewise functions can have distinct segments.
First, consider when \( 4 - t \geq 0 \) (or \( t \leq 4 \)). In this interval, \(|4-t| = 4-t\), which simplifies the function to \( f(t) = 4 - t - (4 - t) = 0 \). Hence, the graph is a horizontal line at \( f(t) = 0 \) for \( 2 \leq t \leq 4 \).
Next, consider when \( 4 - t < 0 \) (or \( t > 4 \)). Here, \(|4-t| = t - 4\), which simplifies the function to \( f(t) = 4 - t - (t - 4) = 8 - 2t \). This part of the graph is a straight line that decreases since its slope is negative. Plot this line from \( t = 4 \) to \( t = 5 \).
By combining these two parts, you've sketched the complete piecewise graph of \( f(t) \). The function changes behavior at \( t = 4 \), remembering that piecewise functions can have distinct segments.
Slope Analysis
The slope in a function tells us how steep the line is, whether it rises or falls, and by how much for each unit increase in \( t \).
- For \( 2 \leq t \leq 4 \), we've determined \( f(t) = 0 \), which is a constant. A constant line has a slope of 0 because it does not rise or fall as \( t \) changes.
- For \( 4 < t \leq 5 \), \( f(t) = 8 - 2t \). This is a linear function with a slope of \(-2\). The negative slope indicates the line is decreasing. Specifically, it falls 2 units down for every unit increase in \( t \).
These slope changes are essential for analyzing how a function behaves over different intervals. It's crucial to identify where the changes occur to understand the full characteristics of the function.
- For \( 2 \leq t \leq 4 \), we've determined \( f(t) = 0 \), which is a constant. A constant line has a slope of 0 because it does not rise or fall as \( t \) changes.
- For \( 4 < t \leq 5 \), \( f(t) = 8 - 2t \). This is a linear function with a slope of \(-2\). The negative slope indicates the line is decreasing. Specifically, it falls 2 units down for every unit increase in \( t \).
These slope changes are essential for analyzing how a function behaves over different intervals. It's crucial to identify where the changes occur to understand the full characteristics of the function.
Finding Range
The range of a function is the set of possible output values (\( f(t) \) values) as \( t \) varies within the given domain.
Examining our function, for the first piece \( 2 \leq t \leq 4 \), \( f(t) = 0 \). So, the range here is just \{0\}. It's a single value because the line doesn't vary over this interval.
For the interval \( 4 < t \leq 5 \), we have \( f(t) = 8 - 2t \). If you substitute the endpoints:
Therefore, the complete range of \( f(t) \) over the interval \( 2 \leq t \leq 5 \) is \{0, [-2, 0)\}. This accounts for the piece at the start remaining constant and the decreasing values in the second interval.
Examining our function, for the first piece \( 2 \leq t \leq 4 \), \( f(t) = 0 \). So, the range here is just \{0\}. It's a single value because the line doesn't vary over this interval.
For the interval \( 4 < t \leq 5 \), we have \( f(t) = 8 - 2t \). If you substitute the endpoints:
- At \( t = 4.1 \) (just above 4), \( f(t) \) is nearly 0, showing it approaches 0 closely, though never quite reaching it.
- At \( t = 5 \), \( f(t) = 8 - 10 = -2 \).
Therefore, the complete range of \( f(t) \) over the interval \( 2 \leq t \leq 5 \) is \{0, [-2, 0)\}. This accounts for the piece at the start remaining constant and the decreasing values in the second interval.
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