Chapter 1: Problem 64
Use a computer or a graphing calculator Let \(f(x)=2 \sqrt{x}-2 x+0.25 x^{2}\). Using the same axes, draw the graphs of \(y=f(x), y=f(1.5 x)\), and \(y=\) \(f(x-1)+0.5\), all on the domain \([1,5]\).
Short Answer
Expert verified
Graph all three functions within the domain \\( [1,5] \\); check transformations: \( f(1.5x) \) stretches and \( f(x-1) + 0.5 \) shifts.
Step by step solution
01
Understand Function Transformations
First, we recognize the three functions we need to graph: (1) The original function \(y = f(x) = 2 \sqrt{x} - 2x + 0.25x^2\), (2) The function stretched horizontally by multiplying \(x\) with 1.5, i.e., \(y = f(1.5x)\), and (3) The function shifted to the right by 1 unit and up by 0.5 units, i.e., \(y = f(x-1) + 0.5\).
02
Evaluate Points for y = f(x)
For the original function \(y = f(x)\), choose values of \(x\) in the domain \([1, 5]\) and substitute them into \(f(x) = 2 \sqrt{x} - 2x + 0.25x^2\). Calculate \(f(x)\) for each value to use as points for plotting. For example, \(x = 1, 2, 3, 4, 5\).
03
Evaluate Points for y = f(1.5x)
For the function \(y = f(1.5x)\), choose values of \(x\) in such a way that \(1.5x\) still remains in the domain \([1, 5]\). For example, take values like \(x = \frac{2}{3}, \frac{4}{3}, \frac{2}{3}, \frac{8}{3}, \frac{10}{3}\). Substitute into \(f(1.5x) = 2\sqrt{1.5x} - 2(1.5x) + 0.25(1.5x)^2\).
04
Evaluate Points for y = f(x-1) + 0.5
For the function \(y = f(x-1) + 0.5\), take values of \(x\) like \(x = 2, 3, 4, 5, 6\) which when substituted into \(f(x-1)\) remain in the domain \([1,5]\). This involves calculating \(f(x-1) = 2\sqrt{x-1} - 2(x-1) + 0.25(x-1)^2 + 0.5\).
05
Plot and Compare Graphs
Use a graphing tool or calculator to plot the points obtained for each function \(y = f(x)\), \(y = f(1.5x)\), and \(y = f(x-1) + 0.5\) on the same axes. Pay attention to the horizontal stretch in \(f(1.5x)\) and the shift in \(f(x-1) + 0.5\).
06
Analyze Graph Behavior
Observe and compare how the transformations affect the shape and position of the original graph. Specifically, check for amplitude changes, shifts, and stretches.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graphing Calculators
Graphing calculators are powerful tools that help visualize and analyze mathematical functions. They allow students to see the effects of transforming a function in real-time. In this exercise, we used a graphing calculator to plot different versions of the function \( f(x) = 2\sqrt{x} - 2x + 0.25x^2 \).
- Using these calculators, you can easily input complex functions and observe their graphs on the screen.
- Graphing calculators often provide features to calculate specific values, evaluate derivatives, and even solve equations.
- By using technology, students can quickly plot multiple functions on the same axes for comparison.
Domain and Range
Understanding the domain and range of a function is crucial when dealing with transformations. The domain of a function is all the possible input values (\(x\)-values) it can accept, while the range is all the possible output values (\(y\)-values).
For the original function \( f(x) = 2\sqrt{x} - 2x + 0.25x^2 \), the domain was restricted to \([1,5]\).
For the original function \( f(x) = 2\sqrt{x} - 2x + 0.25x^2 \), the domain was restricted to \([1,5]\).
- For \( y = f(x) \), the domain directly follows the given limits.
- When examining \( y = f(1.5x) \), the domain changes are subtle. The input \( 1.5x \) must still abide by the original domain, so we solve \( 1 \, \leq \, 1.5x \, \leq \, 5 \) to find appropriate \( x \) values.
- Shifting the function, as in \( y = f(x-1) + 0.5 \), shifts the domain as well. Here, \( x-1 \) must fit within \([1,5]\), affecting which \( x \) values are usable.
Function Evaluation
Function evaluation is the process of finding the output of a function for a specific input value. It serves as a foundational pillar in graphing and transformation exercises.
Consider the function \( f(x) = 2\sqrt{x} - 2x + 0.25x^2 \). Evaluating this means substituting chosen \( x \) values, such as 1, 2, or 3, and calculating the corresponding \( y \) values.
Consider the function \( f(x) = 2\sqrt{x} - 2x + 0.25x^2 \). Evaluating this means substituting chosen \( x \) values, such as 1, 2, or 3, and calculating the corresponding \( y \) values.
- For the original function, this calculation gives us points like \((1, f(1))\), \((2, f(2))\), useful for plotting.
- The transformation \( y = f(1.5x) \) adjusts by inputting values of \( 1.5x \) and calculating the result. This ensures the function is appropriately evaluated over its new adjusted domain.
- Lastly, \( y = f(x-1) + 0.5 \) assesses the altered structure, adding an extra constant to account for vertical shifts.