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Chapter 4: Problem 1

Use a graphing utility to graph the function. $$f(x)=\arctan (2 x-3)$$

Short Answer

Expert verified
The graph of the function \(f(x)=\arctan (2 x-3)\) is, effectively, the graph of \(y = \arctan(x)\) horizontally compressed by a factor of 2 and shifted 3/2 units to the right.

Step by step solution

01

Understand the Base Function

Firstly, the function \(f(x) = \arctan(x) \) should be understood. It is a function whose range is \(-\pi/2\) to \(\pi/2\), it crosses the x-axis at x = 0, and it has horizontal asymptotes at y = \( \pi/2 \) and y = \( -\pi/2 \).
02

Identify the horizontal shift and vertical stretch

The function \(f(x)=\arctan (2 x-3)\) is based on \(f(x)=\arctan (x)\) but has been modified by a horizontal compression by a factor of 2 (which makes the function narrower), and a shift of 3/2 to the right.
03

Plot the function using a graphing utility

Enter \(f(x) = \arctan (2 x-3)\) into a graphing utility to observe the transformation. Compare this graph with the base function to verify your understanding of how the function has been modified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Shift
When graphing functions, a horizontal shift occurs when the entire graph of a function moves left or right along the x-axis.
In the equation \( f(x) = \arctan (2x - 3) \), the term \(-3\) inside the parenthesis is responsible for the horizontal shift.
This term moves the entire graph of the \(\arctan(x)\) function.
  • To determine the extent of the horizontal shift, set the inside of the parenthesis equal to zero: \(2x - 3 = 0\).
  • Solving for \(x\), we find \(x = \frac{3}{2}\).
  • This tells us that the graph shifts to the right by \(\frac{3}{2}\) units.
Notice that shifting the graph changes the point at which it crosses the x-axis. Remember, the graph aligns itself accordingly by shifting every point along with it.
Vertical Stretch
A vertical stretch happens when a function's graph becomes taller or shorter, depending on how the function's values are multiplied.
In our equation \( f(x) = \arctan (2x - 3) \), the coefficient of \(x\), which is 2, affects this transformation.
  • This coefficient results in what's known as a horizontal compression (or vertical stretch from a different perspective) rather than a traditional vertical stretch.
  • This compression causes the function to become narrower on the graph, as the value of \(x\) climbs faster.
The range on the y-axis remains from \(-\pi/2\) to \(\pi/2\), but the points ascend and descend more steeply around the critical x-values due to this compression. This can be influential in identifying and analyzing more rapid changes within the function.
Graphing Utility
Graphing utilities are powerful tools that help visualize mathematical functions and their transformations.
By using a graphing utility for the function \( f(x) = \arctan (2x - 3) \), you can easily see how it differs from the standard \( \arctan(x) \).To get started:
  • Input \( f(x) = \arctan (2x - 3) \) into the utility.
  • Compare the output graph with that of \( \arctan(x) \) to observe changes.
These tools are especially useful for quickly validating analytical predictions made by solving equations or transformations manually.
Reviewing the graphical output helps to ensure your understanding of shifts and stretches, and further consolidates your learning.

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Most popular questions from this chapter

For the simple harmonic motion described by the trigonometric function, find (a) the maximum displacement, (b) the frequency, (c) the value of \(d\) when \(t=5,\) and (d) the least positive value of \(t\) for which \(d=0 .\) Use a graphing utility to verify your results. $$d=\frac{1}{2} \cos 20 \pi t$$

For the simple harmonic motion described by the trigonometric function, find (a) the maximum displacement, (b) the frequency, (c) the value of \(d\) when \(t=5,\) and (d) the least positive value of \(t\) for which \(d=0 .\) Use a graphing utility to verify your results. $$d=\frac{1}{64} \sin 792 \pi t$$

Sketch a graph of the function. $$f(x)=\arccos \frac{x}{4}$$

Fill in the blank. If not possible, state the reason. $$\text { As } x \rightarrow 1^{-}, \text {the value of } \arcsin x \rightarrow$$$\square$.

A ship leaves port at noon and has a bearing of \(\mathrm{S} 29^{\circ} \mathrm{W}\). The ship sails at 20 knots. (a) How many nautical miles south and how many nautical miles west will the ship have traveled by 6: 00 P.M.? (b) At 6: 00 P.M., the ship changes course to due west. Find the ship's bearing and distance from the port of departure at 7: 00 P.M.
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