/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 Sketch the graph of the equation... [FREE SOLUTION] | 91Ó°ÊÓ

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Sketch the graph of the equation. $$y=\tan ^{-1}(x-\pi)$$

Short Answer

Expert verified
The graph of \(y = \tan^{-1}(x-\pi)\) is \(y = \tan^{-1}(x)\) shifted right by \(\pi\) units.

Step by step solution

01

Understand the Function

The function given is \(y = \tan^{-1}(x-\pi)\). This is an inverse tangent function which has been shifted horizontally due to the \((x-\pi)\) term.
02

Identify Important Characteristics

The inverse tangent function \(y = \tan^{-1}(x)\) has an output range of \((-\pi/2, \pi/2)\). This means that as \(x\) approaches negative infinity, \(y\) approaches \(-\pi/2\), and as \(x\) approaches positive infinity, \(y\) approaches \(\pi/2\).
03

Determine Effects of Horizontal Shift

The replacement of \(x\) with \(x-\pi\) shifts the entire graph right by \(\pi\) units. This means the graph of \(y = \tan^{-1}(x)\) is moved right so its axis of symmetry is at \(x = \pi\).
04

Plot Key Points

To sketch the graph, note some key points on \(y = \tan^{-1}(x):\) the origin is a significant point. For \(y = \tan^{-1}(x-\pi)\), this point shifts to \((\pi, 0)\). Additionally, observe that when \(x = \pi - 1\), \(y\) will equal \(\tan^{-1}(-1)\), which is approximately \(-\pi/4\). Similarly, at \(x = \pi + 1\), \(y = \tan^{-1}(1)\), located at roughly \(\pi/4\).
05

Sketch the Graph

Start by plotting the point \((\pi, 0)\) on your graph. Mark the horizontal asymptotes at \(y = \pi/2\) and \(y = -\pi/2\). Sketch the curve starting below \(x = \pi\) approaching \(-\pi/2\) as \(x\) decreases and rising through the point \((\pi, 0)\), leveling out towards \(\pi/2\) as \(x\) increases.

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Key Concepts

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

Horizontal Shift
The function \(y = \tan^{-1}(x-\pi)\) indicates a horizontal shift applied to the basic inverse tangent function. A horizontal shift occurs when there's a change in the typical positioning of a function's graph along the x-axis through the replacement of \(x\) with \(x-c\), where \(c\) is the shift amount. In this scenario, we have \(x-\pi\), which implies a rightward shift of \(\pi\) units.

Think of it like moving everything in the graph \(\pi\) steps to the right. By doing this, every point on the graph has adjusted its position accordingly. The axis of symmetry for the inverse tangent function, \(y = \tan^{-1}(x)\), originally at \(x = 0\), is now at \(x = \pi\). This shift helps redefine the graph's position while maintaining its shape and basic features.
Graph Characteristics
The graph of the inverse tangent function, \(y = \tan^{-1}(x)\), has some unique and identifying characteristics.

The most notable characteristic is its S-shaped curve, which smoothly transitions between its horizontal asymptotes. The output of \(y = \tan^{-1}(x)\) ranges between \(-\pi/2\) and \(\pi/2\), meaning it approaches these values as \(x\) tends to \(-\infty\) and \(+\infty\) respectively. The graph never actually touches these lines, as they represent asymptotes.

With the horizontal shift to \(y = \tan^{-1}(x-\pi)\), the graph still maintains its S-shaped characteristics but starts its journey from this new alignment. The graph remains symmetrical around its center, which has now moved horizontally to \(x = \pi\).
Key Point Plotting
For effectively sketching the graph of \(y = \tan^{-1}(x-\pi)\), plotting key points really helps in understanding its movement.

First, consider the central point of the curve, which, for the inverse tangent function, is typically the origin \((0, 0)\). After the horizontal shift, this pivotal point moves to \((\pi, 0)\).

Next, identify additional points that reveal the slope and curvature across the x-axis. For example, at \(x = \pi - 1\), the function evaluates to \(\tan^{-1}(-1)\), which approximates to \(-\pi/4\). For \(x = \pi + 1\), it evaluates to \(\tan^{-1}(1)\), approximating \(\pi/4\). These points, placed accurately, illustrate how the curve meanders between the asymptotes.
  • Origin shifts to \((\pi, 0)\)
  • \((\pi - 1, -\pi/4)\)
  • \((\pi + 1, \pi/4)\)
Using these reference points ensures a clear depiction of the curve and highlights the effect of the horizontal transformation.
Asymptotes
Asymptotes are fundamentally significant in the graph of an inverse tangent function, representing lines that the graph approaches but never actually meets. For \(y = \tan^{-1}(x)\), there are two horizontal asymptotes at \(y = -\pi/2\) and \(y = \pi/2\). These represent the boundaries for the function’s range.

These asymptotes provide guidance on how the graph behaves as \(x\) progresses toward the infinite regions, both negatively and positively. No matter the horizontal shift, the asymptotes remain at these values because they are defined by the nature of the inverse tangent function itself.
  • As \(x\) reaches highs, the graph edges ever closer to \(y = \pi/2\).
  • As \(x\) dwells in the negatives, it dips toward \(y = -\pi/2\).
Recognizing these parameters not only aids in accuracy while sketching but also provides a deep understanding of the function’s behavior across its entire domain.

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Most popular questions from this chapter

If a tuning fork is struck and then held a certain distance from the eardrum, the pressure \(p_{1}(t)\) on the outside of the eardrum at time \(t\) may be represented by \(p_{1}(t)=A \sin \omega t,\) where \(A\) and \(\omega\) are positive constants. If a second identical tuning fork is struck with a possibly different force and held a different distance from the eardrum (see the figure on the next page), its effect may be represented by the equation \(p_{2}(t)=B \sin (\omega t+\tau)\) where \(B\) is a positive constant and \(0 \leq \tau \leq 2 \pi .\) The total pressure \(p(t)\) on the eardrum is given by $$p(t)=A \sin \omega t+B \sin (\omega t+\tau)$$ (a) Show that \(p(t)=a \cos \omega t+b \sin \omega t,\) where $$a=B \sin \tau \quad \text { and } \quad b=A+B \cos \tau$$ (b) Show that the amplitude \(C\) of \(p\) is given by $$C^{2}=A^{2}+B^{2}+2 A B \cos \tau$$

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