Question Statement

Draw the graph of the following functions and answer the questions given at the end of each question:

(i) $y={\tan }^{-1}(x-1)+\frac{\pi }{4}$ for $x\in \mathbb{R}$ Describe how the graph has been translated from the basic graph of $y={\tan }^{-1}x$.

(ii) $y={\sin }^{-1}(x-2)$ Indicate the domain, range and key points on the graph.

(iii) $y={\cos }^{-1}(2x)+\frac{\pi }{3}$ State the domain and range of the function and describe the transformation applied to the basic graph of $y={\cos }^{-1}(x)$.

(iv) $y={\sin }^{-1}(2x)+\frac{\pi }{6}$ Identify any vertical or horizontal shifts and discuss how the graph differs from the graph of $y={\sin }^{-1}x$.

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Background and Explanation

To graph inverse trigonometric functions, we must consider their standard domains and ranges. Transformations such as $y=f(x-h)+k$ result in a horizontal shift of $h$ units and a vertical shift of $k$ units, while $y=f(ax)$ results in a horizontal compression or stretch.

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Solution

Part (i)

To understand the transformation of $y={\tan }^{-1}(x-1)+\frac{\pi }{4}$, we compare it to the parent function $y={\tan }^{-1}x$.

Table of Values:

Transformation Description: The graph of $y={\tan }^{-1}(x-1)$ is shifted to the right side with a value of $\frac{\pi }{4}$ (representing the vertical shift).

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Part (ii)

For the function $y={\sin }^{-1}(x-2)$, we determine the domain by ensuring the argument of the inverse sine stays within $[-1,1]$.

Domain Calculation: $\begin{array}{c}-1\le x-2\le 1\\ -1+2\le x-2+2\le 1+2\\ +1\le x\le 3\end{array}$

• Domain: $[1,3]$
• Range: $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

Key Points Table:

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Part (iii)

For $y={\cos }^{-1}(2x)+\frac{\pi }{3}$, we calculate the domain based on the restriction of the cosine inverse function.

Domain Calculation: $\begin{array}{rl}& -1\le 2x\le 1\\ & \frac{-1}{2}\le \frac{2x}{2}\le \frac{1}{2}\\ & \frac{-1}{2}\le x\le \frac{1}{2}\end{array}$

• Domain: $\left[-\frac{1}{2},\frac{1}{2}\right]=[1,3]$
• Range: $\left[-\frac{2\pi }{3},\frac{\pi }{3}\right]$

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Part (iv)

For $y={\sin }^{-1}(2x)+\frac{\pi }{6}$, we identify the shifts and compressions.

Horizontal and Vertical Shifts: The function involves a horizontal compression (by a factor of 2) and a vertical shift of $+\frac{\pi }{6}$.

Domain Calculation: $\begin{array}{rl}& -1\le 2x\le 1\\ & \frac{-1}{2}\le \frac{2x}{2}\le \frac{1}{2}\end{array}$

Comparison Table:

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Key Formulas or Methods Used

• Domain of Inverse Sine/Cosine: The argument $u$ in ${\sin }^{-1}(u)$ or ${\cos }^{-1}(u)$ must satisfy $-1\le u\le 1$.
• Horizontal Shift: $f(x-h)$ shifts the graph $h$ units horizontally.
• Vertical Shift: $f(x)+k$ shifts the graph $k$ units vertically.
• Horizontal Compression: $f(ax)$ compresses the graph horizontally by a factor of $a$.

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Summary of Steps

1. Determine Domain: Set the inner expression of the inverse function between $-1$ and $1$ and solve for $x$.
2. Identify Transformations: Look for constants added to $x$ (horizontal shift), constants added to the whole function (vertical shift), and coefficients of $x$ (compression/stretch).
3. Calculate Key Points: Select values of $x$ within the domain to find corresponding $y$ values.
4. Compare with Parent Function: Note how the new values differ from the standard ${\sin }^{-1}x,{\cos }^{-1}x,$ or ${\tan }^{-1}x$ values.
5. Sketch Graph: Plot the calculated points and draw the curve respecting the calculated domain and range.