Question Statement

Draw the graph of each of the following functions against the interval mentioned:

(i) $y={\cos }^{-1}\left(x^2\right)$ for $x\in [-1,1]$

(ii) $y={\sin }^{-1}(2x)$ for $x\in \left[-\frac{1}{2},\frac{1}{2}\right]$

(iii) $y={\tan }^{-1}\left(x^2\right)$ for $x\in \left[-\frac{1}{2},\frac{1}{2}\right]$

(iv) $y={\cot }^{-1}(x-2)$ for $x\in \mathbb{R}$

(v) $y={\sec }^{-1}\left(x^2\right)$ for $x\in (-\infty ,-1]\cup [1,+\infty )$

(vi) $y={\csc }^{-1}(-x)$ for $x\in (-\infty ,-1]\cup [1,+\infty )$

(vii) $y={\cos }^{-1}(-2x)$ for $x\in \left[-\frac{1}{2},\frac{1}{2}\right]$

(viii) $y={\sin }^{-1}(-x)$ for $x\in [-1,1]$

(ix) $y={\tan }^{-1}(3x+1)$ for $x\in \mathbb{R}$

(x) $y={\cos }^{-1}(2x-1)$ for $x\in (0,1]$

(xi) $y={\cot }^{-1}\left(x^2-1\right)$ for $x\in \mathbb{R}$

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Background and Explanation

To graph inverse trigonometric functions, we identify the domain and range of the specific function and calculate key coordinates by substituting values of $x$ from the given interval. Transformations such as $x^2$ (making the function even) or horizontal shifts and scaling (like $2x$ or $x-2$) affect the shape and position of the standard inverse trig curves.

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Solution

To draw these graphs, we construct a table of values for each function and plot the resulting points on a Cartesian plane, connecting them with smooth curves consistent with the properties of inverse trigonometric functions.

Part (i): $y={\cos }^{-1}\left(x^2\right)$

Since $x^2$ is always non-negative, the input to ${\cos }^{-1}$ ranges from $0$ to $1$. This results in a graph symmetric about the $y$-axis (an even function).

Part (ii): $y={\sin }^{-1}(2x)$

This is a horizontal compression of the standard ${\sin }^{-1}(x)$ graph by a factor of $2$.

Part (iii): $y={\tan }^{-1}\left(x^2\right)$

The function is even. As $x$ increases, $x^2$ increases, and $y$ approaches the horizontal asymptote $\pi /2$.

Part (iv): $y={\cot }^{-1}(x-2)$

This represents a horizontal shift of the ${\cot }^{-1}(x)$ graph to the right by $2$ units.

Part (v): $y={\sec }^{-1}\left(x^2\right)$

Defined for $|x^2|\ge 1$, which is $|x|\ge 1$. The graph is symmetric about the $y$-axis.

Part (vi): $y={\csc }^{-1}(-x)$

Using the property ${\csc }^{-1}(-x)=-{\csc }^{-1}(x)$, this is a reflection of the standard cosecant inverse graph.

Part (vii): $y={\cos }^{-1}(-2x)$

This involves a horizontal compression and a reflection across the $y$-axis.

Part (viii): $y={\sin }^{-1}(-x)$

Since ${\sin }^{-1}(-x)=-{\sin }^{-1}(x)$, this is the standard sine inverse graph reflected across the $x$-axis.

Part (ix): $y={\tan }^{-1}(3x+1)$

This is a transformation of ${\tan }^{-1}(x)$ involving a horizontal shift and compression.

Part (x): $y={\cos }^{-1}(2x-1)$

The domain is restricted to $[0,1]$ because $-1\le 2x-1\le 1$.

Part (xi): $y={\cot }^{-1}\left(x^2-1\right)$

An even function symmetric about the $y$-axis.

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Key Formulas or Methods Used

• Inverse Trigonometric Functions: ${\sin }^{-1}x,{\cos }^{-1}x,{\tan }^{-1}x,{\cot }^{-1}x,{\sec }^{-1}x,{\csc }^{-1}x$.
• Transformations:
  • $f(ax)$: Horizontal scaling.
  • $f(x+c)$: Horizontal shift.
  • $f(x^2)$: Symmetry about the $y$-axis (even function).
  • $f(-x)$: Reflection across the $y$-axis.
• Point Plotting: Calculating $y$ for specific $x$ values within the domain.

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Summary of Steps

1. Determine Domain: Identify the valid input range for $x$ based on the inverse trigonometric function's definition.
2. Create Table of Values: Substitute key $x$ values (including boundaries and midpoints) into the function to find corresponding $y$ values.
3. Identify Symmetry: Check if the function is even ($x^2$) or odd ($-x$) to simplify plotting.
4. Plot Points: Mark the calculated $(x,y)$ coordinates on a graph.
5. Draw Curve: Connect the points with a smooth curve, ensuring it respects the asymptotic behavior (for ${\tan }^{-1},{\cot }^{-1},{\sec }^{-1},{\csc }^{-1}$).