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Chapter 12: Problem 18

Find the domain of the following functions. $$f(x, y)=\sin ^{-1}\left(y-x^{2}\right)$$

Short Answer

Expert verified
Answer: The domain of the function $$f(x, y) = \sin^{-1}(y - x^2)$$ is: $$x^2 - 1 \leq y \leq x^2 + 1$$. This includes all points (x, y) that lie between the parabolas $$y = x^2 - 1$$ and $$y = x^2 + 1$$, for any value of x.

Step by step solution

01

Identify acceptable range of input values

Since the function inside the arcsin is \(y-x^2\), we need this value to be between -1 and 1. So, we can write the inequality as: $$-1 \leq y-x^2 \leq 1$$
02

Break down inequality into two parts

To find the domain, we can break the inequality into two separate inequalities: 1. $$y - x^2 \geq -1$$ 2. $$y - x^2 \leq 1$$
03

Solve the first inequality

Rearrange the first inequality to get y in terms of x: $$y \geq x^2 - 1$$ The graph of this inequality represents the region above the parabola $$y = x^2 - 1$$.
04

Solve the second inequality

Rearrange the second inequality to get y in terms of x: $$y \leq x^2 + 1$$ The graph of this inequality represents the region below the parabola $$y = x^2 + 1$$.
05

Combine the inequalities

The domain of the function will be the set of all points that satisfy both inequalities, which represents the region between the parabolas $$y = x^2 - 1$$ and $$y = x^2 + 1$$. So, the domain of the function $$f(x, y) = \sin^{-1}(y - x^2)$$ is: $$x^2 - 1 \leq y \leq x^2 + 1$$ This domain includes all points (x, y) that lie between the parabolas $$y = x^2 - 1$$ and $$y = x^2 + 1$$, for any value of x.

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Most popular questions from this chapter

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