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Magazine Chapter 14: Problem 19
Find and sketch the domain of the function. $$ f(x, y)=\frac{\sqrt{y-x^{2}}}{1-x^{2}} $$
Short Answer
Expert verified
The domain is \( y \geq x^2 \) and \( x \neq \pm 1 \).
Step by step solution
01
Understand the Function's Components
The function is given by \( f(x, y) = \frac{\sqrt{y-x^2}}{1-x^2} \). It consists of a square root \( \sqrt{y-x^2} \) in the numerator and \( 1-x^2 \) in the denominator. Each part will have specific domain constraints.
02
Determine the Domain for the Square Root
For \( \sqrt{y-x^2} \) to be defined, the expression inside the square root must be non-negative. Therefore, we require \( y - x^2 \geq 0 \), or equivalently \( y \geq x^2 \).
03
Identify Domain Restrictions from the Denominator
The denominator \( 1-x^2 \) must not be zero, which implies \( x^2 eq 1 \). Hence, \( x eq 1 \) and \( x eq -1 \).
04
Combine the Domain Conditions
The domain of \( f(x, y) \) requires both conditions from Steps 2 and 3 to be satisfied simultaneously. Therefore, \( y \geq x^2 \) and \( x eq \pm 1 \).
05
Sketch the Domain
On the xy-plane, sketch the region where \( y = x^2 \) is a parabola opening upwards. The domain is the area above this parabola. Ensure \( x = \pm 1 \) are vertical lines excluded from the domain.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Square root function
In mathematics, the square root function plays a crucial role in defining feasible domains for functions. Whenever you encounter a square root, it's essential to ensure that the value inside the root is non-negative. This is because the square root of a negative number is not defined in the set of real numbers.
In the given function, we have a square root component \( \sqrt{y-x^2} \). To find the domain, you need to make sure that the expression \( y-x^2 \) is greater than or equal to zero. Consequently, this turns into an inequality \( y \geq x^2 \). This inequality reveals that for each value of \( x \), \( y \) must be on or above the curve defined by \( y = x^2 \).
In the given function, we have a square root component \( \sqrt{y-x^2} \). To find the domain, you need to make sure that the expression \( y-x^2 \) is greater than or equal to zero. Consequently, this turns into an inequality \( y \geq x^2 \). This inequality reveals that for each value of \( x \), \( y \) must be on or above the curve defined by \( y = x^2 \).
- A square root requires the argument to be non-negative.
- Set up the inequality \( y \geq x^2 \).
- This condition must be met to ensure the square root is defined.
Denominator constraints
The denominator constraints are an important concept when dealing with rational functions. For a rational function \( \frac{p(x)}{q(x)} \), it is crucial that \( q(x) eq 0 \) since division by zero is undefined.In the function given, the denominator is \( 1 - x^2 \). To avoid division by zero, you must ensure that \( 1 - x^2 eq 0 \). Solving \( 1 - x^2 = 0 \) gives \( x^2 = 1 \), which means \( x = \pm 1 \).
Therefore, these values are excluded from the domain.
Therefore, these values are excluded from the domain.
- The denominator \( 1-x^2 \) must not equal zero.
- Find when \( 1 - x^2 = 0 \) to identify domain restrictions.
- Exclude \( x = 1 \) and \( x = -1 \) from the domain.
Parabola sketching
Sketching a parabola is a fundamental skill in visualizing the domain of functions involving inequalities. The expression \( y = x^2 \) forms a standard upward-opening parabola.To sketch it on the xy-plane:- Begin by plotting the vertex at the origin (0,0).- Note that it is symmetric about the y-axis.- As \( x \) moves away from the origin, \( y \) increases as \( x^2 \) increases.For the given function, \( y = x^2 \) is a boundary line. Therefore, the area of interest, or the domain, lies on or above this parabola, except at the points where our denominator constraints apply (\( x = \pm 1 \)).
This means you need to imagine a parabola "cut off" at these points.
This means you need to imagine a parabola "cut off" at these points.
- Identify the parabola shape from \( y = x^2 \).
- The domain is above this parabola.
- Consider the constraints that exclude the lines \( x = \pm 1 \) through sketching.
Inequality constraints
Handling inequalities is crucial when determining the domain of a function. These constraints tell us where the function is valid and can be graphed.In our function, the inequality \( y \geq x^2 \) governs part of the domain. It means that for any value of \( x \), \( y \) should be equal to or higher than the value determined by the parabola \( y = x^2 \).These constraints help define a region above the parabola, but also bear in mind the earlier exclusion of \( x = 1 \) and \( x = -1 \) due to the denominator.
- Inequality \( y \geq x^2 \) restricts \( y \) to non-negative values compared to \( x^2 \).
- Exclusions occur at potential zeroes of the denominator.
- Combine these to fully visualize the acceptable region on the xy-plane.
