Problem 17 - Calculus Early Transcendentals

Question

Find and sketch the domain of the function. $$f(x, y)=\frac{\sqrt{y-x^{2}}}{1-x^{2}}$$

Step-by-Step Solution

Answer

Domain: \(y \geq x^2\), excluding \(x = \pm 1\).

Step 1: Identify Restrictions from the Square Root

The square root function \(\sqrt{y-x^2}\) requires the expression inside to be non-negative. Thus, we need \(y-x^{2} \geq 0\), which simplifies to \(y \geq x^2\). This means the points \((x, y)\) must lie on or above the parabola \(y = x^2\).

Step 2: Identify Restrictions from the Denominator

The denominator \(1-x^2\) cannot be zero, or otherwise, the function becomes undefined. This leads us to the inequality \(1-x^2 eq 0\), which simplifies to \(x^2 eq 1\). Consequently, \(x eq 1\) and \(x eq -1\).

Step 3: Combine Restrictions

The domain of the function is defined by the combination of the inequalities from Steps 1 and 2. Therefore, the domain consists of all points \((x, y)\) such that \(y \geq x^2\) and \(x eq \pm 1\). This means we include all points above the parabola except those where \(x\) is 1 or -1.

Step 4: Sketch the Domain

To sketch the domain, draw the parabola \(y = x^2\) on the xy-plane. Shade the region above the parabola, but exclude the vertical lines at \(x = 1\) and \(x = -1\). These vertical lines are represented as dashed lines to indicate that points along them are not included in the domain. The shaded region represents the domain of the function.

Key Concepts

Function Restrictions

When we talk about a function's domain in calculus, we are essentially talking about all the possible input values that a function can accept without running into any mathematical issues. Common restrictions arise from

• Square roots - to avoid negative numbers under the root;
• Denominators - to prevent division by zero.

For the given function, both these issues need consideration to accurately determine the domain.
Function restrictions ensure that the operations within a function, like square roots and divisions, are mathematically valid.

Square Root Inequalities

Square root functions come with a key requirement: the value inside the square root must always be non-negative. This is because square roots of negative numbers are not real numbers in the context of standard calculus.
For our function, the expression within the square root is \(y-x^2\). Thus, we set up the inequality:

• \(y-x^2 \geq 0\)

This tells us that \(y\) should be greater than or equal to \(x^2\). Essentially, this means any point \((x, y)\) needs to lie on or above the curve described by \(y = x^2\).
This forms a sort of 'ceiling' that our valid points cannot dip below.

Denominator Restrictions

Denominators in a fraction cannot be zero, because division by zero is undefined in mathematics. In the given function, the denominator is \(1-x^2\). Therefore, we must solve:

• \(1-x^2 eq 0\)

Solving this yields \(x^2 eq 1\), which simplifies further to \(x eq 1\) and \(x eq -1\).
These points \(x=1\) and \(x=-1\) cause the function to be undefined, so we have to exclude these vertical lines from our domain.
This is why these points are left out when determining the range of \(x\) values you can input into your function.

Combining Inequalities

To find the full domain of the function, we need to combine the restrictions from both the square root and the denominator. Both conditions must simultaneously be satisfied:

• Points \((x, y)\) need to lie above the parabola \(y = x^2\)
• We exclude \(x = 1\) and \(x = -1\)

Combining these conditions, our domain includes all points above (or on) the parabola except for the vertical lines \(x = 1\) and \(x = -1\).
This process is like piecing together a puzzle, making sure all operations are mathematically valid everywhere within the specified domain.