Question

Challenge domains Find the domain of the following functions. Specify the domain mathematically, and then describe it in words or with a sketch. $$g(x, y, z)=\frac{10}{x^{2}-(y+z) x+y z}$$

Short answer

Question: Determine the domain of the function $$g(x, y, z) = \frac{x + y + z}{x^{2} - (y + z) x + yz}$$. Answer: The domain of the function $$g(x, y, z)$$ is the set of all real numbers $$x, y, z$$, such that the square of the difference between $$y$$ and $$z$$ is negative, represented mathematically as $$\left\{(x,y,z) \in \mathbb{R}^3 : y^2 - 2yz + z^2 < 0 \right\}$$.

Step-by-step solution

Identify the Denominator

The denominator of the given function is $$x^{2} - (y + z) x + yz$$. The function will be undefined when this denominator equals zero.

Solve for the Denominator Equals to Zero

Now, we need to find the values of $$x, y, z$$ for which the denominator equals zero: $$x^{2} - (y + z) x + yz = 0$$

Apply Quadratic Formula

Note that this is a quadratic equation in $$x$$, so we can apply the quadratic formula: $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ In this case, $$a = 1, b = -(y+z), \text{ and } c = yz$$: $$x=\frac{y + z \pm \sqrt{(-y - z)^2 - 4(1)(yz)}}{2(1)}$$

Determine the Discriminant

The discriminant ($$b^2 - 4ac$$) determines whether the quadratic equation in $$x$$ has real roots. If it's negative, the equation has no real solutions: $$(-y - z)^2 - 4(1)(yz) = y^2 + 2yz + z^2 - 4yz$$ The discriminant simplifies to: $$D = y^2 - 2yz + z^2$$

Define the Domain

The function is undefined where the discriminant $$D$$ is non-negative (i.e., when the quadratic equation in $$x$$ has real solutions). Thus, the domain of the function is given by: $$\left\{(x,y,z) \in \mathbb{R}^3 : y^2 - 2yz + z^2 < 0 \right\}$$

Description in Words

The domain of the function $$g(x, y, z)$$ is the set of all real numbers $$x, y, z$$, such that the square of the difference between $$y$$ and $$z$$ is negative.

Key concepts

Quadratic Formula

The quadratic formula is a formidable tool for solving quadratic equations, which are polynomials of the second degree, typically written in the form \( ax^2 + bx + c = 0 \). The formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) elegantly expresses the solutions to these equations.

The importance of the quadratic formula lies not just in its ability to solve quadratic equations efficiently, but also in its clarity in conveying the relationship between the coefficients of the equation and the solutions. By simply knowing the values of \( a, b, \) and \( c \), one can determine the roots of the quadratic equation using this formula.

For example, if a function \( f(x) \) is given by a quadratic equation, the quadratic formula can be used to find the values of \( x \) for which \( f(x) \) is undefined, essentially revealing potential restrictions on the domain of \( f(x) \).

Discriminant of a Quadratic Equation

Within the quadratic formula, the expression under the square root, \( b^2 - 4ac \), is known as the discriminant. It plays a crucial role because it dictates the nature and number of roots the quadratic equation will have.

If the discriminant is positive, the quadratic equation has two distinct real roots. When the discriminant is zero, there is exactly one real root, which is also a repeated root. However, if the discriminant is negative, no real roots exist, as the square root of a negative number is not real in the context of real numbers.

Understanding the discriminant is elemental when identifying the domain of functions involving quadratic expressions, especially in cases where the expression is in the denominator—as seen in the exercise. The condition for the function to remain defined is that the discriminant must be negative, ensuring the absence of real roots that would make the denominator zero, causing the function to become undefined.

Multivariable Calculus

Multivariable calculus is the extension of single-variable calculus to functions of multiple variables. In the realm of multivariable functions, such as \( g(x, y, z) \), the concepts of domain, range, continuity, and differentiability expand into a higher dimensional space.

When dealing with the domain of a multivariable function, one has to consider the set of all possible input values for which the function is well-defined. For instance, in the exercise, the presence of a denominator containing a quadratic in \( x \) with coefficients involving \( y \) and \( z \) requires careful exclusion of values that would render the denominator and thus the function undefined.

Conversely, the domain in single-variable calculus—while conceptually similar—is often simpler to determine, as it involves restrictions along only one dimension. Multivariable calculus requires an added layer of scrutiny to ensure that all combinations of \( x, y, \) and \( z \) are considered, further illustrating the complexity and beauty of functions with multiple variables.