Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Describe the largest set \(S\) on which it is correct to say that \(f\) is continuous. \(f(x, y, z)=\frac{1+x^{2}}{x^{2}+y^{2}+z^{2}}\)

Step-by-Step Solution

Answer

The largest set \( S \) is \( \mathbb{R}^3 \setminus \{ (0, 0, 0) \} \).

Step 1: Understanding Continuity

The function \( f(x, y, z) = \frac{1+x^2}{x^2+y^2+z^2} \) is a multivariable function. For \( f \) to be continuous, the denominator \( x^2 + y^2 + z^2 \) must not be zero.

Step 2: Identify Points of Discontinuity

The function will be undefined where the denominator is zero. Solve \( x^2 + y^2 + z^2 = 0 \). This equation only holds when \( x = 0 \), \( y = 0 \), and \( z = 0 \). Thus, the function \( f \) is discontinuous at the point \( (0, 0, 0) \).

Step 3: Determine Largest Domain for Continuity

The largest set \( S \) where \( f \) is continuous is the set of all points except where the denominator is zero, i.e., \( S = \{ (x, y, z) \, | \, x^2 + y^2 + z^2 eq 0 \} \). This is the set of all points in \( \mathbb{R}^3 \) except \( (0, 0, 0) \).

Step 4: Conclusion

Since the function is continuous on all points except the origin, the largest set \( S \) is \( \mathbb{R}^3 \setminus \{ (0, 0, 0) \} \).

Key Concepts

Function Continuity

In multivariable calculus, function continuity is an important concept that helps in understanding how functions behave in different regions of their domain. A function is said to be continuous at a point if its value approaches the value of the function at that point from all directions in its domain. For simpler terms, imagine drawing the graph of the function without lifting your pencil off the paper and without any breaks or jumps. In mathematical terms, a function \( f(x) \) is continuous at a point \( c \) if \[ \lim_{{x \to c}} f(x) = f(c) \]. This concept extends to functions of multiple variables. For the given function \( f(x, y, z) = \frac{1+x^2}{x^2+y^2+z^2} \), the function is continuous at all points where the denominator \( x^2 + y^2 + z^2 \) is not zero. Whenever the denominator reaches zero, the function cannot be continuous as it would lead to undefined behavior.

Domain of a Function

The domain of a function refers to all possible input values for which the function is defined. For multivariable functions such as \( f(x, y, z) = \frac{1+x^2}{x^2+y^2+z^2} \), determining the domain involves identifying all permissible combinations of \( x, y, \) and \( z \). For this function, the primary restriction is that the denominator, \( x^2 + y^2 + z^2 \), must not be zero to avoid division by zero—which is undefined.
Therefore, the domain of \( f \) includes all points in three-dimensional space \( \mathbb{R}^3 \), excluding the origin \( (0, 0, 0) \). When analyzing functions, always consider whether each mathematical operation is valid for the given inputs. If any input leads to division by zero, take square roots of negative numbers, or other undefined operations, these must be excluded from the domain.

Points of Discontinuity

Points of discontinuity are locations within a function's domain where the function does not behave continuously. Identifying these points is crucial for understanding where a function may not adhere to expected behavior patterns, such as abrupt changes or undefined values. In our function \( f(x, y, z) = \frac{1+x^2}{x^2+y^2+z^2} \), discontinuity occurs when the denominator equals zero. Solving \( x^2 + y^2 + z^2 = 0 \) reveals the only point of discontinuity: \( (0, 0, 0) \). This is because the expression \( x^2 + y^2 + z^2 \) equals zero only when all variables are zero simultaneously. Hence, at this point, the function is undefined and experiences a discontinuity.
To deal with points of discontinuity, it is common practice to either exclude these points from the domain or apply techniques such as limits to analyze the function's behavior near these problematic points.