Question

In Exercises \(47-52,\) discuss the continuity of the function. \(f(x, y, z)=\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\)

Short answer

The function \(f(x, y, z)=\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\) is discontinuous at the origin (0,0,0) and continuous everywhere else.

Step-by-step solution

Identify the denominator

Start by identifying the denominator in the given function \(f(x, y, z)=\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\). The denominator is \(\sqrt{x^{2}+y^{2}+z^{2}}\). Determine the values of (x, y, z) that make it equal to 0.

Solve the equation

Solve the equation \(x^{2}+y^{2}+z^{2} = 0\). The only solution to this equation is when x=y=z=0.

Discuss the continuity

Based on the solution from step 2, it is seen that the denominator becomes 0 only at the point (0,0,0). Hence, the function \(f(x, y, z)=\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\) is not defined at the origin. Hence, the function \(f(x, y, z)\) is discontinuous at the origin, but is continuous everywhere else in its domain.