Problem 67 - College Algebra and Calculus: An Applied Approach

Question

Consider \(f(x)=\sqrt{x-2}\) and \(g(x)=\sqrt[3]{x-2}\). Why are the domains of \(f\) and \(g\) different?

Step-by-Step Solution

Answer

The domains of the functions \(f(x)=\sqrt{x-2}\) and \(g(x)=\sqrt[3]{x-2}\) are different due to the mathematical restrictions of the respective root types. The domain of the square root function \(f(x)\) requires the argument to be greater than or equal to zero (hence \(x \geq 2\)), while the cube root function \(g(x)\) can accept all real numbers as input.

Step 1: Find the domain of the function \(f(x)=\sqrt{x-2}\)

The function \(f(x)=\sqrt{x-2}\) represents a square root function. It is important to remember that a square root function is only defined for values that are greater than or equal to zero, since attempting to take the square root of a negative number yields an imaginary result, which is outside of the set of real numbers. Consequently, to find the domain of this function, set the expression under the square root to be greater than or equal to zero, \(x-2 \geq 0\). Solving for \(x\), we find that \(x \geq 2\). Thus, the domain of \(f(x)\) is \(x \geq 2\).

Step 2: Find the domain of the function \(g(x)=\sqrt[3]{x-2}\)

The function \(g(x)=\sqrt[3]{x-2}\) represents a cube root function. Differently from square root functions, cube root functions can take negative values and still yield real number solutions. Hence, there are no restrictions on the domain and it encompasses all real numbers.

Step 3: Compare the domains of \(f(x)\) and \(g(x)\)

The reason for the difference in the domains between \(f(x)\) and \(g(x)\) is due to the mathematical restrictions associated with their respective root types. A square root function requires its argument to be non-negative (greater than or equal to 0), while a cube root function can accept all real numbers (including negative numbers) as input. Hence, the domain of \(f(x)\) is \(x \geq 2\) and the domain of \(g(x)\) is all real numbers.