Question

Multiple Choice Which of the following gives the domain of $f(x)=\frac{x}{\sqrt{9-x^{2}}} ?$ (A) $x \neq\pm 3$ (B) (-3,3) (C) [-3,3] (D) $(-\infty,-3) \cup(3, \infty)$ (E) $(3, \infty)$

Name: Problem 59, Chapter 1: Calculus: Graphical, Numerical, Algebraic
Uploaded: 2021-08-26T22:37:55-08:00
Duration: 1 min 2 s
Channel: James Kiss
Description: Multiple Choice Which of the following gives the domain of f(x)=(x)/(√(9-x^2)) ? (A) x ≠±3 (B) (-3,3) (C) [-3,3] (D) (-∞,-3) ∪(3, ∞) (E) (3, ∞)

   Multiple Choice Which of the following gives the domain of $f(x)=\frac{x}{\sqrt{9-x^{2}}} ?$
(A) $x \neq\pm 3$
(B) (-3,3)
(C) [-3,3]
(D) $(-\infty,-3) \cup(3, \infty)$
(E) $(3, \infty)$

Calculus: Graphical, Numerical, Algebraic

Ross L. Finney,… 5th Edition

Chapter 1, Problem 59

Instant Answer

Solved by Expert James Kiss

08/26/2021

Step 1

We know that the denominator of a fraction cannot be zero, and the square root function is only defined for non-negative values. Therefore, we need to find the values of $x$ for which $9-x^{2}$ is greater than zero. Show more…

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Multiple Choice Which of the following gives the domain of $f(x)=\frac{x}{\sqrt{9-x^{2}}} ?$ (A) $x \neq\pm 3$ (B) (-3,3) (C) [-3,3] (D) $(-\infty,-3) \cup(3, \infty)$ (E) $(3, \infty)$

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Key Concepts

Domain of a Function

The domain of a function is the complete set of input values (usually x values) for which the function is defined. This involves identifying any restrictions that result from operations within the function, such as square roots, logarithms, or denominators that could become zero.

Square Root Function Restrictions

Functions involving square roots require that the expression inside the square root (the radicand) is non-negative, because the square root of a negative number is not defined in the real number system. When the square root appears in the denominator, the radicand must be strictly positive to avoid division by zero.

Avoiding Division by Zero

In any function where an expression is in the denominator, such as a square root, it is critical to ensure that the denominator does not become zero. This adds an extra condition on the domain of the function beyond those imposed by the operations within the numerator or the radicand.

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