Let the function $f$ be defined by the equation $y = f ( x )$ ,where $x$ and $f ( x )$ are real numbers.Find the domain of the function $f ( x ) = \sqrt { 6 x ^ { 2 } + 16 }$
A)domain: $\left( - \infty , - \frac { 16 } { 36 } \right) \cup \left( \frac { 16 } { 36 } , + \infty \right)$
B)domain: $\left( - \infty , - \frac { \sqrt { 16 } } { 6 } \right)$
C)domain: $\left( \frac { 16 } { 36 } , + \infty \right)$
D)domain: $\left( - \frac { \sqrt { 16 } } { 6 } , \frac { \sqrt { 16 } } { 6 } \right)$
E)domain: $\left( - \infty , - \frac { \sqrt { 16 } } { 6 } \right] \cup \left[ \frac { \sqrt { 16 } } { 6 } , + \infty \right)$

Question

Let the function $f$ be defined by the equation $y = f ( x )$ ,where $x$ and $f ( x )$ are real numbers.Find the domain of the function $f ( x ) = \sqrt { 6 x ^ { 2 } + 16 }$ 
A)domain: $\left( - \infty , - \frac { 16 } { 36 } \right) \cup \left( \frac { 16 } { 36 } , + \infty \right)$
B)domain: $\left( - \infty , - \frac { \sqrt { 16 } } { 6 } \right)$
C)domain: $\left( \frac { 16 } { 36 } , + \infty \right)$
D)domain: $\left( - \frac { \sqrt { 16 } } { 6 } , \frac { \sqrt { 16 } } { 6 } \right)$
E)domain: $\left( - \infty , - \frac { \sqrt { 16 } } { 6 } \right] \cup \left[ \frac { \sqrt { 16 } } { 6 } , + \infty \right)$

Quizplus · Accepted Answer

The domain of $f(x) = \sqrt{6x^2 + 16}$ includes all $x$ values for which the expression under the square root is non-negative. Since $6x^2 + 16$ is always positive for all real numbers $x$, the domain is all real numbers, $(- \infty, +\infty)$. However, the choices seem to misinterpret the domain's description. The correct interpretation, based on the given options, is that the square root function can take all real numbers as input since the expression inside the square root is always non-negative. None of the options perfectly matches this, but option E is the closest in spirit, suggesting a misunderstanding in the formulation of the options. The correct domain should be all real numbers, but given the options, E seems to attempt to express the range of $x$ values that make the expression under the square root non-negative, which is indeed all real numbers.

Let the Function fff Be Defined by the Equation y=f(x)y = f ( x )y=f(x)

Let the Function $f$ Be Defined by the Equation $y = f ( x )$