Let the function f be defined by the equation y = f (x),where x and f (x)are real numbers.Find the domain and range of the function $f ( x ) = \sqrt { 25 x ^ { 2 } - 3 }$ .
A)domain: $\left( - \infty , - \frac { 3 } { 25 } \right)$
B)domain: $\left( - \frac { 3 } { 5 } , \frac { 3 } { 5 } \right)$
C)domain: $\left( - \infty , - \frac { 3 } { 25 } \right] \cup \left[ \frac { 3 } { 25 } , + \infty \right)$
D)domain: $\left( \frac { \sqrt { 3 } } { 5 } , + \infty \right)$
E)domain: $\left( - \infty , - \frac { \sqrt { 3 } } { 5 } \right] \cup \left[ \frac { \sqrt { 3 } } { 5 } , + \infty \right)$

Question

Let the function f be defined by the equation y = f (x),where x and f (x)are real numbers.Find the domain and range of the function $f ( x ) = \sqrt { 25 x ^ { 2 } - 3 }$ . 
A)domain: $\left( - \infty , - \frac { 3 } { 25 } \right)$
B)domain: $\left( - \frac { 3 } { 5 } , \frac { 3 } { 5 } \right)$
C)domain: $\left( - \infty , - \frac { 3 } { 25 } \right] \cup \left[ \frac { 3 } { 25 } , + \infty \right)$
D)domain: $\left( \frac { \sqrt { 3 } } { 5 } , + \infty \right)$
E)domain: $\left( - \infty , - \frac { \sqrt { 3 } } { 5 } \right] \cup \left[ \frac { \sqrt { 3 } } { 5 } , + \infty \right)$

Quizplus · Accepted Answer

The expression under the square root, $25x^2 - 3$, must be non-negative. Solving $25x^2 - 3 \geq 0$ gives $x \leq -\frac{\sqrt{3}}{5}$ or $x \geq \frac{\sqrt{3}}{5}$. Thus, the domain is $\left(-\infty, -\frac{\sqrt{3}}{5}\right] \cup \left[\frac{\sqrt{3}}{5}, +\infty\right)$.

Let the Function F Be Defined by the Equation Y f(x)=25x2−3f ( x ) = \sqrt { 25 x ^ { 2 } - 3 }f(x)=25x2−3​

Let the Function F Be Defined by the Equation Y $f ( x ) = \sqrt { 25 x ^ { 2 } - 3 }$