What is the domain of the function defined by

f(x) = \sqrt [ 3 ]{ \begin{matrix} { 16x }^{ 2 }- & 1 \end{matrix} }\ \ \ \ ?

(A) -1/4 ≤ x ≤ 1/4

(B) -1/4 ≤ x or x ≥ 1/4

(C) x ≥ 0

(D) All Real Numbers

Solution:

**Choice D is correct.**

The radicand in a cube root can either be positive of negative.

Therefore, there is no restriction on the value of x.

So, the domain of the function defined by

f(x) = \sqrt [ 3 ]{ \begin{matrix} { 16x }^{ 2 }- & 1 \end{matrix} }\ \ \ \

is the set of all real numbers.

It is different from radicand in a square root.

If the problem would have asked for the domain of

f(x) = \sqrt [ ]{ \begin{matrix} { 16x }^{ 2 }- & 1 \end{matrix} }\ \ \ \ ,

Then

\ { \begin{matrix} { 16x }^{ 2 }- & 1 \end{matrix} }\ \ \ \

should be greater than or equal to zero.

For example,

\sqrt [ 3 ]{ -27 } = -3

However,

\sqrt{ -16 }

is not defined for any real number.