\[R\left( 0 \right) = \sqrt {3 + 0} - \frac{4}{{0 + 1}} = \sqrt 3 - 4\]

\[R\left( 6 \right) = \sqrt {3 + 6} - \frac{4}{{6 + 1}} = \sqrt 9 - \frac{4}{7} = 3 - \frac{4}{7} = \frac{{17}}{7}\]

\[R\left( -9 \right)=\sqrt{3+\left( -9 \right)}-\frac{4}{-9+1}=\sqrt{-6}-\frac{4}{-8} \Huge \times\]

In this class we only deal with functions that give real values as answers. Therefore, because we have the square root of a negative number in the first term this function is not defined.

Note that the fact that the second term is perfectly acceptable has no bearing on the fact that the function will not be defined here. If any portion of the function is not defined upon evaluation, then the whole function is not defined at that point. Also note that if we allow complex numbers this function will be defined.

\[R\left( {x + 1} \right) = \sqrt {3 + \left( {x + 1} \right)} - \frac{4}{{\left( {x + 1} \right) + 1}} = \sqrt {4 + x} - \frac{4}{{x + 2}}\]

\[R\left( {{x^4} - 3} \right) = \sqrt {3 + \left( {{x^4} - 3} \right)} - \frac{4}{{\left( {{x^4} - 3} \right) + 1}} = \sqrt {{x^4}} - \frac{4}{{{x^4} - 2}} = {x^2} - \frac{4}{{{x^4} - 2}}\]

\[R\left( {\frac{1}{x} - 1} \right) = \sqrt {3 + \left( {\frac{1}{x} - 1} \right)} - \frac{4}{{\left( {{\textstyle{1 \over x}} - 1} \right) + 1}} = \sqrt {2 + \frac{1}{x}} - \frac{4}{{{\textstyle{1 \over x}}}} = \sqrt {2 + \frac{1}{x}} - 4x\]