The standard form of an Irrational function is given as:
![\begin{gathered} y=\sqrt{a(x-p)}+q \\ \text{where:} \\ y\geqslant q\text{ and x}\geqslant p \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%3D%5Csqrt%7Ba%28x-p%29%7D%2Bq%20%5C%5C%20%5Ctext%7Bwhere%3A%7D%20%5C%5C%20y%5Cgeqslant%20q%5Ctext%7B%20and%20x%7D%5Cgeqslant%20p%20%5Cend%7Bgathered%7D)
In the given function:
![\begin{gathered} g(x)=\sqrt{6-x} \\ \text{Domain: }\left\lbrace x\in R\colon x\leqslant6\right\rbrace \\ \text{Range: }\left\lbrace y\in R,\text{ }y\geqslant0\right\rbrace \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20g%28x%29%3D%5Csqrt%7B6-x%7D%20%5C%5C%20%5Ctext%7BDomain%3A%20%7D%5Cleft%5Clbrace%20x%5Cin%20R%5Ccolon%20x%5Cleqslant6%5Cright%5Crbrace%20%20%5C%5C%20%5Ctext%7BRange%3A%20%7D%5Cleft%5Clbrace%20y%5Cin%20R%2C%5Ctext%7B%20%7Dy%5Cgeqslant0%5Cright%5Crbrace%20%20%5Cend%7Bgathered%7D)
Therefore, the standard form of g(x) is:
Answer:
In mathematics, a cube root of a number x is a number y such that y = x. All nonzero real numbers, have exactly one real cube root and a pair of complex conjugate cube roots, and all nonzero complex numbers have three distinct complex cube roots
(Not sure if this is what your looking for)
Answer:
No, because it does not have a constant rate of change.
Step-by-step explanation:
On the <em>x</em><em> </em>side of the table, there is a constant rate of change (+1). However, on the y side, it is not. The first change is +7, the next +5, and the last +6. The rate of change has to be constant on both sides for the table to be considered linear.