Q. 22 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 15–26, determine which functions are polynomial functions. For those that are, state the degree. For those that are not, tell

why not.

$h\left(x\right)=\sqrt{x}(\sqrt{x}-1)$

Step-by-Step Solution

Answer

h is not a polynomial function because $h\left(x\right)=\sqrt{x}(\sqrt{x}-1)$$=x-x^{\frac{1}{2}}$, so the variable x is raised to $\frac{1}{2}$ power, which is not a non negative integer.

Step 1: Given information

$h\left(x\right)=\sqrt{x}(\sqrt{x}-1)$

Step 2: Calculation

$$\begin{gathered} h\left(x\right)=\sqrt{x}(\sqrt{x}-1) \\ =x-\sqrt{x} \\ =x-x^{\frac{1}{2}} \end{gathered}$$

h is not a polynomial function because $h\left(x\right)=\sqrt{x}(\sqrt{x}-1)$$=x-x^{\frac{1}{2}}$, so the variable x is raised to $\frac{1}{2}$ power, which is not a non negative integer.