## Precalculus (6th Edition) Blitzer

The given function $f\left( x \right)=\frac{{{x}^{2}}+7}{{{x}^{3}}}$is not a polynomial.
A function is said to be a polynomial function if $g\left( x \right)={{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+\cdots +{{a}_{2}}{{x}^{2}}+{{a}_{1}}x+{{a}_{0}}$ , where ${{a}_{n}},{{a}_{n-1}},\ldots ,{{a}_{2}},{{a}_{1}},{{a}_{0}}$ are any real numbers with ${{a}_{n}}\ne 0$ and n is any non-negative integer. The function g(x) is called a polynomial function of degree n; its coefficients called the leading coefficient. The given function $f\left( x \right)=\frac{{{x}^{2}}+7}{{{x}^{3}}}$ , simplifies to $f\left( x \right)={{x}^{-1}}+7{{x}^{-3}}$. The powers of $x$ are −1 and –3, which give a negative value for $n$ and hence the function does not satisfy all the conditions to be a polynomial. Hence, the given function $f\left( x \right)=\frac{{{x}^{2}}+7}{{{x}^{3}}}$ is not a polynomial.