Answer
The given function \[f\left( x \right)=\frac{{{x}^{2}}+7}{{{x}^{3}}}\]is not a polynomial.
Work Step by Step
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.
