Answer
The given function \[h\left( x \right)=7{{x}^{3}}+2{{x}^{2}}+\frac{1}{x}\]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 coefficient is called the leading coefficient.
Now, the given function $h\left( x \right)=7{{x}^{3}}+2{{x}^{2}}+\frac{1}{x}$ , simplifies to $h\left( x \right)=7{{x}^{3}}+2{{x}^{2}}+{{x}^{-1}}$.
The power of $x$ is −1, which gives a negative value for $n$ , and hence the function does not satisfy all the conditions to be a polynomial.
Hence, the given function $h\left( x \right)=7{{x}^{3}}+2{{x}^{2}}+\frac{1}{x}$ is not a polynomial.
