## Precalculus (6th Edition) Blitzer

The given function $f\left( x \right)={{x}^{\frac{1}{3}}}-4{{x}^{2}}+7$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. Now, the function g(x) is called a polynomial function of degree n; its coefficient is called the leading coefficient. Now, the given function $f\left( x \right)={{x}^{\frac{1}{3}}}-4{{x}^{2}}+7$ gives a fractional power of $x$ that is $\frac{1}{3}$ ; Thus, the value of $n$ is not an integer and since the function does not satisfy all the conditions to be a polynomial this function is not a polynomial. Hence, the given function $f\left( x \right)={{x}^{\frac{1}{3}}}-4{{x}^{2}}+7$ is not a polynomial.