Question

for each function, determine whether it is a polynomial function.
function | is the function a polynomial?
| yes | no
(a) $g(x) = -\frac{1}{x^2}$ | $\circ$ | $\circ$
(b) $h(x) = x^8 - 2 + 8x^{-7}$ | $\circ$ | $\circ$
(c) $v(x) = 2x^7$ | $\circ$ | $\circ$
(d) $u(x) = 8\sqrt{x} - 6x^4$ | $\circ$ | $\circ$

Explanation:

Step1: Recall Polynomial Definition

A polynomial function is of the form \( f(x) = a_nx^n + a_{n - 1}x^{n - 1}+\dots+a_1x + a_0 \), where \( n \) is a non - negative integer and \( a_i \) are constants. The exponents of \( x \) must be non - negative integers, and the variable \( x \) cannot be in the denominator or under a square root (except when the root is a perfect power corresponding to a non - negative integer exponent).

Step2: Analyze \( g(x)=-\frac{1}{x^2} \)

Rewrite \( g(x) \) as \( g(x)=-x^{-2} \). The exponent of \( x \) is \( - 2 \), which is a negative integer. So, it does not satisfy the polynomial definition. So, for \( g(x) \), the answer is No.

Step3: Analyze \( h(x)=x^8 - 2+8x^{-7} \)

The term \( 8x^{-7} \) has an exponent of \( - 7 \) (a negative integer) for \( x \). Since a polynomial cannot have negative exponents for the variable \( x \), \( h(x) \) is not a polynomial. So, the answer for \( h(x) \) is No.

Step4: Analyze \( v(x)=2x^7 \)

The function \( v(x) = 2x^7\) is in the form of a polynomial \( f(x)=a_nx^n \) with \( n = 7 \) (a non - negative integer) and \( a_7=2 \). So, it satisfies the polynomial definition. So, the answer for \( v(x) \) is Yes.

Step5: Analyze \( u(x)=8\sqrt{x}-6x^4 \)

Rewrite \( \sqrt{x} \) as \( x^{\frac{1}{2}} \). The exponent of \( x \) in the term \( 8x^{\frac{1}{2}} \) is \( \frac{1}{2} \), which is not an integer. So, \( u(x) \) does not satisfy the polynomial definition. So, the answer for \( u(x) \) is No.

Answer:

(a) For \( g(x)=-\frac{1}{x^2} \): No
(b) For \( h(x)=x^8 - 2 + 8x^{-7} \): No
(c) For \( v(x)=2x^7 \): Yes
(d) For \( u(x)=8\sqrt{x}-6x^4 \): No