Question

given the function ( g(x)=sqrt3{x}+8 ), what happens as ( x ) approaches negative infinity? (1 point)
( y ) approaches 8
( y ) approaches ( -8 )
( y ) approaches positive infinity
( y ) approaches negative infinity

Explanation:

Step1: Analyze the limit of the cube - root function

We know that the cube - root function \(y = \sqrt[3]{x}\) has the property \(\lim_{x
ightarrow-\infty}\sqrt[3]{x}=-\infty\). This is because for any negative number \(a=-b\) (\(b > 0\)), \(\sqrt[3]{a}=-\sqrt[3]{b}\), and as \(b
ightarrow+\infty\) (when \(a
ightarrow-\infty\)), \(-\sqrt[3]{b}
ightarrow-\infty\).

Step2: Analyze the limit of the given function \(g(x)\)

We are given \(g(x)=\sqrt[3]{x}+8\). Using the limit property \(\lim_{x
ightarrow a}(f(x)+c)=\lim_{x
ightarrow a}f(x)+c\) (where \(c\) is a constant), we find \(\lim_{x
ightarrow-\infty}g(x)=\lim_{x
ightarrow-\infty}(\sqrt[3]{x}+8)\).
Since \(\lim_{x
ightarrow-\infty}\sqrt[3]{x}=-\infty\), then \(\lim_{x
ightarrow-\infty}(\sqrt[3]{x}+8)=-\infty + 8=-\infty\)

Answer:

\(y\) approaches negative infinity.