Question

using the table prompt, which values of x make y = $x^3$ + 1 negative?
no x values
only x = 0
all x values
large negative x values
large positive x values

Explanation:

Step1: Analyze the function \( y = x^3 + 1 \)

We want to find when \( y<0 \), so we solve the inequality \( x^3 + 1<0 \).

Step2: Solve the inequality

Rewrite the inequality as \( x^3<- 1 \). Taking the cube - root of both sides (since the cube - root function is a one - to - one function and preserves the inequality sign for real numbers), we get \( x <\sqrt[3]{-1}=-1 \).
This means that for \( x \) values that are large negative (values much less than - 1), the inequality \( x^3 + 1<0 \) will hold. For example, if \( x=-2 \), then \( y=(-2)^3 + 1=-8 + 1=-7<0 \). If \( x = 0 \), \( y=0^3+1 = 1>0 \). If \( x = 2 \), \( y=2^3 + 1=9>0 \). And there are solutions (e.g., \( x=-2 \)) so it's not "No x values" and not "All x values" or "Only \( x = 0 \)" or "Large positive x values".

Answer:

Large negative x values