Question

the range of $y = x^3$ is
$y > 0$.
$y \geq 0$.
all real numbers.

Explanation:

Step1: Recall the function \( y = x^3 \)

The function \( y = x^3 \) is a cubic function. For any real number \( x \) (positive, negative, or zero), when we cube it, we can get any real number \( y \). For example, if \( x = -2 \), then \( y = (-2)^3=-8 \); if \( x = 0 \), \( y = 0^3 = 0 \); if \( x = 2 \), \( y = 2^3=8 \).

Step2: Analyze the graph

Looking at the given graph, the curve extends from the bottom left (where \( y \) can be very negative) to the top right (where \( y \) can be very positive), passing through the origin. This shows that \( y \) can take any real - number value. The options \( y>0 \) is wrong because when \( x \) is negative, \( y \) is negative. The option \( y\geq0 \) is also wrong for the same reason (negative \( x \) gives negative \( y \)).

Answer:

The option "all real numbers" (the third option)