Question

use the image to answer the question.
what happens to the range of the function when $f(x)$ is replaced by $f(x + 2)$ in $f(x) = \sqrt3{x}$?
(1 point)
\bigcirc the range contracts to exclude all negative numbers.
\bigcirc the range contracts to exclude all non - negative numbers.
\bigcirc the range remains the same.
\bigcirc the range contracts to exclude all positive numbers.

Explanation:

1. Recall the range of the cube root function \( f(x) = \sqrt[3]{x} \). The cube root of any real number (positive, negative, or zero) is a real number. So, the range of \( f(x) = \sqrt[3]{x} \) is all real numbers, \( (-\infty, \infty) \).
2. Analyze the transformation \( f(x) \to f(x + 2) \). This is a horizontal shift of the graph of \( f(x) \) to the left by 2 units. Horizontal shifts do not affect the range of a function. The range depends on the set of possible output values (y - values), and shifting the graph left or right does not change the set of y - values the function can take. For \( f(x)=\sqrt[3]{x} \), the function \( f(x + 2)=\sqrt[3]{x + 2} \) still has the property that for any real number \( y \), we can find an \( x \) such that \( y=\sqrt[3]{x + 2} \) (by solving \( x=y^{3}-2 \), which is a real number for any real \( y \)). So the range remains all real numbers.

Answer:

The range remains the same.