Question

lesson 12 root functions cubic & cube root functions review f(x) + 5 f(x) -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 click for long desc what happens to the range of the function when f(x) is replaced by f(x) + 5 in f(x) = \sqrt3{x} (1 point) the range remains the same the range contracts to exclude all non-negative numbers. the range contracts to exclude all negative numbers. the range contracts to exclude all positive numbers

Explanation:

Step1: Recall the range of \( f(x)=\sqrt[3]{x} \)

The cube - root function \( y = \sqrt[3]{x} \) has a range of all real numbers, because for any real number \( y \), we can find an \( x=y^{3} \) such that \( \sqrt[3]{x}=y \). Mathematically, the range of \( f(x)=\sqrt[3]{x} \) is \( (-\infty,\infty) \) or \( \mathbb{R} \).

Step2: Analyze the transformation \( y = f(x)+5 \)

When we have a function transformation of the form \( y = f(x)+k \) (in this case \( k = 5 \)), this is a vertical shift. A vertical shift of a function does not change the "spread" of the function's output values in terms of the set of real numbers it can take. It just shifts all the output values up (if \( k>0 \)) or down (if \( k < 0 \)) by \( k \) units.

For the function \( y=f(x)+5=\sqrt[3]{x}+5 \), if we consider the range, let \( y=\sqrt[3]{x}+5 \). Let \( t=\sqrt[3]{x} \), where \( t\in(-\infty,\infty) \). Then \( y=t + 5 \). As \( t \) takes on all real values, \( t + 5 \) also takes on all real values. For example, if \( t\to-\infty \), \( y=t + 5\to-\infty \); if \( t\to\infty \), \( y=t + 5\to\infty \). So the range of \( y = f(x)+5 \) is also all real numbers.

So the range remains the same as the range of the original function \( f(x)=\sqrt[3]{x} \).

Answer:

The range remains the same