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Question
graphing the cube root function practice
complete this assessment to review what youve learned. it will not count toward your g
which of the following is the range of the function $g(x) = -\sqrt3{x + 2} - 5$?
option #1: $-\infty < x < \infty$
option #2: $-\infty \leq y \leq \infty$
option #3: $0 < x < \infty$
option #4: $-\infty < y < \infty$
(1 point)
Step1: Recall Cube Root Function Range
The parent cube root function \( y = \sqrt[3]{x} \) has a range of all real numbers (\( -\infty < y < \infty \)) because the cube root of any real number (positive, negative, zero) is a real number, and as \( x \) approaches \( \pm\infty \), \( \sqrt[3]{x} \) also approaches \( \pm\infty \).
Step2: Analyze Transformations
For the function \( g(x) = -\sqrt[3]{x + 2} - 5 \), we have two transformations: a reflection over the \( x \)-axis (due to the negative sign) and a vertical shift down by 5 units. Reflections over the \( x \)-axis and vertical shifts do not restrict the range of a cube root function—they just move or flip the graph. The domain of the inside \( x + 2 \) is all real numbers, and the cube root operation still outputs all real numbers after reflection and shifting. So the range remains all real numbers, \( -\infty < y < \infty \).
Step3: Match with Options
- Option #1: Describes the domain (since \( x \) can be any real number), not range.
- Option #2: Uses \( \leq \), but the range of a cube root function (even with transformations) is all real numbers, not including endpoints in the sense of a closed interval (it's open because \( y \) can take any real value, not just within a bounded interval).
- Option #3: Describes a restricted domain or a different range, incorrect.
- Option #4: \( -\infty < y < \infty \) matches the range of the transformed cube root function.
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Option #4: \( -\infty < y < \infty \)