Question
What is the Range of the following cube root function \( f(x)=2 \sqrt[3]{x+1}-4 \)
Ask by Chandler Harris. in the United States
Jan 23,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The range of the function \( f(x) = 2 \sqrt[3]{x+1} - 4 \) is all real numbers.
Solution
The function in question is:
\[ f(x) = 2 \sqrt[3]{x+1} - 4 \]
Let's analyze the range step by step:
1. **Cube Root Function**: The basic cube root function \( \sqrt[3]{x} \) has a domain of all real numbers (\( \mathbb{R} \)) and a range of all real numbers (\( \mathbb{R} \)) as well. This is because for every real number input, there is a real number output.
2. **Scaling by 2**: Multiplying the cube root by 2 (\( 2\sqrt[3]{x+1} \)) vertically stretches the graph but does not change the fact that the range is still all real numbers.
3. **Shifting Down by 4**: Subtracting 4 shifts the entire graph downward by 4 units (\( 2\sqrt[3]{x+1} - 4 \)). This transformation also does not restrict the range; it's simply a vertical translation.
Since none of these transformations limit the possible output values, the range of \( f(x) \) remains all real numbers.
**Final Answer:**
All real numbers are possible; the function’s range is (–∞, +∞).
Reviewed and approved by the UpStudy tutoring team
Like
error msg
Explain 
Simplify this solution Extra Insights
The cube root function has a unique property: its range is all real numbers! So, for the function \( f(x)=2 \sqrt[3]{x+1}-4 \), as \( x \) takes on any value from negative to positive infinity, the cube root will also span all values. After scaling by 2 and shifting down by 4, you'll still cover all real numbers. In simpler terms, no matter what, \( f(x) \) can output every possible value. If you were to visualize this function, you'd notice that it approaches negative infinity as \( x \) tends to negative infinity, and it moves towards positive infinity as \( x \) goes to positive infinity. That’s how the magic of cube roots works alongside linear transformations!
