Answer :
To determine the end behavior of the function [tex]\( f(x) = -2 \sqrt[3]{x} \)[/tex], let's analyze the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches positive infinity ([tex]\( +\infty \)[/tex]) and negative infinity ([tex]\( -\infty \)[/tex]).
1. As [tex]\( x \rightarrow +\infty \)[/tex]:
- [tex]\( \sqrt[3]{x} \)[/tex]: When [tex]\( x \)[/tex] becomes very large and positive, the cube root of [tex]\( x \)[/tex] ([tex]\( \sqrt[3]{x} \)[/tex]) also becomes very large and positive.
- Now, considering [tex]\( f(x) = -2 \sqrt[3]{x} \)[/tex]:
[tex]\[ f(x) \rightarrow -2 \times \text{(a large positive number)} \rightarrow -\infty \][/tex]
Therefore, as [tex]\( x \rightarrow +\infty \)[/tex], [tex]\( f(x) \rightarrow -\infty \)[/tex].
2. As [tex]\( x \rightarrow -\infty \)[/tex]:
- [tex]\( \sqrt[3]{x} \)[/tex]: When [tex]\( x \)[/tex] becomes very large and negative, the cube root of [tex]\( x \)[/tex] ([tex]\( \sqrt[3]{x} \)[/tex]) becomes very large and negative, because taking the cube root of a negative number yields a negative result.
- Now, considering [tex]\( f(x) = -2 \sqrt[3]{x} \)[/tex]:
[tex]\[ f(x) \rightarrow -2 \times \text{(a large negative number)} \rightarrow +\infty \][/tex]
Therefore, as [tex]\( x \rightarrow -\infty \)[/tex], [tex]\( f(x) \rightarrow +\infty \)[/tex].
Based on this analysis, the correct end behavior for the function [tex]\( f(x) = -2 \sqrt[3]{x} \)[/tex] is:
- As [tex]\( x \rightarrow -\infty \)[/tex], [tex]\( f(x) \rightarrow +\infty \)[/tex].
- As [tex]\( x \rightarrow +\infty \)[/tex], [tex]\( f(x) \rightarrow -\infty \)[/tex].
Thus, the answer is:
- As [tex]\( x \rightarrow -\infty, f(x) \rightarrow \infty \)[/tex].
- As [tex]\( x \rightarrow \infty, f(x) \rightarrow -\infty \)[/tex].
1. As [tex]\( x \rightarrow +\infty \)[/tex]:
- [tex]\( \sqrt[3]{x} \)[/tex]: When [tex]\( x \)[/tex] becomes very large and positive, the cube root of [tex]\( x \)[/tex] ([tex]\( \sqrt[3]{x} \)[/tex]) also becomes very large and positive.
- Now, considering [tex]\( f(x) = -2 \sqrt[3]{x} \)[/tex]:
[tex]\[ f(x) \rightarrow -2 \times \text{(a large positive number)} \rightarrow -\infty \][/tex]
Therefore, as [tex]\( x \rightarrow +\infty \)[/tex], [tex]\( f(x) \rightarrow -\infty \)[/tex].
2. As [tex]\( x \rightarrow -\infty \)[/tex]:
- [tex]\( \sqrt[3]{x} \)[/tex]: When [tex]\( x \)[/tex] becomes very large and negative, the cube root of [tex]\( x \)[/tex] ([tex]\( \sqrt[3]{x} \)[/tex]) becomes very large and negative, because taking the cube root of a negative number yields a negative result.
- Now, considering [tex]\( f(x) = -2 \sqrt[3]{x} \)[/tex]:
[tex]\[ f(x) \rightarrow -2 \times \text{(a large negative number)} \rightarrow +\infty \][/tex]
Therefore, as [tex]\( x \rightarrow -\infty \)[/tex], [tex]\( f(x) \rightarrow +\infty \)[/tex].
Based on this analysis, the correct end behavior for the function [tex]\( f(x) = -2 \sqrt[3]{x} \)[/tex] is:
- As [tex]\( x \rightarrow -\infty \)[/tex], [tex]\( f(x) \rightarrow +\infty \)[/tex].
- As [tex]\( x \rightarrow +\infty \)[/tex], [tex]\( f(x) \rightarrow -\infty \)[/tex].
Thus, the answer is:
- As [tex]\( x \rightarrow -\infty, f(x) \rightarrow \infty \)[/tex].
- As [tex]\( x \rightarrow \infty, f(x) \rightarrow -\infty \)[/tex].
