To analyze the end behavior of the function [tex]\( f(x) = -2 \sqrt[3]{x} \)[/tex], let's look at how the function acts as [tex]\( x \)[/tex] approaches positive and negative infinity.
1. As [tex]\( x \rightarrow \infty \)[/tex]:
We need to determine what happens to [tex]\( f(x) = -2 \sqrt[3]{x} \)[/tex] as [tex]\( x \)[/tex] becomes very large and positive. The cube root function, [tex]\( \sqrt[3]{x} \)[/tex], grows without bound but at a slower rate compared to a linear function. Therefore,
[tex]\[ f(x) = -2 \sqrt[3]{x} \quad \text{as} \quad x \rightarrow \infty, \][/tex]
since [tex]\( \sqrt[3]{x} \)[/tex] approaches [tex]\( \infty \)[/tex], multiplying by -2 will yield,
[tex]\[ f(x) \rightarrow -2 \cdot \infty = -\infty. \][/tex]
Thus, as [tex]\( x \rightarrow \infty \)[/tex], [tex]\( f(x) \rightarrow -\infty \)[/tex].
2. As [tex]\( x \rightarrow -\infty \)[/tex]:
Now we examine what happens to [tex]\( f(x) = -2 \sqrt[3]{x} \)[/tex] as [tex]\( x \)[/tex] becomes very large and negative. The cube root function of a negative number is also negative and grows without bound in the negative direction. Therefore,
[tex]\[ f(x) = -2 \sqrt[3]{x} \quad \text{as} \quad x \rightarrow -\infty, \][/tex]
since [tex]\( \sqrt[3]{x} \)[/tex] approaches [tex]\( -\infty \)[/tex], multiplying by -2 will yield,
[tex]\[ f(x) \rightarrow -2 \cdot (-\infty) = \infty. \][/tex]
Thus, as [tex]\( x \rightarrow -\infty \)[/tex], [tex]\( f(x) \rightarrow \infty \)[/tex].
So, summarizing the end behavior:
- As [tex]\( x \rightarrow \infty \)[/tex], [tex]\( f(x) \rightarrow -\infty \)[/tex].
- As [tex]\( x \rightarrow -\infty \)[/tex], [tex]\( f(x) \rightarrow \infty \)[/tex].
Therefore, the correct statement about the end behavior of [tex]\( f(x) = -2 \sqrt[3]{x} \)[/tex] is:
As [tex]\( x \rightarrow -\infty, f(x) \rightarrow \infty \)[/tex], and as [tex]\( x \rightarrow \infty, f(x) \rightarrow -\infty \)[/tex].
