At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
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].
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.