To analyze the behavior of the function [tex]\( m(x) = \frac{1}{3} x^3 + 6 \)[/tex] as [tex]\( x \)[/tex] approaches positive and negative infinity, let's consider the dominant term in the function, which is [tex]\( \frac{1}{3} x^3 \)[/tex].
1. As [tex]\( x \)[/tex] approaches positive infinity ([tex]\( x \to +\infty \)[/tex]):
- The dominant term [tex]\( \frac{1}{3} x^3 \)[/tex] grows rapidly.
- Since [tex]\( x^3 \)[/tex] becomes very large, adding 6 to it will not significantly affect its behavior.
- Therefore, [tex]\( \frac{1}{3} x^3 \)[/tex] will dominate, and as [tex]\( x \)[/tex] increases without bound, the term [tex]\( \frac{1}{3} x^3 \)[/tex] will also increase without bound.
So, as [tex]\( x \to +\infty \)[/tex], [tex]\( m(x) \to +\infty \)[/tex].
2. As [tex]\( x \)[/tex] approaches negative infinity ([tex]\( x \to -\infty \)[/tex]):
- The dominant term [tex]\( \frac{1}{3} x^3 \)[/tex] will become very large in the negative direction because [tex]\( x^3 \)[/tex] for negative [tex]\( x \)[/tex] is also negative and grows without bound.
- Similarly, the constant 6 will not counterbalance the behavior of the term [tex]\( \frac{1}{3} x^3 \)[/tex].
So, as [tex]\( x \to -\infty \)[/tex], [tex]\( m(x) \to -\infty \)[/tex].
Therefore, the correct answers are:
- As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( m(x) \)[/tex] approaches [tex]\( +\infty \)[/tex].
- As [tex]\( x \)[/tex] approaches negative infinity, [tex]\( m(x) \)[/tex] approaches [tex]\( -\infty \)[/tex].
