To understand what happens to the function [tex]\( m(x) = \frac{1}{3} x^3 + 6 \)[/tex] as [tex]\( x \)[/tex] approaches positive and negative infinity, we need to analyze the behavior of the terms involved.
1. As [tex]\( x \)[/tex] approaches positive infinity:
[tex]\[
\begin{aligned}
m(x) &= \frac{1}{3} x^3 + 6
\end{aligned}
\][/tex]
- The term [tex]\( \frac{1}{3} x^3 \)[/tex] will become very large, much larger than the constant term 6, as [tex]\( x \)[/tex] increases.
- Therefore, [tex]\( m(x) \)[/tex] is dominated by [tex]\( \frac{1}{3} x^3 \)[/tex] when [tex]\( x \)[/tex] is very large.
Thus, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( m(x) \)[/tex] approaches positive infinity.
2. As [tex]\( x \)[/tex] approaches negative infinity:
[tex]\[
\begin{aligned}
m(x) &= \frac{1}{3} x^3 + 6
\end{aligned}
\][/tex]
- For very large negative values of [tex]\( x \)[/tex], the cubic term [tex]\( \frac{1}{3} x^3 \)[/tex] will become very large in the negative direction.
- Again, this term will dominate over the constant 6 as [tex]\( x \)[/tex] becomes very negative.
Thus, as [tex]\( x \)[/tex] approaches negative infinity, [tex]\( m(x) \)[/tex] approaches negative infinity.
So, 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].
