Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
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].
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].
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.