91Ó°ÊÓ

A derivative is a central concept in calculus and provides information about the rate at which a function is changing at any given point. In the context of sequences, finding the derivative can help us understand whether a sequence is increasing or decreasing.
To calculate the derivative for the function related to the sequence given, use the quotient rule. The rule states that for a function \( f(x) = \frac{u(x)}{v(x)} \), the derivative \( f'(x) \) is given by:

• \( f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \)

For the sequence \( a_n = \frac{n}{n^2 + 10} \), we let \( u(x) = x \) and \( v(x) = x^2 + 10 \). After applying the quotient rule, we find:

• \( f'(x) = \frac{-x^2 + 10}{(x^2 + 10)^2} \)

This derivative tells us how the function, and thus the sequence, behaves as \( x \) changes. It aids in determining intervals where the sequence is increasing or decreasing, providing critical insights into its overall behavior.

A sequence is considered strictly increasing when each successive term is greater than the preceding one. Mathematically, for a sequence \( \{a_n\} \), it is strictly increasing if \( a_{n+1} > a_n \) for all \( n \).
In our given task, we check if the sequence \( a_n = \frac{n}{n^2 + 10} \) is ever strictly increasing. To do this, we look at the derivative \( f'(x) = \frac{-x^2 + 10}{(x^2 + 10)^2} \).

• The derivative is positive for \( x < \sqrt{10} \), indicating the sequence might be increasing there.
• However, after \( x > \sqrt{10} \), the derivative becomes negative.

This implies that the sequence does not remain strictly increasing as \( n \) exceeds a certain point, specifically beyond \( \sqrt{10} \), making it only potentially increasing for small values up to that threshold.

A sequence is described as strictly decreasing if each term is less than the previous term, i.e., \( a_{n+1} < a_n \) for all \( n \).
Examining the behavior of the sequence \( a_n = \frac{n}{n^2 + 10} \) using its derivative provides insight into whether it becomes strictly decreasing.
By analyzing \( f'(x) = \frac{-x^2 + 10}{(x^2 + 10)^2} \):

• The expression \(-x^2 + 10\) becomes negative when \( x > \sqrt{10} \).
• Once this happens, the derivative \( f'(x) \) is less than zero.

This indicates that beyond \( x = \sqrt{10} \), the terms of the sequence start decreasing. Hence, for values of \( n \) greater than \( \sqrt{10} \), \( a_n \) eventually becomes strictly decreasing.This understanding supports the conclusion drawn in the solution, that the sequence is eventually strictly decreasing as \( n \) grows large.