Need to find the intervals where the function increase, decrease & concavity

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SUMMARY

The discussion centers on analyzing the function f(x) = x/(x^2 - 9) to determine its intervals of increase, decrease, and concavity. Participants identified the horizontal asymptote at y=0 and vertical asymptotes at x=3 and x=-3. The first derivative, f'(x) = (-x^2 - 9)/(x^2 - 9)^2, indicates that the function is decreasing in the intervals (-∞, -3), (-3, 3), and (3, ∞). The second derivative, after simplification, is f''(x) = (2x(x^2 + 27))/(x^2 - 9)^3, which is crucial for determining concavity.

PREREQUISITES
  • Understanding of calculus concepts such as derivatives and asymptotes.
  • Familiarity with the quotient rule for differentiation.
  • Knowledge of how to analyze the sign of a function's derivative.
  • Ability to simplify and analyze higher-order derivatives.
NEXT STEPS
  • Learn how to apply the quotient rule in calculus for complex functions.
  • Study the implications of horizontal and vertical asymptotes on function behavior.
  • Explore methods for determining intervals of increase and decrease using first derivatives.
  • Investigate concavity and inflection points through second derivative tests.
USEFUL FOR

Students and educators in calculus, mathematicians analyzing rational functions, and anyone seeking to deepen their understanding of function behavior through derivatives.

  • #31
so... what is left now is to try to simplify and evaluate the second derivative right?

if I'm not mistaken, after simplifying, the derivative ends like this:
f"(x)=(2x(x^2 + 27))/(x^2 - 9)^3
 
Last edited:
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  • #33
yay! now i can sketch the graph!
 

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