Finding the Values of x for Increasing Functions: A Simple Q&A

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SUMMARY

The discussion focuses on determining the values of x for which the function f(x) = (x^2 + 3)/(4x + 1) is increasing. The derivative f'(x) = (4x - 6)(x + 2)/(4x + 1)^2 is analyzed, leading to the conclusion that f'(x) > 0 for x < -2 and x > 2/3. However, the correct interpretation includes the boundaries, resulting in x <= -2 and x >= 2/3, as f'(x) must be non-negative for the function to be increasing. The distinction between increasing and strictly increasing functions is also clarified.

PREREQUISITES
  • Understanding of calculus, specifically derivatives
  • Knowledge of rational functions and their properties
  • Familiarity with the concepts of increasing and strictly increasing functions
  • Ability to analyze inequalities
NEXT STEPS
  • Study the properties of rational functions and their derivatives
  • Learn about the Mean Value Theorem and its implications for increasing functions
  • Explore the differences between increasing and strictly increasing functions in depth
  • Practice solving inequalities involving derivatives to find intervals of increase and decrease
USEFUL FOR

Students studying calculus, mathematics educators, and anyone interested in understanding the behavior of functions through their derivatives.

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to cut it short, was asked to find the values of x for which f(x) is an increasing function
[tex]\dfrac{(x^2 + 3)}{4x + 1} = f(x)[/tex]
[tex]\dfrac{(4x-6)(x+2)}{(4x+1)^2} = f'(x)[/tex] so setting this to be greater than zero I get the values of x < -2, and x > 2/3 however in the answers they got x <= -2 and x >= 2/3, and they set f'(x) to be >= 0. I thought with increasing functions f'(x) is > 0?
 
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phospho said:
to cut it short, was asked to find the values of x for which f(x) is an increasing function
[tex]\dfrac{(x^2 + 3)}{4x + 1} = f(x)[/tex]
[tex]\dfrac{(4x-6)(x+2)}{(4x+1)^2} = f'(x)[/tex] so setting this to be greater than zero I get the values of x < -2, and x > 2/3 however in the answers they got x <= -2 and x >= 2/3, and they set f'(x) to be >= 0. I thought with increasing functions f'(x) is > 0?
Some texts distinguish between increasing and strictly increasing. For an increasing function, if a < b, then f(a) ≤ f(b). For a function that is strictly increasing, if a < b, then f(a) < f(b).

And similar for decreasing/strictly decreasing.
 
Mark44 said:
Some texts distinguish between increasing and strictly increasing. For an increasing function, if a < b, then f(a) ≤ f(b). For a function that is strictly increasing, if a < b, then f(a) < f(b).

And similar for decreasing/strictly decreasing.

oh I see, thank you!
 

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