MHB Increase/Decrease of Rational Function | Jillian's Yahoo Answers

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To determine the intervals of increase and decrease for the function F(x) = (x^2 - 1) / (x^2 + 1), the first derivative f'(x) is calculated as f'(x) = 4x / (x^2 + 1)^2. The denominator is always positive, so the sign of the derivative depends solely on the numerator, which is 4x. Thus, the function is decreasing on the interval (-∞, 0) and increasing on (0, ∞). This analysis provides a clear understanding of the behavior of the rational function.
MarkFL
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Hello Jillian,

I am assuming we have:

$$f(x)=\frac{x^2-1}{x^2+1}$$

To investigate where the function is increasing/decreasing, we need to compute the first derivative, and find where it is positive (function increasing) and where it is negative (function decreasing).

Using the quotient and power rules of differentiation, we find:

$$f'(x)=\frac{(x^2+1)(2x)-(x^2-1)(2x)}{(x^2+1)^2}=\frac{2x(x^2+1-x^2+1)}{(x^2+1)^2}=\frac{4x}{(x^2+1)^2}$$

Now, we see the denominator is positive for all real $x$, so we need only concern ourselves with the sign of the numerator, and we see this simply has the sign of $x$ itself. Hence:

$$(-\infty,0)$$ $f(x)$ is decreasing.

$$(0,\infty)$$ $f(x)$ is increasing.

To Jillian and any other guests viewing this topic, I invite and encourage you to post other calculus questions here in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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