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Jillian's question at Yahoo! Answers regarding increasing/decreasing behavior of rational function

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MarkFL

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Feb 24, 2012
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Jillian,

I am assuming we have:

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

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:

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

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:

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

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

To Jillian and any other guests viewing this topic, I invite and encourage you to post other calculus questions here in our Calculus forum.

Best Regards,

Mark.
 
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