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imagator2 · Mar 25, 2007, 06:26 PM

Hello I need Calculus help!
Given the function: f(x)= 2x^2(1-x^2)
I need to analyze the function to determine if it has vertical or horzintal asymptotes?? Then find where it is pos? And negative?
then find the f'(x) the derviative? Figure out from that where the fcn is increasing and decreasing/ max and min points? Please help!

galactus · Mar 25, 2007, 06:42 PM

I assume you have a typo and mean:

$\frac{2x^{2}}{1-x^{2}}$

To find the vertical asymptotes, what x makes the denominator equal 0?

To find the derivative, use the quotient rule. $\frac{(1-x^{2})(4x)-2x^{2}(-2x)}{(1-x^{2})^{2}}=\frac{4x}{(x^{2}-1)^{2}}$ .

Set to 0 and solve for x to find the extrema.

A nice clue about the horizontal asymptote is to notice that if the power of the numerator is equal to the power of the denominator, then the line $y=\frac{a_{n}}{b_{n}}$ (ratio of leading coefficients) is the horizontal asymptote. The leading coefficients are $a_{n}=2$ and $b_{n}=-1$ . In this case, 2/(-1)=-2.

The line y=-2 is the horizontal asymptote.