# question on curve, points of inflection, stationary points and max and min points

• Aug 24th 2011, 11:57 AM
MaggieNF
question on curve, points of inflection, stationary points and max and min points
For the function: y=2x^2/x^2-1

(i) Find all stationary points.

(ii) Find the intervals where y is increasing and the intervals where y is decreasing.

(iii) Find all local minima and local maxima.

(iv) Find all inflection points.

(v) Sketch the graph, indicating all maxima and minima, and all in
ection points.
• Aug 24th 2011, 12:05 PM
Siron
Re: question on curve, points of inflection, stationary points and max and min points
First, start differentiating the function.
• Aug 24th 2011, 12:17 PM
waqarhaider
Re: question on curve, points of inflection, stationary points and max and min points
1st find $\frac{dy}{dx}=-\frac{4x}{(x^2)-1}$
putting $\frac{dy}{dx}=0$ gives (0,0) as stationary point
Now for x < 0, y is increasing and for x > 0, y is decreasing
apply second derivative test to find (0,0) as maxima
curve has no point of inflection
• Aug 24th 2011, 12:35 PM
Siron
Re: question on curve, points of inflection, stationary points and max and min points
You forgot a square in the denominator:
$\frac{dy}{dx}\left(\frac{2x^2}{x^2-1}\right)=\frac{4x(x^2-1)-2x^2(2x)}{(x^2-1)^2}=\frac{-4x^3-4x-4x^3}{(x^2-1)^2}=\frac{-4x}{(x^2-1)^2}$

Which is offcourse don't make any difference in calculating the stationary points, but maybe for the inflection points it does.