From this picture I know that the parabolic asymptote is $x^2$ just by dividing. In the original graph, the point I am going to be looking at is a local maximum I think, so do I find the first derivative and set it equal to 0 and then calculate the y coordinate. then I can plug in that x coordinate into $x^2$ to get the y value of that and then just subtract the two values?
I'm confused because I'm not sure if the original graph is a local min or max. It kind of looks like an inflection point but I haven't learned inflection points or second derivatives in class. How can I solve this?
Best Answer
We are looking for the maximum of $$g(x)=\frac {x^4-x}{x^2+1} - (x^2-1) = \frac {x^4-x}{x^2+1} -\frac{x^4-1}{x^2+1}=\frac {1-x}{1+x^2}.$$
To find critical points, set $$g'(x)=\frac {-(1+x^2)-(1-x)2x}{(1+x^2)^2}=\frac{-1-2x+x^2}{(1+x^2)^2}=0.$$
Setting the numerator equal to zero and solving the quadratic equation yields $x=\frac{2\pm \sqrt{8}}2=1\pm \sqrt2.$
Note that $g(0)=1$ and $g(1)=0$ and $1-\sqrt2<0<1<1+\sqrt2,$
so $g(1-\sqrt2)$ is a maximum and $g(1+\sqrt2)$ is a minimum.
The maximum value is thus $g(1-\sqrt2)=\frac{\sqrt2}{4-2\sqrt2}=\frac {\sqrt2+1}2=1.207...$