analysis of rational function

May 2008
400
82
Ottawa, Canada
for: \(\displaystyle f(x)=\frac{(x-1)^2}{x^2+1}\) determine:

1/x and y intercepts
2/co-ordinates of all critical values
3/classify the critical values using the second derivative test
4/determine the coordinates of all possible points of inflection
5/check to make sure there is a change in concavity at these points
6/determine all increasing and decreasing intervals
7/determine the intervals of concavity
8/determine the equation of the horizontal asymptote
9/provide a sketch and label all parts of the graphical analysis determined above

so for f'(x) i get \(\displaystyle f'(x)=\frac{2(x^2-1)}{(x^2+1)^2}\) and for f''(x) i get \(\displaystyle f''(x)=\frac{-4x(x^2-3)}{(x^2+1)^3}\) both of which i got using the quotient rule, although very unfamiliar as to how to work out each step
 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
Well here's some advice

1/x and y intercepts
x-intercepts make y=0, what do you get?

y-intercepts make x=0, what do you get?

2/co-ordinates of all critical values
Find where \(\displaystyle f'(x)=0\)

3/classify the critical values using the second derivative test
for the points found in the previous question, plug them into \(\displaystyle f''(x)\) . Then if \(\displaystyle f''(x)>0 \implies \text{min}, f''(x)<0 \implies \text{max}, f''(x)=0 \implies \text{pt of inflection}\)

4/determine the coordinates of all possible points of inflection
I think I have already answered that in my previous explanation

5/check to make sure there is a change in concavity at these points
Pick a point either side of each critical point. Then check the gradients of these points with \(\displaystyle f'(x)\), they should have opposite signs. This implies a change in concavity.

That is enough for now, have a go, let me know what you think.