# Thread: analysis of rational function

1. ## analysis of rational function

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

Originally Posted by euclid2

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

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

Originally Posted by euclid2

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

Originally Posted by euclid2

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

Originally Posted by euclid2

4/determine the coordinates of all possible points of inflection

Originally Posted by euclid2

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 $f'(x)$, 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.