Hmm, the 2nd derivitve is good for finding concavity
let's find the max and min points that is where the first derivitive is equal to 0 remember the difference quotient
so f'(x)=(x^2-2x)/(x^2-2x+1) find where it equals 0 set numerator equal to 0 0=x^2-2x 0=x(x-2) 0=x 0=x-2 2=x
so at 0 and 2 are the min and max find if the signs go from negative to positive (min) or from positive to negative (max) at those points
f'(-1)>0 f'(1.5)<0 f'(3)>0
so at x=0, the sign go from positive to negative (local maximum) at x=2, the sign go from negative to positive (local minimum)
we can take the 2nd derivitive to see the inflection points f''(x)=2/((x-1)^3) where does it equal 0? it doesn't so no inflection point but, we can test it at x=0 and x=2 at x=0, we get f''(0)<0 so it is concave down. that means that x=0 being a max makes sense at x=2, we get f''(2)>0 so it is concave up. that means that x=2 being a max make sense
local max is at x=0 (the point (0,0)) local min is at x=2 (the point (2,4))
