differential calculus

Use (xyz)'=xyz'+xy'z+x'yz

f(x)=x(x-1)/(x-2)^2

f'(x)=(x-1)/(x-2)^2+x(x-1)/(x-2)^2+(-2)x(x-1)/(x-2)^3

after simplification this becomes

f'(x)=-3x+2/(x-2)^3

maxima, minima and points of inflection are when f'(x)=0

-3x+2/(x-2)^2=0 or 3x=2 (x=2/3)

There is a single solution at x=2/3. So the slope of f(x) is 0 only at x=2/3 Note that at x=2 the function f(x) is undefined..looks like it blows up to infinity as x approaches 2

To determine whether this point of zero slope is maxima, minima or inflection point it is necessary to determine the second derivative

f"(x)=df'(x)/dx=d[-3x+2/(x-2)^3]/dx=-3/(x-2)^2+(-3)(-3x+2)/(x-2)^4

simplifying

f"(x)=6x/(x-4)^4

putting in x=2/3 f"(x)=4/(-10/3)^4, which is positive

and a positive second derivative indicates a minima.

So there is a single minimum

Rap

That just made me wish I knew calc...