its been too long since i took differential equations classes
mmm ok..
what exactly does this part mean?
Let f and g be twice differentiable function
But anyhow, im not sure you can differentiate how you are currently.
If h(x)= f(g'(x)) and h'(3)= -2
then h'(3) = f'(g'(3)) = -2
now f'(x)>/= 0 for all x in the domain of f seems to directly contradict this equation in that its saying the value of g'(x) will yield a result contrary to it
the second one is fairly easy (unless im making a boo boo)
x^2-2x+3/(x-1)^2= 3/2
so multiply by (x-1)^2 to get
x^2 - 2x + 3 = 3/2(x-1)^2
now make the right side long hand
x^2 - 2x + 3 = 3/2x^2 -3x +3/2
then make it equal to zero
so 1/2x^2 - x - 3/2 = 0
now multply by 2 to simplify
x^2 - 2x - 3 = 0
now solve that by some math term i forget
and get (x-3)(x+1) = 0
or x = 3, -1
[ 01-19-2004, 06:12 PM: Message edited by: Gabrielles blades ]