I'm thinking that dismembering my fingers will give me more satisfaction than trying to finish this calculus. This for example:

Let f and g be twice differentiable function such that f'(x)>/= 0 for all x in the domain of f. If h(x)= f(g'(x)) and h'(3)= -2, then at x= 3
&nbsp&nbsp&nbsp&nbsp&nbspa. h is concave down
&nbsp&nbsp&nbsp&nbsp&nbspb. g is decreasing
&nbsp&nbsp&nbsp&nbsp&nbspc. f is concave down
&nbsp&nbsp&nbsp&nbsp&nbspd. g is concave down
&nbsp&nbsp&nbsp&nbsp&nbspe. f is decreasing

I don't even know where to start with that. Let f and g be twice differentiable function; ok, so that's like the second derivative. If h(x)= f(g'(x)) and h'(3)= -2; that means that:

f'(g'(3)) + f(g"(3))= -2, right?

All this math.. I'm going to die young, I think by now that's almost a guarantee. Then, the problem I posted last night. You guys showed me where I went wrong, but then I'm stuck with this:

x²-2x+3/(x-1)²= ³/₂

And I have to reduce all that to make x= 3. :(

I tried getting out of this class last semester, but they said dropping out might effect my acceptance into college. So that's kind of like saying, "You have two options: go crazy or don't get into college."

Okay, now I'm pretty sure you're a maths teacher/lawyer, or are going to be one. You think up some really hard problems.

:( 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 ]

Back in college I was three classes away from an engineering degree. Calculus was one of those classes. Needless to say I won't ever be an engineer... :\