01-19-2004, 11:02 AM | #1 |
Thoth - Egyptian God of Wisdom
Join Date: March 1, 2001
Location: NC
Age: 38
Posts: 2,890
|
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      a. h is concave down      b. g is decreasing      c. f is concave down      d. g is concave down      e. 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." |
01-19-2004, 11:03 AM | #2 |
Quth-Maren
Join Date: February 17, 2003
Location: Portsmouth
Age: 34
Posts: 4,145
|
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.
|
01-19-2004, 06:10 PM | #3 |
Baaz Draconian
Join Date: April 26, 2002
Location: florida
Age: 42
Posts: 761
|
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 ] |
01-19-2004, 08:21 PM | #4 |
Zhentarim Guard
Join Date: December 31, 2003
Location: SE Tornado Belt
Age: 63
Posts: 341
|
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... :\
|
Currently Active Users Viewing This Thread: 1 (0 members and 1 guests) | |
|
|
Similar Threads | ||||
Thread | Thread Starter | Forum | Replies | Last Post |
Iraq on the brink of anarchy | Dreamer128 | General Discussion | 2 | 04-07-2004 03:02 AM |
Moments of Hysteria within the Spire | CerebroDragon | Wizards & Warriors Forum | 2 | 10-19-2003 12:33 PM |
The Tavern on the Brink of sanity hill. | Kothoses | General Conversation Archives (11/2000 - 01/2005) | 9 | 08-10-2001 12:04 AM |