Nabs preparing for an analysis test (you may can help)

[Naboki]

yo like that the topic said i have in a week analysis test and i have some questions:

my first one

f in I(a,b) is a differentiable function prove:
a) f`>= 0 is f is a monoton function (raising)
a) f`=0 is  f a constant

yeah i know its simple to equations to find minimas and maximas of a given function - but how can i prove it?

my try was useing the mean value theorem for diff. functions

so i got for a) f(a)<=f(b) ---> f(b)-f(a) / b-a >=0 ---> what leads back to f(a)<=f(b)
and for the b) i have done the same

[Tannar]

I am not sure what exactly you are supposed to prove. Equivalence or only implication?

I would got with proof by contradiction with mean value theorem as you did. Assume f is not <property> then mvt gives you something about f' contradicting the precondition. Easiest to write down quickly imho.

[Zruty]

for a) you can use Lagrange's theorem: f(y) - f(x) = f'(z) * (y-x), for all x<z<y.
Apply the theorem for each x and y between a and b, note that f' >= 0 and you're set

actually, for (b) you can use the same: f(y) - f(x) = f'(z)*(y-x) = 0 because f'(z) = 0.

I'm not sure if you can use the theorem though: maybe its proof itself relies on (b), need to check.

[Zruty]

I am not sure what exactly you are supposed to prove. Equivalence or only implication?

Hmm,  I went for implication.
For backwards proof you need something else.

[Zruty]

Yea, the mean value theorem (= Lagrange's theorem) is proved without those facts. You can use it to prove the => of (a) and the <=> of (b).

For <= of (a) (if f(x) is raising then f'(x) >= 0), it seems you can't use this theorem, because it doesn't allow you to pick the point where you calculate the derivative f'(x).
But you can use the definition of the derivative:
f'(x) = lim_{D-->+0} ( f(x+D) - f(x) / D ).
Here f(x+D) - f(x) > 0 because f is raising, and D>0 because we pick it as positive.
so, f'(x) = limit of some POSITIVE expression, hence it can't be negative.

[lexaal]

look like my work is done...   

[skutch]

Im sooooo happy I got my Degree 15 years ago =)
Good luck Nab, a good education is really valuable.

[Naboki]

yo thx for the help

i have a second one

Its about Eigenwerte, Eigenvalues and the characteristal polynom
bla bla bla

figured out the polynoms and have 3 zero points from my 3x3matrix. now i want to calculate the eigenvectors and for that i use the formular (A-lambda * I) - I.. the basic matrix (i am currently missing the right word for it the one of the 1,0,0 - 0,1,0......). After solving that euqation system i get the same vector for all 3 different eigenvalues, is that possible?
-----
Matrix A
0  -1   1
-3 -2   3
-2  -2  3
-------
nabs polynom
t*(t²-t-6)=0
eigenvalues
t1=0
t2=3
t3=-2
eigenvector
all 3 are (0,0,0) ?????

[lexaal]

Look for non-zero vectors which solve (A-lam*I).

prefered method: good looking: for 0 it is the vector (1,1,3)'.

[Zruty]

yea naboki, zero vector is always a solution for (A-lambda*I)v = 0, it's obvious.
you need non-zero solution.
Basically, it's always a system of linear algebraic equations, and you can also use Gauss method to solve it.

[Naboki]

ok did my last 2 math tests at uni, and now i have to wait until end of february ;( (and i am already so nervous - cos i am really not good at written tests (nearly all 4) - all my practicals and oral tests i got a 1 or 2).

thx again

[Dangeruss]

im sure you aced it nab