| - Let c be such that f'(c)=0.
- If f(c) is also 0, that's it. Otherwise, w.l.g f(c)>0.
- Because f is continguous, there is an interval (s,t) around c where f(x)>0,
and f(s)=f(t)=0 (may be s=a and/or t=b).
- By the definition of f', we get f'(s)>0 and f'(t)<0, thus g(s)>0 and g(t)<0.
- g is continuous (?) so there exist a point c2 in (s,t) such that g(c2)=0. []
ND
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| SOLUTION182 V1.1
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If f(x)=0 all over --> finished. Otherwise, there exist an interval (s,t) such
that f(s)=f(t)=0 and w.l.g.(*) f(x)>0 inside. In (s,t) define h(x)=ln(f(x))+kx,
which is cont. and diff. in (s,t). lim h(x)=-inf in s,t (since f(s)=f(t)=0), so
there exist s<u<v<t such that h(u)=h(v). In [u,v] h is cont. and diff., so by
Roll's Thm there exist u<c<v such that h'(c)=0 ==> f'(c)/f(c)+k=0 ==>
==> f'(c)+kf(c)=0. []
[ (*) If f(x)<0 in (s,t), then def h(x)=ln(-f(x))+kx ]
|
| I think it can be generalized even more, by the same proof:
-----------------------------------------------------------
Instead of defining h(x)=ln(f(x))+kx, define h(x)=ln(f(x))+B(x), where B(x) is
cont.diff.function, BOUNDED (both sides) in (a,b), so by the same argument:
h'(c)=0 ==> f'(c)/f(c)+B'(c)=0 ==> f'(c)+f(c)B'(c)=0.
In Roll's: B(x)=constant.
In our example: B(x)=kx.
NITSAN
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