[maths] Re: more of calculus
- From: "Ned Granic" <ngranic@xxxxxxx>
- To: <maths@xxxxxxxxxxxxx>
- Date: Sat, 2 Dec 2006 04:54:08 -0700
Oh thank you Nelson so much; it really helps a lot.
I'll post these theorems for you, and for the knowledge-starving audience
<smile>
but I think I was right in finding the solution to problems related to them.
The Rolle's Theorem
Let f be a function that satisfies the following three hypotheses:
1. f is continuous on the closed interval [a,b];
2. f is differentiable on the open interval (a,b);
3. f(a) = f(b).
Then, there is a number c on the open interval (a,b) such that f prime of c
equals 0.
There are three cases:
case 1. f(x) = k -- a constant;
then f^'(x) = 0, so the number c can be taken to be any number on the open
interval (a,b).
case 2. f(x) > f(a) for some x in the open interval (a,b).
By the extreme value theorem,
<< the extreme value theorem:
if f is continuous on a closed interval [a,b], then f attains an absolute
value f(c) and an absolute minimum value f(d) at some numbers c and d in the
closed interval [a,b].
>>
which we can apply by hypothesis 1,
-- f is continuous on the closed interval [a,b] --
f has a maximum value somewhere in the closed interval [a,b].
Since f(a) = f(b), it must attain this maximum value at a number c in the open
interval (a,b).
Then f has a local maximum at c, and by hypothesis 2,
-- f is differentiable on the open interval (a,b) --
it is differentiable at c.
Therefore, f'(c) = 0 by Fermat's Theorem,
<< Fermat's theorem:
if f has a local maximum or a minimum at c, and if f^'(c) exists, then f^'(c) =
0.
But there is a warning that caussions us not to read to much into Fermat's
Theorem since we can't expect to locate extreme values simply by setting f^'(x)
= 0 and solving for x.
An example of that would be if f(x) = x^3, then f^'(x) = 3x^2, so f^'(0) = 0.
But f has no maximum or minimum at 0.
Also, x^3 > 0 for x > 0, and x^3 < 0 for x < 0.
The fact that f(0) = 0 simply means that the curve y = x^3 has a horizontal
tangent at (0,0), and instead of having a maximum or a minimum, it crosses its
horizontal tangent there.
>>
case 3. f(x) < f(a) for some x in the open interval (a,b).
By the extreme value theorem, f has a minimum value on the closed interval
[a,b], and since f(a) = f(b), it attains this minimum value at a number c in
the open interval (a,b).
Again, f^'(c) = 0 by Fermat's Theorem.
The Mean Value Theorem
Let f be a function that satisfies the following hypotheses:
1. f is continuous on the closed interval [a,b];
2. f is differentiable on the open interval (a,b).
Then, there is a number c in the open interval (a,b) such that:
f^'(c) = [f(b)-f(a)] / [(b-a)], or:
f(b)-f(a) = f^'(c)(b-a).
Many thanks again for your time!
Ned
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