Mean Value Theorem - f(x) = f(a)+(x-a)f'(u)

[rhia]

Hi,
If f is continuous in [a,b] and differentiable in (a,b), and if xE(a,b), then there exists u in (a,b) such that
f(x) = f(a)+(x-a)f'(u)


What's wrong if i state :
(i)If f is continuous in [a,b] and differentiable in [a,b], and if xE[a,b]
or
(ii)If f is continuous in (a,b) and differentiable in (a,b), and if xE(a,b)

Thanks!

 

[TenaliRaman]

usually we write,
f'(u) = (f(x)-f(a))/(x-a) .. (1)

(i)If f is continuous in [a,b] and differentiable in [a,b], and if xE[a,b]

here x can take value of a ..
see what happens to (1) when x takes value of a?

(ii)If f is continuous in (a,b) and differentiable in (a,b), and if xE(a,b)

it is easy to note that if f is continuous in (a,b) then it is continuous in [a,b]
cuz f(a) = lim_{x->a+}f(x)
(we don't bother about x->a-)
similarly,
f(b) = lim_{x->b-}f(x)
(we don't bother about x->b+)

-- AI

 

[Wong]

In the second case, f may tend to infinity at a or at b. So f being continuous on [a, b] is required.

 

[chronon]

it is easy to note that if f is continuous in (a,b) then it is continuous in [a,b]
cuz f(a) = lim_{x->a+}f(x)
-- AI

f(a)=lim{x->a} f(x) is the definition of continuity at a, so this is just assuming what you are trying to show.

If you don't assume continuity at a then f(a) can have whatever value you like, unrelated to the rest of f(x). e.g.
f(0)=1, f(1)=1,f(x)=0 for x in (0,1).

 

[TenaliRaman]

In the second case, f may tend to infinity at a or at b. So f being continuous on [a, b] is required.

in the second case,
f(x) is defined over (a,b) which excludes a and b
so f(a) and f(b) are actually not defined originally and hence cannot be infinity
so i can define continuity extending to both extremities and still the theorem would be true

[Edit]eeps! Wong is right on one thing here
say the function is 1/x and the interval is (0,1) the limx->0+ f(x) would be infinity ... So i suppose we could add a condition that lim x->a+ f(x) and limx->b-f(x) must exist and must be finite... then it is extendible as i said earlier[/Edit]

f(a)=lim{x->a} f(x) is the definition of continuity at a, so this is just assuming what you are trying to show.

No
definition of continuity requires that,
f(a) = limx->a- f(x) = limx->a+ f(x)
(this is a strict necessity)
But if u see what i said in my earlier post, when we talk of an interval like (a,b) i can skip one of the conditions since my viewport now is restricted to interval (a,b) alone.

-- AI

 

[Hurkyl]

so f(a) and f(b) are actually not defined originally and hence cannot be infinity
so i can define continuity extending to both extremities and still the theorem would be true

In other words, you're assuming f is a function continuous on [a, b].

 

[HallsofIvy]

There is nothing WRONG with assuming f is differentiable on [a,b] except that it is not as general. The theorem is true even if f is NOT differentiable at a or b.
(Consider the function f(x)= |x| and apply the mean value theorem on [0, 1].)

On the other hand you MUST have f continuous at a and b since the values used in the theorem are calculated there.

For example, if f(x)= x for x NOT equal to 0 or 1 but f(0)= -1, f(1)= 2. Can you find c between 0 and 1 such that f'(c)= f(1)- f(0) ?

 

[TenaliRaman]

In other words, you're assuming f is a function continuous on [a, b].

Well i did edit my post indicating that fact
Though i am not exactly assuming it

 

[JasonRox]

For example, if f(x)= x for x NOT equal to 0 or 1 but f(0)= -1, f(1)= 2. Can you find c between 0 and 1 such that f'(c)= f(1)- f(0) ?

If I got this right, you can't.

This is because it is not continuous at 0 and 1.

Continuity does not imply differentiability.

When they say it is continuous on the interval, they are saying:

f(a)=the limit, as x approaches a, is equal to f(a). Again, it has been mentionned earlier that the right-hand limit must equal the left-hand limit. That is continuity. It may be continuous on the left or right, but the theorem has mentionned nothing about it, so it shall be excluded in this case.

Feel free to write your own theorem.