on what interval(s) is the function f(x) = x^3 increasing?
Kenneth Bull
a) (-oo, 0) U (0, oo)
b) (-oo, oo)
[Can someone prove their answer?]
I pick a because:
f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is increasing on
that interval
f'(0) = 0 so f is not increasing at 0.
Agreed?
David Moran
"Kenneth Bull" <kennet...@gmail.com> wrote in message
news:1137363568.7...@g44g2000cwa.googlegroups.com...
Looks good to me.
Dave
Rob
f is increasing on this interval.
At 0, it could be said to be increasing; it depends on your definition.
Here are what I think are usual definitions:
A function is said to be increasing on some interval A if for every x,y
in A, x < y implies f(x) <= f(y). A function is said to be strictly
increasing on some interval A if for every x,y in A, x < y implies f(x)
< f(y).
Similarly, on points themselves, a function is said to be increasing on
some interval A if for every x in A, f'(x) >= 0. Similarly, a function
is said to be strictly increasing on some interval A if for every x in
A, f'(x) > 0.
These two definitions are equivalent (see Bartle, R. G., Sherbert, D.
R., Introduction to Real Analysis, 3rd Edition for proofs)
JeeBee
>
> "Kenneth Bull" <kennet...@gmail.com> wrote in message
> news:1137363568.7...@g44g2000cwa.googlegroups.com...
>> On what interval(s) is f(x) = x^3 increasing?
>>
>> a) (-oo, 0) U (0, oo)
>>
>> b) (-oo, oo)
>>
>> [Can someone prove their answer?]
>>
>> I pick a because:
>>
>> f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is increasing on
>> that interval
Ehm, no, f'(x) = 3 * x^2.
But the rest seems ok :)
A Aitken
b)
Proof:
Let p,q be any real numbers, such that p < q.
p < q <==> p^3 < q^3 <==> f(p) < f(q)
Thus for any p < q, f(p) < f(q).
++++++++++++++
Andre Aitken
A Aitken
Rob wrote:
> f'(x) = 3x^2, not just x^2, but ya, f'(x) > 0 on (-oo,0)U(0,oo), and so
> f is increasing on this interval.
>
> At 0, it could be said to be increasing; it depends on your definition.
> Here are what I think are usual definitions:
>
> A function is said to be increasing on some interval A if for every x,y
> in A, x < y implies f(x) <= f(y). A function is said to be strictly
> increasing on some interval A if for every x,y in A, x < y implies f(x)
> < f(y).
>
using this definition the OP's function is strictly increasing because
p < q <==> p^3 < q^3 directly (no need for <=)
>
> Similarly, on points themselves, a function is said to be increasing on
> some interval A if for every x in A, f'(x) >= 0. Similarly, a function
> is said to be strictly increasing on some interval A if for every x in
> A, f'(x) > 0.
>
using this definition the function is not strictly increasing
>
>
> These two definitions are equivalent
>
Not sure about that.
>
>(see Bartle, R. G., Sherbert, D.
> R., Introduction to Real Analysis, 3rd Edition for proofs)
++++++++++++++
Andre Aitken
A Aitken
Rob wrote:
> f'(x) = 3x^2, not just x^2, but ya, f'(x) > 0 on (-oo,0)U(0,oo), and so
> f is increasing on this interval.
>
> At 0, it could be said to be increasing; it depends on your definition.
> Here are what I think are usual definitions:
>
> A function is said to be increasing on some interval A if for every x,y
> in A, x < y implies f(x) <= f(y).
>
This makes little sense. f(x) = 0 is increasing based on the
definition.
>
> A function is said to be strictly
> increasing on some interval A if for every x,y in A, x < y implies f(x)
> < f(y).
>
> Similarly, on points themselves, a function is said to be increasing on
> some interval A if for every x in A, f'(x) >= 0. Similarly, a function
> is said to be strictly increasing on some interval A if for every x in
> A, f'(x) > 0.
>
Both cases here still fall under the second section of the first
definition, neither fall under the first secttion. The definitions are
not parellel.
>
> These two definitions are equivalent (see Bartle, R. G., Sherbert, D.
> R., Introduction to Real Analysis, 3rd Edition for proofs)
+++++++++++++++
Andrew Aitken
Stephen Montgomery-Smith
No. f'(x) > 0 implies that f is increasing, but the converse is not
true. (Indeed f(x)=x^3 is the simplest counterexample.)
Aitken did a good job of explaning why (b) is true.
quasi
wrote:
No.
The definition of increasing is not based on derivatives.
f is increasing if x1<x2 implies f(x1)<f(x2).
Since x1<x2 implies x1^3<x2^3, it follows that f is increasing on
(-oo, oo).
The answer is (b).
quasi
[Mr.] Lynn Kurtz
wrote:
>On what interval(s) is f(x) = x^3 increasing?
As others have pointed out, the function is increasing on (-oo,oo).
But if that is a multiple choice question it is poorly stated. Is the
function increasing on (-oo,0)? Yes. Is it increasing on (0,oo)? Yes.
Is it increasing on (-oo, 0) U (0, oo)? Yes, although a point of the
domain is missing. It is also increasing on [1,3] or any other
interval you care to write. Hopefully you have simply paraphrased the
question carelessly. As stated, "yes" to all.
--Lynn
john_r...@sagitta-ps.com
Before seeing the other replies I'd have unhesitatingly agreed
(apart from f'(x) being 3.x^2 ! ), believing that a function is not
increasing at a stationary point; but now I'm confused.
Obviously there's a distinction between "increasing at a point"
and "increasing over an interval", despite the same word being
used in each phrase.
I think we need to wait for twenty or more people to reply, and
take a vote on the number of (a)s versus (b)s ;-P
(Also, Rob's definition of "increasing" is what I'd call
"non-decreasing",
reserving "increasing" for what he calls "strictly increasing"; but
that's
a side-issue.)
Kenneth Bull
You're right. I should have said "what is the greatest set of real
numbers on which f(x) could be considered to be increasing?"
Kenneth Bull
Try pondering on which greatest set of real numbers f(x) = x^4 is
convex on (this is closely related to the original question).
quasi
wrote:
>
Same answer (-oo, oo).
The definition of convexity is also not based on derivatives.
quasi
True Raptor
> Kenneth Bull wrote:
>
>>On what interval(s) is f(x) = x^3 increasing?
>>
>>a) (-oo, 0) U (0, oo)
>>
>>b) (-oo, oo)
>>
>>[Can someone prove their answer?]
>>
>>I pick a because:
>>
>>f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is increasing on
>>that interval
>>
>>f'(0) = 0 so f is not increasing at 0.
>>
>>Agreed?
>
>
> Before seeing the other replies I'd have unhesitatingly agreed
> (apart from f'(x) being 3.x^2 ! ), believing that a function is not
> increasing at a stationary point; but now I'm confused.
>
> Obviously there's a distinction between "increasing at a point"
> and "increasing over an interval", despite the same word being
> used in each phrase.
>
What's more peculiar is that (0,0) is a "stationary point." Someone
named points on a function where the derivative vanishes as
"stationary." The name suggests that if there was such a thing as
"increasing or decreasing at a point" then stationary points would be
neither.
The World Wide Wade
<1137367861.3...@g44g2000cwa.googlegroups.com>,
"Kenneth Bull" <kennet...@gmail.com> wrote:
> Try pondering on which greatest set of real numbers f(x) = x^4 is
> convex on (this is closely related to the original question).
Clearly all of R. What's the definition of convexity?
True Raptor
> f'(x) = 3x^2, not just x^2, but ya, f'(x) > 0 on (-oo,0)U(0,oo), and so
> f is increasing on this interval.
>
> At 0, it could be said to be increasing; it depends on your definition.
> Here are what I think are usual definitions:
>
> A function is said to be increasing on some interval A if for every x,y
> in A, x < y implies f(x) <= f(y). A function is said to be strictly
> increasing on some interval A if for every x,y in A, x < y implies f(x)
> < f(y).
>
g(x) = { 4 on [-4,4]
{ x - 4 on (4, inf)
{ x + 4 on (-inf,-4)
By the first definition you state, g(x) is increasing everywhere.
It would make little sense to call g(x) an increasing function (on it's
domain).
quasi
wrote:
>
Aren't you using a textbook?
How about looking up "increasing function" and "convex function" in
the index?
After you check the definitions, you should be easily able to verify:
If n is an odd positive integer, f(x)=x^n is increasing on (-oo, oo).
If n is an even positive integer, f(x)=x^n is convex on (-oo, oo).
quasi
john_r...@sagitta-ps.com
That I do agree with, and as it's easy to see that a function
need not be convex in an interval in which it is strictly
increasing I'm not entirely sure that Kenneth is hinting at.
quasi
Perhaps the connection he has in mind is this:
Let f(x) be continuous and let F(x)=int(f(t),t=0..x). Then:
f is increasing iff F is convex.
f is convex iff F is increasing.
quasi
Kenneth Bull
f'(x) = 4x^3
My book defines concavity of a function using f' (convex if f' > 0 ,
concave if f' < 0)
john_r...@sagitta-ps.com
I agree with that if your "increasing" means "non-decreasing"
(as opposed to "strictly increasing").
quasi
Thanks for the alert -- something has to be revised.
Ok, here's what I meant to say:
Let f(x) be continuous and let F(x)=int(f(t),t=0..x). Then:
f is increasing iff F is strictly convex.
f is strictly convex iff F is increasing.
For the above claims, increasing means strictly increasing, thus
everything above strict.
A non-strict version would go as follows:
f is non-decreasing iff F is convex.
f is convex iff F is non-decreasing.
In the case of x^3 and x^4, the strict version applies.
quasi
quasi
wrote:
You mean f'', not f', but even so ...
Are you sure you're referencing the actual definition? It might be
that what you are calling the definition is not the definition, but
the following proposition:
If f''>0 for all x, then f is convex.
If f''<0 for all x, then f is concave.
In other words, the implication is in one direction only, not the
direction you're claiming.
Just to be sure, can you tell us which text you're using?
quasi
The World Wide Wade
<1137372499.8...@g14g2000cwa.googlegroups.com>,
"Kenneth Bull" <kennet...@gmail.com> wrote:
> f'(x) = 4x^3
>
> My book defines concavity of a function using f' (convex if f' > 0 ,
> concave if f' < 0)
Try to be careful. You mean f''. While it is true that f'' > 0 in
an interval implies f is convex there, the converse is false
assuming the usual definition of convexity in mathematics. See
Kenneth Bull
Correction, it defines concave up and down (concave and convex) using a
vague sentence "if the graph lies above all its tangents on an
interval, it is concave up on the interval" and similar for concave
down.
Right after this it presents the Concavity Test which uses f'' to test
for concavity.
But I realize what you are saying about the one direction implication.
quasi
wrote:
Ok, so x^4 lies above all its tangents, right?
quasi
Virgil
quasi <qu...@null.set> wrote:
And below all its secant segments (joining two points of the curve).
This test has the advantage of not needing anything about tangents,
which need not exist at all points of convexity.
E.g., f(x) = (x^4)^(1/3) is convex everywhere but has no tangent at x=0.
Luke Wu
f'(x) = [ 4 * x^(1/3) ] / 3
f'(0) = 0
Tangent at x=0 has slope 0
quasi
Well, I prefer the usual algebraic definition but the version you
describe using secants is also fine since it's equivalent.
>E.g., f(x) = (x^4)^(1/3) is convex everywhere but has no tangent at x=0.
Right.
But given that his text defines a convex function as a function whose
graph lies above its tangents, he can at least get the correct answer
for x^4.
quasi
quasi
wrote:
Right.
Ok, how about this example instead:
f(x)=exp(|x|)
f is convex on (-oo, oo) but not differentiable at x=0.
quasi
quasi
For that matter, a simpler example, also convex but not differentiable
at x=0, is |x|.
However exp(|x|) is strictly convex, whereas |x|, though convex, is
not strictly convex.
quasi
Luke Wu
Yes. But the original example given by the other fellow was
differentiable at x=0, that's why I responded to note that. I know
that differentiability is not tied to the definition of convex/concave
or increasing/decreasing. You don't have to convince me.
> >
> >quasi
>
> For that matter, a simpler example, also convex but not differentiable
> at x=0, is |x|.
>
|x| is not convex. If your notion of non-strict convex allows for the
graph to lie on secants (instead of below), then that notion is weird.
That's as weird as the non-strict increasing that was described by
another poster, which allows functions like y = 0.32 to be considered
increasing.
senga...@gmail.com
> On what interval(s) is f(x) = x^3 increasing?
>
> a) (-oo, 0) U (0, oo)
>
> b) (-oo, oo)
>
> [Can someone prove their answer?]
>
> I pick a because:
>
> f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is increasing on
> that interval
>
> f'(0) = 0 so f is not increasing at 0.
>
> Agreed?
The confusion stems from the fact that the idea of
increasing/decreasing is tied to a function, yet you are trying to use
the derivative of the function to investigate whether it is increasing
or decreasing.
There are several ways to use the derivative of a function to gain
insights into the functions. Say function f has derivative f'
it can be proved that if f'(x) > 0 on an interval A, then f is
increasing on A
[using the Mean Value Theorem]
This requires the derivative to exist, and it requires an interval over
which f'(x) > 0 is true. There is no notion of a "point" here because
the MVT has no business with points.
The converse of this is not generally true!!! Try to prove it, and you
will hit major road bumps.
it can also be proved that if f'(x) = 0 on an interval B, then f is
constant on B
[using the Mean Value Theorem]
This requires the derivative to exist, and it requires an interval over
which f'(x) = 0 is true.
There is also no notion of a "point" here. The converse of this result
is true (easily proven). So you cannot apply the logic to a "single
point" where f'(x) = 0. Ex: (0,0) of x^3 doesn't mean the function is
not increasing at or around 0
Both of these results use the Mean Value Theorem, which works on pairs
of points (any general pair of points in the intervals). This is why
you can't gain insights into f when f' or higher derivative have
special qualities at single isolated points - it's invervals that
matter.
Now, it can also be proved that if f'(x) > 0 for all x in it's
domain(which is a single interval) except at the non-adjacent points a,
b, c, d ... etc where f'(a) =f'(b) = f'(c) = f'(d) = ...= 0 then f is
increasing on it's domain.
Your function f(x) = x^3 falls under this.
But the best way to prove that f(x) is increasing it to deal with the
function directly, instead of painfully trying to gain info into f(x)
through f'(x).
Cheers
quasi
wrote:
>
The standard definition of convex function is as follows:
f is convex if for any two points x1, x2 and any t in [0,1],
f(t*x1 + (1-t)*x2) <= t*f(x1) + (1-t)*f*(x2)
In other words, the definition of convex function defaults to a
non-strict version, whereas the definition of increasing defaults to a
strict version.
Thus, by the standard definition, |x| is a convex function.
quasi
Luke Wu
So in the default sense: So y = 0.32 is convex, but not increasing?
True Raptor
That's expensive. Is it heavy too? Good for propping up tables?
quasi
wrote:
>
>quasi wrote:
>> On 15 Jan 2006 19:53:12 -0800, "Luke Wu" <LookSk...@gmail.com>
>> wrote:
>>
>> >
>> >quasi wrote:
>> >> On Sun, 15 Jan 2006 22:31:13 -0500, quasi <qu...@null.set> wrote:
>> >>
>> >> For that matter, a simpler example, also convex but not differentiable
>> >> at x=0, is |x|.
>> >
>> >|x| is not convex. If your notion of non-strict convex allows for the
>> >graph to lie on secants (instead of below), then that notion is weird.
>> >
>> >That's as weird as the non-strict increasing that was described by
>> >another poster, which allows functions like y = 0.32 to be considered
>> >increasing.
>>
>> The standard definition of convex function is as follows:
>>
>> f is convex if for any two points x1, x2 and any t in [0,1],
>>
>> f(t*x1 + (1-t)*x2) <= t*f(x1) + (1-t)*f*(x2)
>>
>> In other words, the definition of convex function defaults to a
>> non-strict version, whereas the definition of increasing defaults to a
>> strict version.
>>
>
>So in the default sense: So y = 0.32 is convex, but not increasing?
Yes.
quasi
Virgil
"Luke Wu" <LookSk...@gmail.com> wrote:
> quasi wrote:
> > >Ok, how about this example instead:
> > >
> > >f(x)=exp(|x|)
> > >
> > >f is convex on (-oo, oo) but not differentiable at x=0.
>
> Yes. But the original example given by the other fellow was
> differentiable at x=0, that's why I responded to note that.
My error. Sorry!
Virgil
"Luke Wu" <LookSk...@gmail.com> wrote:
> quasi wrote:
> > >That's as weird as the non-strict increasing that was described by
> > >another poster, which allows functions like y = 0.32 to be
> > >considered increasing.
> >
> > The standard definition of convex function is as follows:
> >
> > f is convex if for any two points x1, x2 and any t in [0,1],
> >
> > f(t*x1 + (1-t)*x2) <= t*f(x1) + (1-t)*f*(x2)
> >
> > In other words, the definition of convex function defaults to a
> > non-strict version, whereas the definition of increasing defaults
> > to a strict version.
> >
>
> So in the default sense: So y = 0.32 is convex, but not increasing?
>
Right!. But it is non-decreasing and not /strictly/ convex, which would
require that
f(t*x1 + (1-t)*x2) < t*f(x1) + (1-t)*f*(x2)
David C. Ullrich
>
>Kenneth Bull wrote:
>> On what interval(s) is f(x) = x^3 increasing?
>>
>> a) (-oo, 0) U (0, oo)
>>
>> b) (-oo, oo)
>>
>> [Can someone prove their answer?]
>>
>> I pick a because:
>>
>> f'(x) = x^2 and f'(x) > 0 on (-oo,0)U(0,oo) thus f is increasing on
>> that interval
>>
>> f'(0) = 0 so f is not increasing at 0.
>>
>> Agreed?
>
>Before seeing the other replies I'd have unhesitatingly agreed
>(apart from f'(x) being 3.x^2 ! ), believing that a function is not
>increasing at a stationary point; but now I'm confused.
>
>Obviously there's a distinction between "increasing at a point"
>and "increasing over an interval", despite the same word being
>used in each phrase.
I don't recall ever seeing a definition of "increasing at x".
A function is increasing on a set if a < b implies f(a) < f(b).
Honest. Even if there is such a thing as "increasing at x",
the definition of "increasing on S" is positively _not_
"increasing at x, for all x in S."
Now, if I were going to _invent_ a definition of "f is
increasing at x" I think the most reasonable choice would
be "f(y) < f(x) for all y < x in some neighborhood of x
and f(x) < f(y) for all y > x in some neighborhood of x".
With that definition the funtion x^3 is increasing at 0.
What definition of "f is increasing at x" did you have in mind?
>I think we need to wait for twenty or more people to reply, and
>take a vote on the number of (a)s versus (b)s ;-P
>
>(Also, Rob's definition of "increasing" is what I'd call
>"non-decreasing",
>reserving "increasing" for what he calls "strictly increasing"; but
>that's
>a side-issue.)
************************
David C. Ullrich
G.E. Ivey
G.E. Ivey
I don't think there is any definition of "increasing" that makes sense of increasing AT A SINGLE POINT. A function is increasing on an interval is whenever x> y then f(x)> f(y). It's certainly true that if x> y then x
^3> y^3 for x, y any real numbers.
Ronald Bruck
<CB4...@raptorsfans.com> wrote:
> Rob wrote:
> > f'(x) = 3x^2, not just x^2, but ya, f'(x) > 0 on (-oo,0)U(0,oo), and so
> > f is increasing on this interval.
> >
> > At 0, it could be said to be increasing; it depends on your definition.
> > Here are what I think are usual definitions:
> >
> > A function is said to be increasing on some interval A if for every x,y
> > in A, x < y implies f(x) <= f(y). A function is said to be strictly
> > increasing on some interval A if for every x,y in A, x < y implies f(x)
> > < f(y).
> >
>
>
> g(x) = { 4 on [-4,4]
> { x - 4 on (4, inf)
> { x + 4 on (-inf,-4)
>
> By the first definition you state, g(x) is increasing everywhere.
> It would make little sense to call g(x) an increasing function (on it's
> domain).
How do you figure his definition says g is increasing?
g(3) = 4 > g(4.01) = 0.01.
--Ron Bruck
True Raptor
Sorry, should be 0 on [-4, 4]
g(x) = { 0 on [-4,4]