# Increasing and Decreasing function given f'(x)

• Jun 11th 2008, 09:18 PM
Myung
Increasing and Decreasing function given f'(x)
Ok this has been killing me for a while.

Here's a random graph to illustrate my points.
http://www.analyzemath.com/calculus/...ve_graph_1.gif
So this is f'(x). At points x=-3 and x=-2 the function is increasing. What gets me is that my teacher says it's a close function or [-3,-2].

In my opinion it's (-3,-2). Here are the definitions we use:

[Definition 4.1]
A function f is (strictly) increasing on an interval I if for every x1x2 e I with x1 < x2, f(x1) < f(x2) [ie., f(x) gets larger as x gets larger]

Now using this definition I can see how it's a closed interval. But we are given the derivative of the graph, so here's the Theorem for a derivative of a graph.

[Theorem 4.1]
Suppose that f is differentiable on an interval I.

(i) if f'(x) > 0 for all x e I, then f is increasing on I.

So let's use [-3,-2] for I. At the points -3 and -2, it is not differentiable because we can't find the derivative at endpoints. So wouldn't it make this an open interval (-3,-2)? Or if we use an open interval outside to the left os -3 and to the right of -2, the derivative at -3 and -2 is zero, so the theorem these points can't be included because they are not increasing according to theorem 4.1! This has me so puzzled. Which do I use and what about theorem 4.1?
• Jun 11th 2008, 09:21 PM
Mathstud28
Quote:

Originally Posted by Myung
Ok this has been killing me for a while.

Here's a random graph to illustrate my points.
http://www.analyzemath.com/calculus/...ve_graph_1.gif
So this is f'(x). At points x=-3 and x=-2 the function is increasing. What gets me is that my teacher says it's a close function or [-3,-2].

In my opinion it's (-3,-2). Here are the definitions we use:

[Definition 4.1]
A function f is increasing on an interval I if for every x1x2 e I with x1 < x2, f(x1) < f(x2) [ie., f(x) gets larger as x gets larger]

Now using this definition I can see how it's a closed interval. But we are given the derivative of the graph, so here's the Theorem for a derivative of a graph.

[Theorem 4.1]
Suppose that f is differentiable on an interval I.
(i) if f'(x) > 0 for all x e I, then f is increasing on I.

So let's use [-3,-2] for I. At the points -3 and -2, it is not differentiable because we can't find the derivative at endpoints. So wouldn't it make this an open interval (-3,-2)? Or if we use an open interval outside to the left os -3 and to the right of -2, the derivative at -3 and -2 is zero, so the theorem these points can't be included because they are not increasing according to theorem 4.1! This has me so puzzled. Which do I use and what about theorem 4.1?

Increasing And Decreasing
• Jun 11th 2008, 09:30 PM
Myung
Quote:

Originally Posted by Mathstud28
Increasing And Decreasing

Ok, so it's closed interval. I understand why using [Definition 4.1] but what about Theorem [4.1]? The derivative at x=-2 and x=-3 are 0, so it is not increasing or decreasing at that point specifically, so it can't be included right? The point that's confusing me is that contradiction.
• Jun 11th 2008, 10:11 PM
Myung
If I wasn't clear, what I mean is [Theorem 4.1] tells me that if the derivative at a point is greater than 0, it is increasing. But the derivative at -2 and -3 is not greater than 0, so it shouldn't be increasing there. Normally when looking at a derivative graph, I would use this derivative theorem. So why is this theorem null? It feels like I'm completely ignoring it when looking for increasing/decreasing functions.

Edit: Does (strictly) increasing or increasing affect anything?
• Jun 11th 2008, 11:12 PM
Myung
Incidentally, the textbook has it as open interval which confuses me even more. Any real answers? The problem seems simple but this contradiction is killing me. I have a hard time learning new material if I believe the old material I've learned is incorrect so that's why I want to clear this up.
• Jun 11th 2008, 11:52 PM
Reckoner
Quote:

Originally Posted by Myung
If I wasn't clear, what I mean is [Theorem 4.1] tells me that if the derivative at a point is greater than 0, it is increasing. But the derivative at -2 and -3 is not greater than 0, so it shouldn't be increasing there.

Get your logic straight. Just because a theorem says that the truth of a statement $p$ implies the truth of a statement $q$ does not mean that not- $p$ implies not- $q$.

The theorem states: $\text{If }f'(x) > 0\;\forall x\in I, \text{ then }f\text{ is increasing on }I$.

The theorem does not state that $f\text{ is increasing on }I\text{ only if }f'(x) > 0\;\forall x\in I$ nor does it say $f'(x) > 0\;\forall x\in I\text{ if and only if }f\text{ is increasing on }I$. Learn the distinction.

From your definition, you can see that a closed interval is appropriate. The book might use open intervals to avoid confusing students by saying things like $x^2\text{ is decreasing on }(-\infty,\;0]\text{ and increasing on }[0,\;\infty)$ where a student might think that the function is both increasing and decreasing at 0, which is not the case.

Edit: About the use of the word "strictly": sometimes an increasing function will be defined as one for which $f(b)\;{\color{red}\geq}\;f(a)\text{ for }b > a$, and then when you want to talk about an increasing function where $f(b)$ is always greater than but not equal to $f(a)$ (what you normally think of as "increasing") you say that the function is strictly increasing. Sometimes you will hear words like "nondecreasing" to emphasize that the function may increase or stay constant on the interval, but never actually decrease.
• Jun 11th 2008, 11:55 PM
Moo
Hello,

I do not really agree when you say that it does not increase or decrease at the end points. Actually, you only care about the part right to -3, and the part left to -2.
I can't really explain, but it's clearly defined...
Imagine a curve where the tangent at certain points is 0, that is to say an horizontal line. ermm...
I'll try to make an example...

• Jun 12th 2008, 12:03 AM
Myung
Quote:

Originally Posted by Reckoner
Get your logic straight. Just because a theorem says that the truth of a statement $p$ implies the truth of a statement $q$ does not mean that not- $p$ implies not- $q$.

The theorem states: $\text{If }f'(x) > 0\;\forall x\in I, \text{ then }f\text{ is increasing on }I$.

The theorem does not state that $f\text{ is increasing on }I\text{ only if }f'(x) > 0\;\forall x\in I$ nor does it say $f'(x) > 0\;\forall x\in I\text{ if and only if }f\text{ is increasing on }I$. Learn the distinction.

From your definition, you can see that a closed interval is appropriate. The book might use open intervals to avoid confusing students by saying things like $x^2\text{ is decreasing on }(-\infty,\;0]\text{ and increasing on }[0,\;\infty)$ where a student might think that the function is both increasing and decreasing at 0, which is not the case.

I realise this, but at the points -2 and -3, the derivative is 0, which I probably should have stated even though it's implied when I say it's not > 0. So if the derivative at -2 is 0, the derivative of the function f(x) at that specific point is constant, which means it's not increasing at that specific point. Which means it doesn't include the endpoints, right? That's my logic in a nutshell.

And the idea of the textbook throwing out the wrong answer without any clarification seems a bit incredulous to me.
• Jun 12th 2008, 12:05 AM
Moo
Quote:

Originally Posted by Myung
I realise this, but at the points -2 and -3, the derivative is 0, which I probably should have stated even though it's implied when I say it's not > 0. So if the derivative at -2 is 0, the function f(x) is constant, which means it's not increasing at that specific point. Which means it doesn't include the endpoints, right? That's my logic in a nutshell.

You can't talk about a function being constant at one point...

If I say f(5)=14, I would be able to state that the function is the constant 14 at the point 5, therefore, the function is constant at this point.
But your questions are quite interesting in some way...
• Jun 12th 2008, 12:12 AM
Reckoner
Quote:

Originally Posted by Myung
So if the derivative at -2 is 0, the function f(x) is constant, which means it's not increasing at that specific point. Which means it doesn't include the endpoints, right? That's my logic in a nutshell.

We define increasing and decreasing on intervals, not at points. Doesn't $y = x^3$ increase over the entire real line, despite $\frac{dy}{dx} = 0$ when $x = 0$ (look at your definition)?

And if you want to define what it means to be increasing at a point, you have to realize that a function increasing on an interval will not necessarily be increasing at all points in that interval, depending on how you choose your definition.
• Jun 12th 2008, 12:12 AM
Myung
Quote:

Originally Posted by Moo
Hello,

I do not really agree when you say that it does not increase or decrease at the end points. Actually, you only care about the part right to -3, and the part left to -2.
I can't really explain, but it's clearly defined...
Imagine a curve where the tangent at certain points is 0, that is to say an horizontal line. ermm...
I'll try to make an example...

Thx for the drawing. I understand that part, so if the derivative at a certain point is constant, it can still be increasing at that isolated point? I guess this is the part that confounds me.
• Jun 12th 2008, 12:15 AM
Myung
Quote:

Originally Posted by Moo
You can't talk about a function being constant at one point...

If I say f(5)=14, I would be able to state that the function is the constant 14 at the point 5, therefore, the function is constant at this point.
But your questions are quite interesting in some way...

I said derivative of the function at the point, I think you misquoted me.
• Jun 12th 2008, 12:20 AM
Moo
Quote:

Originally Posted by Myung
I said derivative of the function at the point, I think you misquoted me.

The derivative is a function.
I repeat : it doesn't make much sense to say that a function (or its derivative) is constant at one point.

The graphical "translation" of the derivative number (i.e. the value of the derivative at a given point) is that it is the slope of the tangent to the curve (see, the grey lines).
• Jun 12th 2008, 12:27 AM
Myung
This has definitely cleared up a lot. Thx for the responses Moo and Reckoner.