Strictly increasing vs. increasing

[Shackleford]

This is starting to tick me off. I've emailed the professor about this. It seems like contradictory information. According to the textbook:

If the derivative of a function is bigger than or equal to zero on an interval, it is increasing.

If the derivative of a function is bigger than zero on an interval, it is strictly increasing.

Professor,

For HW11 #5(a), should it be show that f is increasing on I, not strictly increasing? The derivative of f is zero at pi/2. Strictly increasing requires the derivative to be greater than zero for all points on the interval.

f(x) = x^3 is strictly increasing on (-5, 5) and f'(0) = 0.

If f'(x) > 0 on I, then f is strictly increasing on I.

If f is strictly increasing on I, then f' is greater than or equal to 0 on I.

The book is making a distinction between strictly increasing and increasing. According to these definitions, #4 is strictly increasing, but #5 is only increasing. Just to clarify,

for #4, f(x) = x + 2(root2), and
for #5, f(x) = x - pi + cos x.

On page 249, problem 26.8 says f is increasing on I iff the derivative is bigger than or equal to 0 for all x in I.

On page 245, Theorem 26.8 says if the derivative is bigger than 0 for all x in I, then f is strictly increasing.

The bottom line is: strictly increasing does not imply f'(x)>0. See my example below.

How does it not imply the derivative is bigger than zero if that's what's in the definition?

 

[LCKurtz]

This is starting to tick me off. I've emailed the professor about this. It seems like contradictory information. According to the textbook:

If the derivative of a function is bigger than or equal to zero on an interval, it is increasing.

If the derivative of a function is bigger than zero on an interval, it is strictly increasing.

How does it not imply the derivative is bigger than zero if that's what's in the definition?

You are asking about the converse of the statement I have bolded which is:

If f is strictly increasing on an interval then f'(x) > 0 on the interval.

Converses of theorems aren't always true, and this converse isn't. The problem is that f'(x) can be zero at a single point and not prevent the function from being strictly increasing. Look at the graph of f(x) = x³ at x = 0. Slope 0 and horizontal tangent line there, but still x < y implies x³ < y³ so the function is strictly increasing.

 

[Shackleford]

You are asking about the converse of the statement I have bolded which is:

If f is strictly increasing on an interval then f'(x) > 0 on the interval.

Converses of theorems aren't always true, and this converse isn't. The problem is that f'(x) can be zero at a single point and not prevent the function from being strictly increasing. Look at the graph of f(x) = x³ at x = 0. Slope 0 and horizontal tangent line there, but still x < y implies x³ < y³ so the function is strictly increasing.

This is silly. Forget the converse and logic. I have to use language first. When I read that sentence, I have to ask myself what does the term strictly increasing mean? Well, I would go to that definition.

Since the definition of increasing includes bigger than or equal to zero, what else would I conclude? Why is not valid to say this is an increasing function?

Oh, I forgot about that last part. It seems the derivative characterization is simply a general tool to determine if a function is strictly increasing, but there are some point-wise exceptions. The essential aspect of it is to determine is if x < y, then f(x) < f(y).

 

[LCKurtz]

This is silly. Forget the converse and logic. I have to use language first. When I read that sentence, I have to ask myself what does the term strictly increasing mean? Well, I would go to that definition.

OK. If my response is silly and you feel the need to explain to me how I should help you, you obviously don't need any more help from me.

 

[Mark44]

The difference between increasing versus strictly increasing is the difference between <= and <.

If a function is increasing on an interval [a, b], and if x < y, then f(x) <= f(y).
If a function is strictly increasing on an interval [a, b], and if x < y, then f(x) < f(y).

The function f(x) = 3 would be considered increasing, but not strictly increasing. (Note that f'(x) = 0 on any interval.)

The function g(x) = 2x + 1 is strictly increasing. Note that g'(x) = 2 > 0 on any interval.

 

[Shackleford]

OK. If my response is silly and you feel the need to explain to me how I should help you, you obviously don't need any more help from me.

No. I didn't mean your response is silly. I was talking about the problem. I forgot about the most important part which is the last thing you mentioned. Your response elucidated it for me. Thanks.

 

[Shackleford]

The difference between increasing versus strictly increasing is the difference between <= and <.

If a function is increasing on an interval [a, b], and if x < y, then f(x) <= f(y).
If a function is strictly increasing on an interval [a, b], and if x < y, then f(x) < f(y).

The function f(x) = 3 would be considered increasing, but not strictly increasing. (Note that f'(x) = 0 on any interval.)

The function g(x) = 2x + 1 is strictly increasing. Note that g'(x) = 2 > 0 on any interval.

Yeah, I was being a bit singleminded in this by just focusing on the derivative condition in the theorem and statement. I forgot that what determines strictly increasing is if x < y, then f(x) < f(y). In the case of x³, even though there is a point which the derivative is zero, the strictly increasing property still holds, like the first guy mentioned.

 

[LCKurtz]

No. I didn't mean your response is silly. I was talking about the problem. I forgot about the most important part which is the last thing you mentioned. Your response elucidated it for me. Thanks.

OK, I can go with that. Thanks for clarifying.