Maybe This Will Spark A Debate
The graph of y = x 3 is always increasing.
Is the above statement true or false?
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9 solutions
While I agree with your point, there is something I must dispute.
The question asks if the function is ALWAYS increasing. A derivative providing a solution of 0 does not technically provide an increasing solution, it merely provides a critical point. Therefore, the rationale is false. If the function had excluded the value of x=0, then yes the function would ALWAYS be increasing but since it was included, the term "always" is false and therefore the answer should be marked false.
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I find this sort of question silly and frustrating as it comes down to a definition. I would not regard f '(0) = 0 increasing as it is clearly stationary. However if increasing means not decreasing then the answer is false. Terrible question in my opinion.
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It's very tempting to say that the function is not increasing at x = 0 because the derivative is 0 there. However, the derivative that we learn in calculus isn't what gives us our intuition of the word "increasing." The concept of increasing exists a priori.
-3, -2, -1, 0, 1, 2, 3, .... Is this sequence increasing? Yes! Do you need a derivative to verify that? No.
Now, let's look at the function
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Is this increasing? Also, yes. No derivative is required to see that every
y
-value is larger than the previous.
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@Andrew Ellinor – y isn't larger than x for values of 1, -1 and 0..........which means that the graph isn't ALWAYS increasing. Not ALWAYS, even though mostly.
One thought is that we can interpret "always increasing" as "never decreasing", in which case the answer would be "true".
Another thought is that, even though f ′ ( 0 ) = 0 , at no point is the function not strictly increasing. So while x = 0 is indeed a critical point (with both f ′ ( 0 ) = 0 and f ′ ′ ( 0 ) = 0 ) we have that f ( − ϵ ) < f ( ϵ ) for any ϵ > 0 . So as f ( x ) is strictly increasing on any neighborhood of x = 0 , at no point is the function not increasing, i.e., f ( x ) = x 3 is "always" increasing, even "at" x = 0 .
As Andrew intended in posting this question, a debate has been sparked. :)
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f
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x
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>
0
is a sufficient, but not necessary condition for a function to be increasing. There are several edge cases that we need to consider. Even for a function to be strictly increasing, it's not sufficient that
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0
. If you want to state it in calculus terms, what we need is:
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{
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does not contain any intervals.
For a similar analogy, when asked to find the local maximum and minimum of a function, it is not sufficient to only look at f ′ ( x ) = 0 . We also have to consider f ′ ′ ( x ) , in case we have an inflection point.
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@Calvin Lin – What you wrote is wrong, "sufficient but not sufficient"!?! It should have been, "sufficient but not necessary ..." I am surprised nobody has pointed this out for over two days.
Real simply put: if you pick an x extremely close to zero, but still negative, say -0.0001, then y is -0.0000000001. Then at x = 0, you get y = 0. Next at x = 0.0001, y = 0.0000000001. Always increasing no matter how close to 0 you choose an x to be.
Yes, a critical point exist at x = 0, but the function is increasing on both sides of the critical point and it is a critical point with a real value in the original function, therefore, the function is still considered to bet increasing at that critical point.
Hi Timothy, I like your argument and I used to think that a derivative of 0 means that a function is not increasing.
But let me put it to you this way: Can you find values of a and b such that a < b , but f ( a ) > f ( b ) ?
I agree. I thougt the same way
I agree Timothy. A stationary point has no gradient and thus there is no increase at that point
Yes, according to the formal definition, even the constant function y = 2 is technically an increasing function over R .
It's interesting to note that f ( x ) = x 3 can also be considered to be a "strictly increasing" function over R , even though f ′ ( 0 ) = 0 . (For a function f ( x ) to be strictly increasing on an interval I we must have f ( a ) < f ( b ) for all a < b , where a , b , ∈ I . ) To prove this, we can look at b 3 − a 3 given a < b ⟹ b − a > 0 . We have that
b 3 − a 3 = ( b − a ) ( b 2 + a b + a 2 ) = ( b − a ) ( 2 b 2 + a 2 + ( a + b ) 2 ) .
We are given that ( b − a ) > 0 , and ( b 2 + a 2 + ( a + b ) 2 ) > 0 except when a = b = 0 . Thus b 3 > a 3 whenever b > a , qualifying f ( x ) = x 3 as a strictly increasing function over all reals.
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The definition, f ( b ) ≥ f ( a ) ∀ b > a is the definition for non reducing function and does not seem to be the definition for an increasing function.
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I'm assuming that "non-reducing" is equivalent to "non-decreasing", in which case the definitions, according to Wolfram MathWorld, of an increasing function and a non-decreasing function are identical.
That said, there do seem to be different conventions when applying these terms. Sometimes "increasing" is used for "strictly increasing", (which I suppose is closer to "non-math" usage), for example. In this convention, a function meeting the f ( b ) ≥ f ( a ) ∀ b > a condition would be referred to as "non-decreasing", and would only be considered as "increasing" if f ( b ) > f ( a ) ∀ b > a . So it would appear as though we are using different conventions; neither one is better than the other, but on this site I find that whatever convention Wolfram uses is considered to be the "standard" one. :)
The problem is misleading, what is being discussed has very little to do with the particular function, it is about the definition of what "always increasing" means. As an engineer I can say that, at zero, the stuff-thing-device-spacecraft-cableway is flat and does not increase. Depending on the scale used, a ball placed at that point would be in unstable equilibrium. Therefore, the answer is in the eyes of the beholder.
The debate here seems to be between those who say "false, the derivative at x=0 is 0 " and those who say "true, there is no interval where the function fails to increase." You defend the latter with the formal definition of "increasing". It is unfortunate that the author of the question used the English phrase "ALWAYS increasing" (emphasis mine), introducing the ambiguity. I suggest that unless there is a precise mathematical definition of "always increasing" versus "increasing over every interval" (leading to "true") and versus "for every value x" (leading to false), then the debate is pointless.
I am a really simple bloke, but does the graph accompanying the question not show that it is not always increasing?
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Haha, no worries Ian! It certainly looks like it goes flat at x = 0 , but if you could magnify the center of the graph, you would find something interesting... All of the points are arranged so that ANY two points you choose, the point on the right is higher than the point on the left. Isn't that what it means to be increasing?
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For an infinitely small period, there is no increase.
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@Landon Johnson – That would be the smae for any function wouldn't it?
Yes I suppose I was thinking that at point 0 the line is not increasing, so for a single value it neither increase or decreases, however I accept your argument.
What if one says:
limit(x approaches 0 from the left) of f ' is 0 and limit(x approaches 0 from the left) of f ' is 0.
They obviously approach the same value and as shown, the derivative is also 0 from both sides.
Will this not mean that between the limit from the left and limit from the right, the function is not increasing?
Perhaps it should be explicitly stated that the interval is (-inf, inf) because this would not hold true for (inf, -inf) and the definition of "always" is debatable
I refute your answer as you presented an assumption that the function as graphed is always going from left to right (quadrant 3 to quadrant 1). What if the values of X move the graph in the opposite direction?
You really need the distinction between monotone increasing and strictly increasing. By this logic a constant function is increasing.
- so there are some mistaqe because for x= - 1,0 and 1 the graph of this function dont increasing so the right answer not can being ,,TRUE"
- the above reply hope so much that is more understandably
It seems to me that the question needs to add the the phrase "as x increases " to be accurate. Remember that derivative limits may be approached from left or right. A positive slope implies that the value for y increases as x increases, or decreases as x decreases.
But what about when x approaches negative infinity? I didn't think it was implied that x is approaching positive infinity.
As Andrew points out, we need to show that f ( b ) ≥ f ( a ) whenever b > a . By the mean value theorem, we have b − a f ( b ) − f ( a ) = f ′ ( c ) = 3 c 2 ≥ 0 for some c on the interval ( a , b ) , so that f ( b ) − f ( a ) ≥ 0 and f ( b ) ≥ f ( a ) as claimed.
To show that f ( x ) = x 3 is strictly increasing, as Brian points out, we can refine the argument above. By the mean value theorem, f ( x ) is strictly increasing for x ≥ 0 and also for x ≤ 0 , so that it is strictly increasing for all reals.
Moderator note:
I like appealing to the mean value theorem to show that we have an increasing function. In fact, this is the reason why
f is an increasing function if and only if f ′ ≥ 0 .
An alternative is to directly show that for k > 0 , ( a + k ) 3 − a 3 = k ( k 2 + 3 a k + 3 a 2 ) > 0 .
I wrote this with some trickery in mind, thinking that students who are new to calculus might think the function isn't actually increasing at x = 0 . Admittedly, even I fell victim to this trap back when I was first learning calculus. Do you or @Brian Charlesworth have any light to shed on encountering those who had this same failing of intuition?
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There's always the obvious one of "A continuous function can be drawn without lifting the pen form the page".
I was a victim. I forgot we must look at two points, because we are deciding whether or not the function is always increasing over any given interval. Thanks for ruining my day (lol)
f(x) = c ; f'(x) = 0 so f'(x) >= 0 for all x. But f is not an increasing function. so your theorem is false.
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According to the definition we use here, increasing means that f ( a ) ≤ f ( b ) whenever a < b . According to this definition, a constant function is indeed "increasing", somewhat counter-intuitively.
Simple understanding is to look forward on its increasing trend. So, the graph is always an increasing function. Mathematical proof is fine but with your intuition is simple enough.
For x=0 y=0, x=-1 y=-1 and for x=-2 y=-8, x=-3 y=-29 the results of y are geting smaller and smaller related to x, where x is smaller than -1. How could we say it is an increasing function ALWAYS??
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The answer has been well explained by these guys. Thanks. :)
cos x is decreasing
It's conventional to say that -3 < -2 < -1, and so on. Although |x| is decreasing as we move from x = -3 to x = -2 to x = -1, the actual VALUE of x is said to be increasing.
Rule of thumb: when comparing two numbers, the number that is either "more positive" or "less negative" is said to be greater.
Let's imagine you draw out the number line, with negative infinity to the left, 0 at center, and positive infinity to the right. For any two values A and B, if B appears further to the right than A, then B > A. This is why using absolute value alone can get confusing when discussing questions of "greater" or "lesser".
The entire question comes down to exponents in a graphing situation. And since the graphing is a progression of y/x (otherwise known as the slope) and the origin (0,0) is only one set of coordinate pairs, we can discard the notion that existence of the x-coordinate as 0 would lead to no growth because that is only one x coordinate in a set.
A positive odd monomial will always be constantly increasing, whether it's x, x^3, or x^67000001. It's only a visual effect of the numbers having such small absolute values that makes the graph appear to flatten between the x-values of -1 and 1. If you think about it mathematically, it makes sense that x^3 would increase forever without incident.
I think y=x^3 is always increasing. And y=-x^3 is decreasing. So the answer is true.
I remembered when I watched mean girls. "The limit does not exist!"
Calculus wise, the first derivative is always positive except at x=0, so yes it is increasing (except for a brief horizontal slope at 0).
y = x^3 => y' = 3x^2 => at x = 0 and y = 0; gemotrically P(x,y)= (0, 0); therefore, the graph provides an increasing function.
Students who are new to calculus might immediately seek to take a derivative (and rightly so!) because the derivative gives insight about whether a function is increasing or decreasing. Indeed, the derivative of y = x 3 is y ′ = 3 x 2 , which takes on a value of 0 at x = 0 . But does this mean the function isn't always increasing?
Before we start answering this question from the perspective of calculus, we should actually know what it means for a function to be increasing. We say a function is increasing on the interval I if f ( a ) ≤ f ( b ) for all a < b in I . Indeed, we have that for every choice of a < b on the real number line, it is guaranteed that a 3 ≤ b 3 , that is f ( a ) ≤ f ( b ) . This is true even if one of a or b is 0! Therefore, it is true that the function y = x 3 is always increasing.
So now we must defuse the argument that says, "We know that y ′ ( 0 ) = 0 , which implies that the function y = x 3 is not increasing at x = 0 ." Simple enough. This argument is a misrepresentation of what the derivative of a function can tell us. The derivative merely tells us that if f ′ ( x ) > 0 for all x on a given interval I , then f is increasing on I . That does not imply that f is not increasing if f ′ ( x ) = 0 for a particular x in I .