People always make up nonsense terms instead of using terms which already exist, causing more confusion for themselves and others around them. It's hard to teach these people who refuse to use the correct terms.
I've also heard them called singularities, but I think those apply specifically to complex analysis where you can get removeable discontinuities and non-removeable discontinuities (poles).
I think it would be easier to explain by first factoring numerator and denominator, and then showing the cases as: a) factors in denominator that become zero (remove those from domain), b) common factors in numerator and denominator (point breaks), and c) factors in denominator (and not in numerator) that become 0 (vertical asymptotes).
The after reduction in vertical asymptotes and point breaks would be cases like (x + 3)^2 or higher in the denominator. It would still be a vertical asymptote for that function because there was only a (x + 3)^1 in the numerator.
For such an easy function there is only one case where the domain value is NOT valid. It's when the denominator equals 0. This doesn't tell VA apart from point breaks. In order to tell them apart precisely you need calculus knowledge. lim left f(x)→x equals lim right f(x)→x while f(x) on x is undefined that leads to point breaks. If they don't equal this is not a point break something like f(x)=x+abs(x)/x when x=0
For a VA you need lim left f(x)→x=+inf and lim right f(x)→x=-inf OR lim left f(x)→x=-inf and lim right f(x)→x=+inf But I am not sure what VA means in Chinese so I don't know if it still is when both sides are + or -
No, you don't need calculus for this problem. The numerator and denominator have a factor of x+3. When x is not -3, these cancel out and the function will look like 2x/x+4. The only thing stopping these from being equal is the x+3. Thus, a hole. It's not the most rigorous explanation, but it's enough to see what's going on here.
@@robertlunderwood yep this is enough to prove there is a hole in this graph. But in case of other functions probably irrelevant to division you need the limit.
I did not know how to exactly find the hole so, I differentiated it and equated it to 0 as there is no point there in the graph so the slope will be 0, and it actually came to be -3!
It was just dumb luck if that worked here, and essentially wrong. When you solve the equation "slope = 0", you get the result x = -3, yes - but -3 is not part of the domain of the function, so x = -3 is actually _not_ a solution of the equation "slope = 0". "there is no point there in the graph so the slope will be 0" That does not follow at all. What _actually_ follows is that the function itself and hence also its slope is not defined at that point. "is not defined" is different from "is 0".
@@bjornfeuerbacher5514 oh, right thank you very much, but still how did it still come to be -3, there has to be some other factor right, which brought the answer?
@@abhaychordiya6489 Both the numerator and the denominator of the initial function contain a factor (x+3). Hence the numerator of the derivative will also contain such a factor. However, the denominator of the derivative then contains a factor (x+3)². So x=-3 is a zero of the numerator, but _not_ of the whole derivative.
I don’t like working with not equals “equations”. There are no properties of not equals. If you have x + 3 != 0, then 5 is a solution to this “equation”. I would reason: It cannot be zero. Find when it is zero. For example, x+3=0, so x=-3 Now we can negate that and say x does not equal -3. This follows from the rules of logic. The other method works in this situation, but might not if more complex situations.
because when we work with inequalities (like with ≠) we are not trying to find some of the solutions (like x=5) but we are trying to find all of the solutions, (x≠3)
in general, we are going to find what the inequality is equivalent to (like x≠3), not simply what implies the inequality (like x=5) this is true for equations too
You should think of equations as a type of statement, and each step of your process you derive a new statement whose truthfulness is the same as the one preceding. For example when you solve 3x-2=x+2, you assert that this is a statement you don’t know is true or not, or that it is true for an unknown x. Then you can subtract x from both sides and get 2x-2=2, another statement which is as true as the previous if you handle stuff like extraneous solutions or complex numbers or domain issues. Then you can add 2 and get 2x=4, another statement which is as true as the previous, meaning it’s as true as the first. Then you can divide by 2 to get x=2, which is as true as the preceding once again. If X is a bound variable, then check if x is 2. If it is, every preceding statement, including the original, is true, thereby solving the problem. If it is a free variable, then we have found the x such that the original statement is true. The same framework applies to inequalities. The “solution” is not always that which makes the statement true, because outside elementary mathematics there are problems which do not ask for the value of a free variable (example: proofs); the “solution” is in general a “more helpful” statement or a series of statements that are all equal to the original statement. As is often said in math, the goal is not always to solve a problem; reducing it to a previously solved problem eliminates the need to solve it completely because the solution can be derived from the previously solved problem, so why reinvent the wheel. The same applies to the inequalities. When he solves the inequality in the video, he starts with a statement stating that a polynomial quadratic in x cannot be equal to zero. He ends with the statement that X cannot be equal to -3 and X cannot be equal to -4. The latter statement is more helpful than the original, so it is the “solution” to the problem. Re-phrase it for yourself: I want to plug in x into this quadratic but I can’t have it equal zero. What numbers am I not allowed to put in? Ah, I cannot put in -3 and I cannot put in -4.
x^2 - 4 != 0, factor (x + 2)(x - 2) != 0 Where do you go from here? Is it (x - 2) != 0 OR (x + 2) != 0 like you have with equals 0. Actually you have to replace OR with AND. You get the same result, but I think it is better to think, when is it 0, and then negate that. So it is equal to 0 at 2 and -2. Then the negation is x is everything except those two numbers. Just my opinion based on the way maths is taught. The axioms and properties are for equalities and inequalities. I don’t think I’ve ever seen a property for not equals. If there is, let me know. Have a great day.
Yes, of course you can. It is called implicit derivative. Edit: I assume you mean something like f(x, y) = g(x, y). If you are talking about an equation with single variable, then answer is no as @DaisyGamer8367 says.
No, it's not the same trying to find where two functions intersect as where their derivatives are the same. For example: x=ln x does not have a solution in the real numbers, but if you differentiate both, there is a solution at x=1 because their derivatives are the same there.
If f(x)=g(x) everywhere, then f'(x)=g'(x) is also true. That is to say, if it's an identity then it's true. sin²(x)+cos²(x)=1 everywhere, so we can differentiate both sides and it will still work. 2sin(x)cos(x)+2cos(x)(-sin(x))=0 is true. But if it's just the intersection of two things, like x=x², then it's not true everywhere. It's not an identity, so differentiating both sides won't do anything useful at all.
