I think I would be tempted not to try out values, but jump straight to visualizing that when top and bottom are both positive, or both negative the quotient is positive. The denomenator is only positive above 20 and below -20, only positive values give a positive numerator, giving >20 and the denomenator is negative between -20 and 20, but the numerator is only negative below 0 so the overlap is -20> to 0.
We do NOT need to consider all 4 cases in solving 18x/(x^2 - 400) ≥ 0. Notice that we only have 2 cases --- the bottom is positive or the bottom is negative (it obviously cannot be 0) Case 1: If x^2 - 400 > 0, the inequality sign keeps, so 18x ≥ 0, i.e. x ≥ 0 x^2 - 400 > 0 ⟺ x^2 > 400 ⟺ x < - 20 or x > 20 Since we want x ≥ 0, we have x > 20 Case 2: If x^2 - 400 < 0, the inequality sign flips, so 18x ≤ 0, i.e. x ≤ 0 x^2 - 400 < 0 ⟺ x^2 < 400 ⟺ - 20 < x < 20 Since we want x ≤ 0, we have - 20 < x ≤ 0 Combining the 2 cases gives x ∈ (- 20, 0] ∪ (20, + ∞)
Another approach would be to graph the rational function (which can be done quickly), and observe the values of x for which the function values are >= 0.
Graphing y=18x/(x^2-400) is quite complex to do in the head for most people. I would graph y = 18x and y= x^2 - 400. One is a increasing line going trough (0,0), the other a parabola open to the top going trough (-20,0) and (20,0). Then the quotient is positive when both lines are on the same side of the x-axis.
We know that at the numbers you "care" about, the sign flips for either the numerator or denominator (and not both) Therefore, once you find the sign of one interval, you can simply alternate positive and negative to deduce the others. This lets us only investigate one value of x, for instance x = 1 since it's easy
You can do that as long as the number we "care" about is not a double root. For instance, if the numerator was 18*x^2, we would have to note that 0 is a double root and the sign wouldn't flip when we cross it from one side to the other. But otherwise, quite a nifty trick, just calculate the sign of the leftmost subdomain and keep flipping the sign as you go right.
@@phoenixUPC Well we already know a function like 18x² is always non-negative, so I wouldn't think of 0 as a number we "care" about in this case in the first place.
@@BigDBrian In terms of determining the sign, you are correct. But you need to mark 0 on that number line, because it is a still number we "care" about while determining the domains, even if just to see whether to include it or not. For instance, if the double(/quadruple/etc..) root was at the denominator, you'd need to mark it, just to remember that you'd have to exclude it from the domain.
If you want the domain for cases where the range is real, then the domain is the union of x = 0, x < -20, and x > 20. If you want the domain for when the range is imaginary, the domain is the intersection of |x| < 20 and x != 0. The domain for a range in the complex plane is |x| != 20.
Calculus basics. In my limited experience as a tutor finding and defining domain is something that murders half of the students. I have shivers when I saw the title.
@@MinecraftMasterNo1 That's why it's called "Definitionsbereich" (definition range) in my language. Meaning, for which values for the variable, does the function define real values.
To my knowledge (and a quick google to double check) in maths, there’s only one use of the word domain. To explain how it is defined, we should define some other terminology. A set can be thought of as a list of “objects” in which we don’t care about order, and we don’t care about repeats. A function or a map f (the two words are completely interchangeable with a few exceptions where someone explicitly defines them to be different) takes an element from a set of inputs X (called the domain) and assigns it a unique output in the set of outputs Y (referred to as either the codomain, or the range). Overall, the domain is the set of valid inputs of a function. In the case of this video, the domain is a subset of the real line, and the codomain is also a subset of the real line. To make the square root a function, it is a convention that we only deal with the positive root (both the positive and negative values of a number x square to the same thing, so if we didn’t have this convention, the square root function would map each positive real number to two different outputs, and hence it wouldn’t be a function). And we have a lot of real numbers which aren’t assigned anywhere in the output, so the question boils down to what is the set of real numbers that give a real number as an output
Ive been stuck on this question for 30 mins please anybody help, Find the equations of the four circles which are tangent to the x-axis, the y-axis, and the line x + y = 2.
This is a really nice interesting question. If you try sketching the graphs, you will observe there are two identical circles + one small circle and one big circle. Let's start with the small one. It is inscribed in a right angled isosceles triangle with base and height 2 (since the line x + y = 2 has intercepts on both axes at 2). The hypotenuse of the triangle, using Pythagoras', is 2sqrt2. Since the lengths of two tangents to a circle from a single external point are equal, it follows that 0.5(2sqrt2) = 2 - r, so r = 2 - sqrt2 = 0.586. Hence, the equation for this first circle is (x - 0.586)^2 + (y - 0.586)^2 = 0.586^2. Next, let's look at the large circle with radius R. If we call the centre of this circle C and drop a perpendicular from C down to the x-axis at A, then triangle OAC is again isosceles, with OA = R. So, the distance OC = Lsqrt2. So R + the length from O to the circumference of the circle = Rsqrt2. This length can be calculated to be sqrt2 (use the results from the smaller circle). So R + sqrt2 = Rsqrt2. Solving, R = 3.41. Hence, (x - 3.41)^2 + (y - 3.41)^2 = 3.41^2. Now for the two identical circles, radius = a. You should observe that the reflection of y + x = 2 in the line y = - x will also be tangent to both these circles. The distance between these two parallel tangents is 2•sqrt2 or simply 2sqrt2, which is also equal to the diameter. So 2a = 2sqrt2 or a = sqrt2 = 1.41. Hence, the final two circles have equations (x +/- 1.41)^2 + (y -/+ 1.41)^2 = 2.
Given: four circles are tangent to both the x & y axes, and the line x + y = 2 Solve the line for y: y = -x + 2 This means, the line has a down-right slope of -1, with intercepts of (2, 0) and (0, 2). Start with a qualitative description of where each of the four circles are: Circle 1: in Quadrant 1, nested within the triangle between the given line, and the two axes. Has its center along the line y=x. Circle 2: in Quadrant 2, with the x-axis on the bottom, the y-axis on the right, the given line on top, and no bounds on its left. Has its center along the line y = -x. Circle 3: in Quadrant 4, mirrored about the line y=x, from circle 2. So once we calculate circle 2, just swap x and y coordinates of its center. Circle 4, the least obvious: In Quadrant 1, this is a big circle, bounded by the x-axis on bottom, the y-axis on the right, the given line on its bottom-left, and no bounds on its top and its right sides. Like Circle 1, its center is also on the line y=x. To be continued in next post.
Circle 1 has a radius equal to its x and y coordinate. Call that r1. This means, its equation has the form: (x - r1)^2 + (y - r1)^2 = r1^2 And we have the constraint that it touches the given line, y= -x + 2, which from symmetry will happen at a location where x=y. This means it will intersect at (1, 1). Apply this point to the circle equation, and solve for r1: (1 - r1)^2 + (1 - r1)^2 = r1^2 2*(1 - r1)^2 = r1^2 2 - 4*r1 + 2*r1^2 = r1^2 r1^2 - 4*r1 + 2 = 0 r1 = -(-4/2) +/- sqrt((4/2)^2 - 2) r1 = 2 +/- sqrt(2) It turns out, that we've solved for both Circle 1 and Circle 4's radius, since circle 4 has the same constraints. The solution with the minus sign is the solution for Circle 1. The solution with the plus sign, is the solution for circle 4. Thus: r1 = 2 - sqrt(2) r4 = 2 + sqrt(2) Thus: Circle 1: (x - 2 + sqrt(2))^2 + (y - 2 + sqrt(2))^2 = (2 - sqrt(2))^2 Circle 4: (x - 2 - sqrt(2))^2 + (y - 2 - sqrt(2))^2 = (2 + sqrt(2))^2 Now we move on to solving for the radius and center of circle 2, which with symmetry, we can get circle 3. Its radius will be r2, and its center will be located at (-r2, +r2). Set up this circle's equation: (x + r2)^2 + (y - r2)^2 = r2^2 And substitute the given line, y = -x + 2 (x + r2)^2 + (-x + 2 - r2)^2 = r2^2 We need one more constraint to lock down this line. From its center, the circle's intersection with the given line, will be at a 45 degree diagonal, with a length of r2. The horizontal projection, will be r2/sqrt(2). The x-position will be -r2 + r2/sqrt(2). Substitute this x-value: (-r2 + r2/sqrt(2) + r2)^2 + (-(-r2 + r2/sqrt(2)) + 2 - r2)^2 = r2^2 Expand and simplify: r2^2 - 2*sqrt(2)*r2 + 4 = r2^2 This quadratic has a repeated solution, r2 = sqrt(2). Thus, we have our solution: Circle 2: (x + sqrt(2))^2 + (y - sqrt(2))^2 = 2 And by symmetry, we have circle 3: Circle 3: (x - sqrt(2))^2 + (y + sqrt(2))^2 = 2
What is lim(0.999...) supposed to mean? lim expressions should have a variable and a target. What is x and y for lim x->y (0.999...) and how are they relevant? What is 1^- supposed to mean?
My brain just didn't really work. I thought you'd get a negative number under the root, if you plug 0 into the equation, because the denominator would be negative.
I don’t like solving an equation with a not equals sign. There are no theorems/rules about what operations you can use. 1 does not equal 2. But I can add 2 to the left side only and 3 does not equal 2, so it is still 2. So I would solve it by reasoning that I want 18x to not be 0. So find out when it is 0 by usual methods. Then go back and say it can’t be those numbers. Most of the time it works both ways, but I think it is safer my way. It is the same reason why he didn’t solve the original inequality by multiplying both sides by the denominator. Sometimes it works ok, but sometimes not.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
Now do the expansion of a domain
DOMAIN EXPANSION
I have been think so hard on how to expand an interval. And then I realize that you are (probably) talking about Jujutsu Kaisen.😂
@@cyrusyeung8096 that is the joke.
Nah I'd not
@@cdkw8254nah I'd win
I think I would be tempted not to try out values, but jump straight to visualizing that when top and bottom are both positive, or both negative the quotient is positive. The denomenator is only positive above 20 and below -20, only positive values give a positive numerator, giving >20 and the denomenator is negative between -20 and 20, but the numerator is only negative below 0 so the overlap is -20> to 0.
We do NOT need to consider all 4 cases in solving 18x/(x^2 - 400) ≥ 0. Notice that we only have 2 cases --- the bottom is positive or the bottom is negative (it obviously cannot be 0)
Case 1:
If x^2 - 400 > 0, the inequality sign keeps, so 18x ≥ 0, i.e. x ≥ 0
x^2 - 400 > 0
⟺ x^2 > 400
⟺ x < - 20 or x > 20
Since we want x ≥ 0, we have x > 20
Case 2:
If x^2 - 400 < 0, the inequality sign flips, so 18x ≤ 0, i.e. x ≤ 0
x^2 - 400 < 0
⟺ x^2 < 400
⟺ - 20 < x < 20
Since we want x ≤ 0, we have - 20 < x ≤ 0
Combining the 2 cases gives
x ∈ (- 20, 0] ∪ (20, + ∞)
You're implicitely doing 4 cases here, and it's not as clear as juste seperating properly
Actually you only need to test 1 of the cases, e.g. x = 100, and the signs alternate as we have no double roots.
@@IvayloHristakiev❤ well said
Another approach would be to graph the rational function (which can be done quickly), and observe the values of x for which the function values are >= 0.
Graphing y=18x/(x^2-400) is quite complex to do in the head for most people. I would graph y = 18x and y= x^2 - 400. One is a increasing line going trough (0,0), the other a parabola open to the top going trough (-20,0) and (20,0). Then the quotient is positive when both lines are on the same side of the x-axis.
We know that at the numbers you "care" about, the sign flips for either the numerator or denominator (and not both)
Therefore, once you find the sign of one interval, you can simply alternate positive and negative to deduce the others. This lets us only investigate one value of x, for instance x = 1 since it's easy
You can do that as long as the number we "care" about is not a double root. For instance, if the numerator was 18*x^2, we would have to note that 0 is a double root and the sign wouldn't flip when we cross it from one side to the other. But otherwise, quite a nifty trick, just calculate the sign of the leftmost subdomain and keep flipping the sign as you go right.
@@phoenixUPC Well we already know a function like 18x² is always non-negative, so I wouldn't think of 0 as a number we "care" about in this case in the first place.
@@BigDBrian In terms of determining the sign, you are correct. But you need to mark 0 on that number line, because it is a still number we "care" about while determining the domains, even if just to see whether to include it or not. For instance, if the double(/quadruple/etc..) root was at the denominator, you'd need to mark it, just to remember that you'd have to exclude it from the domain.
@@phoenixUPC fair point :)
I think doing a sign chart is the clearest way to solve rational inequalities
1:10 "Intel inside" not the answer?
If you want the domain for cases where the range is real, then the domain is the union of x = 0, x < -20, and x > 20. If you want the domain for when the range is imaginary, the domain is the intersection of |x| < 20 and x != 0. The domain for a range in the complex plane is |x| != 20.
If it maps to [0,∞) then (-20,0]U(20,∞)
If it maps to C then C\{20,-20}
where is the calculus? Isn't this algebra 2
Calculus basics. In my limited experience as a tutor finding and defining domain is something that murders half of the students. I have shivers when I saw the title.
More like algebra 1
The big question I have: What is domain in this context?
the range of values for x such that f(x) has a real value.
@@MinecraftMasterNo1
That's why it's called "Definitionsbereich" (definition range) in my language. Meaning, for which values for the variable, does the function define real values.
The interval of numbers you can plug in which result in a valid real answer
To my knowledge (and a quick google to double check) in maths, there’s only one use of the word domain. To explain how it is defined, we should define some other terminology. A set can be thought of as a list of “objects” in which we don’t care about order, and we don’t care about repeats. A function or a map f (the two words are completely interchangeable with a few exceptions where someone explicitly defines them to be different) takes an element from a set of inputs X (called the domain) and assigns it a unique output in the set of outputs Y (referred to as either the codomain, or the range). Overall, the domain is the set of valid inputs of a function. In the case of this video, the domain is a subset of the real line, and the codomain is also a subset of the real line. To make the square root a function, it is a convention that we only deal with the positive root (both the positive and negative values of a number x square to the same thing, so if we didn’t have this convention, the square root function would map each positive real number to two different outputs, and hence it wouldn’t be a function). And we have a lot of real numbers which aren’t assigned anywhere in the output, so the question boils down to what is the set of real numbers that give a real number as an output
Ive been stuck on this question for 30 mins please anybody help,
Find the equations of the four circles which are tangent to the x-axis, the y-axis, and the line x + y = 2.
This is a really nice interesting question.
If you try sketching the graphs, you will observe there are two identical circles + one small circle and one big circle.
Let's start with the small one. It is inscribed in a right angled isosceles triangle with base and height 2 (since the line x + y = 2 has intercepts on both axes at 2). The hypotenuse of the triangle, using Pythagoras', is 2sqrt2. Since the lengths of two tangents to a circle from a single external point are equal, it follows that 0.5(2sqrt2) = 2 - r, so r = 2 - sqrt2 = 0.586. Hence, the equation for this first circle is (x - 0.586)^2 + (y - 0.586)^2 = 0.586^2.
Next, let's look at the large circle with radius R. If we call the centre of this circle C and drop a perpendicular from C down to the x-axis at A, then triangle OAC is again isosceles, with OA = R. So, the distance OC = Lsqrt2. So R + the length from O to the circumference of the circle = Rsqrt2. This length can be calculated to be sqrt2 (use the results from the smaller circle). So R + sqrt2 = Rsqrt2. Solving, R = 3.41. Hence, (x - 3.41)^2 + (y - 3.41)^2 = 3.41^2.
Now for the two identical circles, radius = a. You should observe that the reflection of y + x = 2 in the line y = - x will also be tangent to both these circles. The distance between these two parallel tangents is 2•sqrt2 or simply 2sqrt2, which is also equal to the diameter. So 2a = 2sqrt2 or a = sqrt2 = 1.41. Hence, the final two circles have equations (x +/- 1.41)^2 + (y -/+ 1.41)^2 = 2.
Given: four circles are tangent to both the x & y axes, and the line x + y = 2
Solve the line for y: y = -x + 2
This means, the line has a down-right slope of -1, with intercepts of (2, 0) and (0, 2).
Start with a qualitative description of where each of the four circles are:
Circle 1: in Quadrant 1, nested within the triangle between the given line, and the two axes. Has its center along the line y=x.
Circle 2: in Quadrant 2, with the x-axis on the bottom, the y-axis on the right, the given line on top, and no bounds on its left. Has its center along the line y = -x.
Circle 3: in Quadrant 4, mirrored about the line y=x, from circle 2. So once we calculate circle 2, just swap x and y coordinates of its center.
Circle 4, the least obvious: In Quadrant 1, this is a big circle, bounded by the x-axis on bottom, the y-axis on the right, the given line on its bottom-left, and no bounds on its top and its right sides. Like Circle 1, its center is also on the line y=x.
To be continued in next post.
Circle 1 has a radius equal to its x and y coordinate. Call that r1. This means, its equation has the form:
(x - r1)^2 + (y - r1)^2 = r1^2
And we have the constraint that it touches the given line, y= -x + 2, which from symmetry will happen at a location where x=y. This means it will intersect at (1, 1). Apply this point to the circle equation, and solve for r1:
(1 - r1)^2 + (1 - r1)^2 = r1^2
2*(1 - r1)^2 = r1^2
2 - 4*r1 + 2*r1^2 = r1^2
r1^2 - 4*r1 + 2 = 0
r1 = -(-4/2) +/- sqrt((4/2)^2 - 2)
r1 = 2 +/- sqrt(2)
It turns out, that we've solved for both Circle 1 and Circle 4's radius, since circle 4 has the same constraints. The solution with the minus sign is the solution for Circle 1. The solution with the plus sign, is the solution for circle 4. Thus:
r1 = 2 - sqrt(2)
r4 = 2 + sqrt(2)
Thus:
Circle 1: (x - 2 + sqrt(2))^2 + (y - 2 + sqrt(2))^2 = (2 - sqrt(2))^2
Circle 4: (x - 2 - sqrt(2))^2 + (y - 2 - sqrt(2))^2 = (2 + sqrt(2))^2
Now we move on to solving for the radius and center of circle 2, which with symmetry, we can get circle 3. Its radius will be r2, and its center will be located at (-r2, +r2). Set up this circle's equation:
(x + r2)^2 + (y - r2)^2 = r2^2
And substitute the given line, y = -x + 2
(x + r2)^2 + (-x + 2 - r2)^2 = r2^2
We need one more constraint to lock down this line. From its center, the circle's intersection with the given line, will be at a 45 degree diagonal, with a length of r2. The horizontal projection, will be r2/sqrt(2). The x-position will be -r2 + r2/sqrt(2).
Substitute this x-value:
(-r2 + r2/sqrt(2) + r2)^2 + (-(-r2 + r2/sqrt(2)) + 2 - r2)^2 = r2^2
Expand and simplify:
r2^2 - 2*sqrt(2)*r2 + 4 = r2^2
This quadratic has a repeated solution, r2 = sqrt(2). Thus, we have our solution:
Circle 2: (x + sqrt(2))^2 + (y - sqrt(2))^2 = 2
And by symmetry, we have circle 3:
Circle 3: (x - sqrt(2))^2 + (y + sqrt(2))^2 = 2
Not related to this video. I've got one for ya: Does 0.999... = 1^- ?
What is lim(0.999...) supposed to mean? lim expressions should have a variable and a target. What is x and y for lim x->y (0.999...) and how are they relevant?
What is 1^- supposed to mean?
What the hell is even that!?
@@javier_me2 It's gibberish.
@@Chris-5318 your brain is just too tiny to understand the question. Pity.
@@teelo12000 LOL. I understand your question very well, it is gibberish. The answer to your question is, no.
The domain is strange bc it's when x is all real numbers expect when it's smaller than -20 and when it's between 0 and 20
I'm pretty sure its x in (-20,0]U(20, ♾️)
You got it right.
My brain just didn't really work. I thought you'd get a negative number under the root, if you plug 0 into the equation, because the denominator would be negative.
The denominator would be negative but the numerator would be zero. Zero divided by anything is zero.
I don’t like solving an equation with a not equals sign. There are no theorems/rules about what operations you can use.
1 does not equal 2. But I can add 2 to the left side only and 3 does not equal 2, so it is still 2.
So I would solve it by reasoning that I want 18x to not be 0. So find out when it is 0 by usual methods. Then go back and say it can’t be those numbers. Most of the time it works both ways, but I think it is safer my way. It is the same reason why he didn’t solve the original inequality by multiplying both sides by the denominator. Sometimes it works ok, but sometimes not.