💡Alright, so I have this problem here: "Given that the curve ( y = f(x) ) is symmetric to the curve ( y = x^2 + 1 ) (where ( x < 0 )) about ( y = x ), determine the domain of ( f(x) )." The options are A: ( (0, +infty) ), B: ( (1, +infty) ), C: ( (-infty, 0) ), and D: ( (-infty, 1) ).Hmm, okay. So, first, I need to understand what it means for two curves to be symmetric about the line ( y = x ). I remember that symmetry about ( y = x ) usually implies that the functions are inverses of each other. So, if ( y = f(x) ) is symmetric to ( y = x^2 + 1 ) (for ( x < 0 )) about ( y = x ), then ( f(x) ) should be the inverse function of ( y = x^2 + 1 ) restricted to ( x < 0 ).Let me recall: If two functions are inverses, then their graphs are reflections of each other over the line ( y = x ). That makes sense. So, if I have a function ( g(x) ) and its inverse ( g^{-1}(x) ), then ( g^{-1}(x) ) is the reflection of ( g(x) ) over ( y = x ).So, in this case, ( f(x) ) is the inverse of ( y = x^2 + 1 ) for ( x < 0 ). Therefore, to find ( f(x) ), I need to find the inverse of ( y = x^2 + 1 ) but only considering ( x < 0 ).Alright, let's try to find the inverse function. Starting with ( y = x^2 + 1 ). To find the inverse, I need to solve for ( x ) in terms of ( y ).So, ( y = x^2 + 1 ) implies ( x^2 = y - 1 ), so ( x = pm sqrt{y - 1} ).But wait, in the original function, ( x < 0 ). So, when we take the inverse, we have to consider which branch of the square root to take. Since ( x < 0 ), the inverse function will correspond to the negative square root. So, ( x = -sqrt{y - 1} ).Therefore, the inverse function ( f(x) ) is ( f(x) = -sqrt{x - 1} ).Now, I need to determine the domain of ( f(x) ). The domain of an inverse function is the range of the original function. So, what was the range of the original function ( y = x^2 + 1 ) with ( x < 0 )?Let's think about that. The original function is ( y = x^2 + 1 ) where ( x < 0 ). Since ( x^2 ) is always positive, and ( x < 0 ), as ( x ) becomes more negative, ( x^2 ) increases. So, the smallest value of ( y ) occurs when ( x ) is closest to zero, which is when ( x ) approaches 0 from the negative side. At ( x = 0 ), ( y = 0 + 1 = 1 ), but since ( x < 0 ), ( y ) approaches 1 from above. As ( x ) becomes more negative, ( y ) increases without bound.Therefore, the range of the original function ( y = x^2 + 1 ) (for ( x < 0 )) is ( (1, +infty) ).Since the domain of the inverse function ( f(x) ) is the range of the original function, the domain of ( f(x) ) is ( (1, +infty) ).Let me double-check that. The original function ( y = x^2 + 1 ) with ( x < 0 ) has a range of ( (1, +infty) ). Therefore, its inverse function ( f(x) = -sqrt{x - 1} ) must have a domain of ( (1, +infty) ) because the expression inside the square root must be non-negative, so ( x - 1 geq 0 ) which implies ( x geq 1 ). But since the original function's range was ( (1, +infty) ), the domain of the inverse is ( (1, +infty) ).So, the domain of ( f(x) ) is ( (1, +infty) ), which corresponds to option B.I think that makes sense. Let me visualize it. The original function ( y = x^2 + 1 ) for ( x < 0 ) is a parabola opening upwards, but only the left half (since ( x < 0 )). Its vertex is at ( (0,1) ), and it goes upwards as ( x ) becomes more negative. Reflecting this over ( y = x ) would give us the inverse function, which is ( f(x) = -sqrt{x - 1} ). This inverse function is defined for ( x ) values starting just above 1 and going to infinity, which matches the range of the original function.So, yeah, I'm pretty confident that the domain of ( f(x) ) is ( (1, +infty) ), which is option B.**Final Answer**boxed{B}

💡Alright, so I have this inequality: ((m^{2} - m) cdot 4^{x} - 2^{x} < 0), and it needs to hold true for all (x) in the interval ((-infty, -1]). I need to figure out the range of values for the real number (m) that satisfy this condition. Hmm, okay, let me break this down step by step.First, let me rewrite the inequality to make it a bit clearer. I can factor out (2^{x}) since both terms have that in common. So, ((m^{2} - m) cdot 4^{x} - 2^{x} < 0) can be rewritten as:[(m^{2} - m) cdot 4^{x} < 2^{x}]Now, I can divide both sides by (2^{x}) to simplify the inequality further. Since (2^{x}) is always positive for any real (x), dividing by it won't change the direction of the inequality. So, dividing both sides by (2^{x}) gives:[(m^{2} - m) cdot frac{4^{x}}{2^{x}} < 1]Simplifying (frac{4^{x}}{2^{x}}), I know that (4^{x}) is the same as ((2^{2})^{x} = 2^{2x}). Therefore, (frac{4^{x}}{2^{x}} = frac{2^{2x}}{2^{x}} = 2^{x}). So substituting that back into the inequality, we have:[(m^{2} - m) cdot 2^{x} < 1]Wait, hold on, that seems a bit circular because we just divided by (2^{x}) and then multiplied by (2^{x}) again. Maybe I made a mistake there. Let me double-check.Starting again: The original inequality is ((m^{2} - m) cdot 4^{x} - 2^{x} < 0). Let me factor out (2^{x}):[2^{x} cdot [(m^{2} - m) cdot 2^{x} - 1] < 0]Since (2^{x}) is always positive, the inequality simplifies to:[(m^{2} - m) cdot 2^{x} - 1 < 0]Which can be rewritten as:[(m^{2} - m) cdot 2^{x} < 1]Okay, that makes more sense. So, ((m^{2} - m) cdot 2^{x} < 1). Now, since (x in (-infty, -1]), (2^{x}) is going to be a positive number less than or equal to (2^{-1} = frac{1}{2}). Because as (x) becomes more negative, (2^{x}) approaches zero.So, (2^{x}) is in the interval ((0, frac{1}{2}]) for (x in (-infty, -1]). Therefore, (2^{x}) is always positive but less than or equal to (frac{1}{2}).Given that, let's denote (t = 2^{x}), where (t in (0, frac{1}{2}]). Then, our inequality becomes:[(m^{2} - m) cdot t < 1]We need this inequality to hold for all (t) in ((0, frac{1}{2}]). So, we can think of this as a linear inequality in terms of (t). The left-hand side is linear in (t), and we need it to be less than 1 for all (t) in that interval.To ensure that ((m^{2} - m) cdot t < 1) for all (t in (0, frac{1}{2}]), we need to consider the maximum value of the left-hand side in that interval. Since (t) is positive, the expression ((m^{2} - m) cdot t) will be maximized when (t) is at its maximum, which is (t = frac{1}{2}).Therefore, to satisfy the inequality for all (t) in ((0, frac{1}{2}]), it suffices to ensure that:[(m^{2} - m) cdot frac{1}{2} < 1]Multiplying both sides by 2:[(m^{2} - m) < 2]So, we have the quadratic inequality:[m^{2} - m - 2 < 0]Let's solve this quadratic inequality. First, find the roots of the equation (m^{2} - m - 2 = 0). Using the quadratic formula:[m = frac{1 pm sqrt{1 + 8}}{2} = frac{1 pm 3}{2}]So, the roots are:[m = frac{1 + 3}{2} = 2 quad text{and} quad m = frac{1 - 3}{2} = -1]The quadratic (m^{2} - m - 2) can be factored as:[(m - 2)(m + 1) < 0]To determine where this product is less than zero, we can analyze the sign changes around the roots (m = -1) and (m = 2).- For (m < -1), say (m = -2): ((-2 - 2)(-2 + 1) = (-4)(-1) = 4 > 0)- For (-1 < m < 2), say (m = 0): ((0 - 2)(0 + 1) = (-2)(1) = -2 < 0)- For (m > 2), say (m = 3): ((3 - 2)(3 + 1) = (1)(4) = 4 > 0)So, the inequality ((m - 2)(m + 1) < 0) holds true when (m) is between (-1) and (2). Therefore, the solution is:[-1 < m < 2]But wait, I should verify if this is indeed the case. Let me test a value within the interval, say (m = 0):Original inequality: ((0^{2} - 0) cdot 4^{x} - 2^{x} = 0 - 2^{x} = -2^{x} < 0), which is true because (2^{x}) is positive.Now, test a value outside the interval, say (m = 3):((3^{2} - 3) cdot 4^{x} - 2^{x} = (9 - 3) cdot 4^{x} - 2^{x} = 6 cdot 4^{x} - 2^{x})Since (4^{x} = (2^{x})^{2}), let (t = 2^{x}), so the expression becomes (6t^{2} - t). For (t in (0, frac{1}{2}]), let's see if this is less than zero.At (t = frac{1}{2}): (6 cdot frac{1}{4} - frac{1}{2} = frac{3}{2} - frac{1}{2} = 1 > 0). So, the inequality does not hold for (m = 3), which is consistent with our solution.Similarly, test (m = -2):((-2)^{2} - (-2) = 4 + 2 = 6). So, the expression becomes (6 cdot 4^{x} - 2^{x}). Again, at (t = frac{1}{2}), it's (6 cdot frac{1}{4} - frac{1}{2} = 1 > 0), so the inequality does not hold for (m = -2), which is also consistent.Therefore, the range of (m) is indeed (-1 < m < 2).But just to be thorough, let me consider the endpoints. If (m = -1):((-1)^{2} - (-1) = 1 + 1 = 2). So, the expression becomes (2 cdot 4^{x} - 2^{x}). Let's see if this is less than zero for all (x leq -1).Expressed in terms of (t = 2^{x}), it's (2t^{2} - t). We need (2t^{2} - t < 0). Factor:(t(2t - 1) < 0). So, this is negative when (0 < t < frac{1}{2}). Since (t = 2^{x}) and (x leq -1), (t in (0, frac{1}{2}]). So, at (t = frac{1}{2}), (2 cdot frac{1}{4} - frac{1}{2} = frac{1}{2} - frac{1}{2} = 0). So, the expression equals zero at (t = frac{1}{2}), which is the boundary. Therefore, (m = -1) does not satisfy the strict inequality for all (x leq -1) because at (x = -1), the expression equals zero.Similarly, for (m = 2):(2^{2} - 2 = 4 - 2 = 2). So, the expression is (2 cdot 4^{x} - 2^{x}), same as when (m = -1). Again, at (t = frac{1}{2}), it's zero, so (m = 2) also doesn't satisfy the strict inequality for all (x leq -1).Therefore, the endpoints are excluded, and the solution is indeed (-1 < m < 2).I think that covers all the cases. I've tested values inside and outside the interval, checked the endpoints, and verified the reasoning. It seems solid.**Final Answer**The range of values for (m) is boxed{(-1, 2)}.

💡Alright, so I've got this problem here about systematic sampling in a class of 48 students. The students are numbered from 1 to 48, and we're supposed to draw a sample of 4 students using systematic sampling. We're told that students numbered 6, 30, and 42 are in the sample, and we need to find the number of the fourth student in the sample.Okay, first, I need to recall what systematic sampling is. From what I remember, systematic sampling is a method where you select every kth individual from a population. The value of k is determined by dividing the population size by the desired sample size. So in this case, the population size is 48, and we want a sample of 4. That means k should be 48 divided by 4, which is 12. So, every 12th student should be selected.Wait, but in this case, the sample includes students 6, 30, and 42. Let me check if these numbers fit into the systematic sampling pattern with k=12. Starting from 6, adding 12 would give me 18, then 30, then 42, and so on. Hmm, 6, 18, 30, 42... That seems to fit. So, if the starting point is 6, then the sample would be 6, 18, 30, 42. But in the problem, they've given us 6, 30, and 42, and we need to find the fourth one. So, based on this, the fourth student should be 18.But wait, let me make sure I'm not missing anything. Sometimes, in systematic sampling, especially when the population size isn't perfectly divisible by the sample size, there might be some adjustments or random starts. But in this case, 48 divided by 4 is exactly 12, so there shouldn't be any adjustments needed. The sampling interval is straightforward.Let me double-check the arithmetic. Starting at 6, adding 12 gives 18, adding another 12 gives 30, and adding another 12 gives 42. So, the sequence is 6, 18, 30, 42. Since 6, 30, and 42 are already in the sample, the missing one must be 18.But just to be thorough, let me consider if there's another possible starting point that could result in the given numbers. Suppose the starting point was different. For example, if we started at 6, then we get 6, 18, 30, 42. If we started at 30, adding 12 would give us 42, then 54, but 54 is beyond 48, so we might wrap around or stop. Similarly, starting at 42, adding 12 would take us beyond 48. So, it seems like starting at 6 is the only logical starting point that gives us all four numbers within the range of 1 to 48.Another thing to consider is whether the sampling could be done in reverse. If we started at 42 and subtracted 12, we'd get 30, then 18, then 6. So, in reverse, it's still the same sequence. Therefore, regardless of the direction, the numbers involved are 6, 18, 30, and 42.I also want to make sure that there's no other number that could fit into the sample. Let's see, if we consider the sampling interval of 12, the possible numbers would be 6, 18, 30, 42, 54, etc. But since we only have 48 students, 54 is out of range. So, the only possible numbers are 6, 18, 30, and 42.Wait a second, what if the starting point wasn't 6? Could there be another starting point that still includes 6, 30, and 42? Let's think about that. If the starting point was, say, 18, then the sample would be 18, 30, 42, and 54 (which is out of range). So, 18 is already in the sample, but 54 isn't a valid student number. Similarly, if we started at 30, the next would be 42, then 54, which is invalid. Starting at 42 would only give us 42 and then 54, which is invalid. So, starting at 6 seems to be the only way to get all four valid student numbers within the range of 1 to 48.Just to be absolutely sure, let's list out all possible systematic samples with a sampling interval of 12:1. Starting at 1: 1, 13, 25, 372. Starting at 2: 2, 14, 26, 383. Starting at 3: 3, 15, 27, 394. Starting at 4: 4, 16, 28, 405. Starting at 5: 5, 17, 29, 416. Starting at 6: 6, 18, 30, 427. Starting at 7: 7, 19, 31, 438. Starting at 8: 8, 20, 32, 449. Starting at 9: 9, 21, 33, 4510. Starting at 10: 10, 22, 34, 4611. Starting at 11: 11, 23, 35, 4712. Starting at 12: 12, 24, 36, 48Looking at these, the only sample that includes 6, 30, and 42 is the one starting at 6: 6, 18, 30, 42. All the other starting points don't include 6, 30, and 42 together. Therefore, the fourth student must be 18.I think I've covered all possible angles here. I've considered the systematic sampling method, checked the arithmetic, considered alternative starting points, and even listed out all possible samples to confirm. It all points to 18 being the missing number in the sample.

💡Okay, so I have this problem where I need to find the value of ( x ) for which ( f(x) = f^{-1}(x) ) where ( f(x) = 3x^2 + 2x - 1 ). Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I know that ( f^{-1}(x) ) is the inverse function of ( f(x) ). That means if ( f(x) = y ), then ( f^{-1}(y) = x ). So, if I set ( f(x) = f^{-1}(x) ), it means that ( f(f(x)) = x ). Wait, is that right? Let me think. If ( f(x) = f^{-1}(x) ), then applying ( f ) to both sides gives ( f(f(x)) = f(f^{-1}(x)) ), which simplifies to ( f(f(x)) = x ). Yeah, that makes sense.So, I need to solve the equation ( f(f(x)) = x ). That seems like a good approach. Let me compute ( f(f(x)) ) first.Given ( f(x) = 3x^2 + 2x - 1 ), let's compute ( f(f(x)) ).First, substitute ( f(x) ) into itself:( f(f(x)) = 3(f(x))^2 + 2(f(x)) - 1 ).Now, substitute ( f(x) = 3x^2 + 2x - 1 ):( f(f(x)) = 3(3x^2 + 2x - 1)^2 + 2(3x^2 + 2x - 1) - 1 ).Okay, that looks complicated, but let me expand it step by step.First, compute ( (3x^2 + 2x - 1)^2 ):Let me denote ( A = 3x^2 + 2x - 1 ), so ( A^2 = (3x^2 + 2x - 1)(3x^2 + 2x - 1) ).Multiplying term by term:- ( 3x^2 times 3x^2 = 9x^4 )- ( 3x^2 times 2x = 6x^3 )- ( 3x^2 times (-1) = -3x^2 )- ( 2x times 3x^2 = 6x^3 )- ( 2x times 2x = 4x^2 )- ( 2x times (-1) = -2x )- ( (-1) times 3x^2 = -3x^2 )- ( (-1) times 2x = -2x )- ( (-1) times (-1) = 1 )Now, adding all these terms together:( 9x^4 + 6x^3 - 3x^2 + 6x^3 + 4x^2 - 2x - 3x^2 - 2x + 1 ).Combine like terms:- ( 9x^4 )- ( 6x^3 + 6x^3 = 12x^3 )- ( -3x^2 + 4x^2 - 3x^2 = (-3 + 4 - 3)x^2 = (-2)x^2 )- ( -2x - 2x = -4x )- ( +1 )So, ( A^2 = 9x^4 + 12x^3 - 2x^2 - 4x + 1 ).Now, go back to ( f(f(x)) ):( f(f(x)) = 3A^2 + 2A - 1 ).Substitute ( A^2 ) and ( A ):( f(f(x)) = 3(9x^4 + 12x^3 - 2x^2 - 4x + 1) + 2(3x^2 + 2x - 1) - 1 ).Let's compute each part:First, ( 3(9x^4 + 12x^3 - 2x^2 - 4x + 1) ):- ( 3 times 9x^4 = 27x^4 )- ( 3 times 12x^3 = 36x^3 )- ( 3 times (-2x^2) = -6x^2 )- ( 3 times (-4x) = -12x )- ( 3 times 1 = 3 )So, that part is ( 27x^4 + 36x^3 - 6x^2 - 12x + 3 ).Next, ( 2(3x^2 + 2x - 1) ):- ( 2 times 3x^2 = 6x^2 )- ( 2 times 2x = 4x )- ( 2 times (-1) = -2 )So, that part is ( 6x^2 + 4x - 2 ).Now, combine all parts:( 27x^4 + 36x^3 - 6x^2 - 12x + 3 + 6x^2 + 4x - 2 - 1 ).Combine like terms:- ( 27x^4 )- ( 36x^3 )- ( -6x^2 + 6x^2 = 0 )- ( -12x + 4x = -8x )- ( 3 - 2 - 1 = 0 )So, ( f(f(x)) = 27x^4 + 36x^3 - 8x ).Now, set ( f(f(x)) = x ):( 27x^4 + 36x^3 - 8x = x ).Bring all terms to one side:( 27x^4 + 36x^3 - 8x - x = 0 ).Simplify:( 27x^4 + 36x^3 - 9x = 0 ).Factor out a common term, which is ( 9x ):( 9x(3x^3 + 4x^2 - 1) = 0 ).So, the equation is satisfied if either ( 9x = 0 ) or ( 3x^3 + 4x^2 - 1 = 0 ).First, ( 9x = 0 ) gives ( x = 0 ).Now, let's solve ( 3x^3 + 4x^2 - 1 = 0 ).This is a cubic equation. Let me see if I can find rational roots using the Rational Root Theorem. The possible rational roots are ( pm1, pmfrac{1}{3} ).Testing ( x = 1 ):( 3(1)^3 + 4(1)^2 - 1 = 3 + 4 - 1 = 6 neq 0 ).Testing ( x = -1 ):( 3(-1)^3 + 4(-1)^2 - 1 = -3 + 4 - 1 = 0 ).Oh, ( x = -1 ) is a root.So, we can factor ( (x + 1) ) out of ( 3x^3 + 4x^2 - 1 ).Using polynomial division or synthetic division.Let's use synthetic division with root ( x = -1 ):Coefficients: 3 | 4 | 0 | -1Wait, the polynomial is ( 3x^3 + 4x^2 - 1 ), so the coefficients are 3, 4, 0, -1.Wait, actually, it's ( 3x^3 + 4x^2 + 0x - 1 ).So, set up synthetic division:-1 | 3   4   0   -1Bring down the 3.Multiply -1 by 3: -3. Add to next coefficient: 4 + (-3) = 1.Multiply -1 by 1: -1. Add to next coefficient: 0 + (-1) = -1.Multiply -1 by (-1): 1. Add to last coefficient: -1 + 1 = 0.So, the result is ( 3x^2 + x - 1 ).Therefore, ( 3x^3 + 4x^2 - 1 = (x + 1)(3x^2 + x - 1) ).Now, set ( 3x^2 + x - 1 = 0 ).Use quadratic formula:( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ).Here, ( a = 3 ), ( b = 1 ), ( c = -1 ).Discriminant: ( 1^2 - 4(3)(-1) = 1 + 12 = 13 ).So, roots are:( x = frac{-1 pm sqrt{13}}{6} ).Therefore, the solutions to ( 3x^3 + 4x^2 - 1 = 0 ) are ( x = -1 ), ( x = frac{-1 + sqrt{13}}{6} ), and ( x = frac{-1 - sqrt{13}}{6} ).So, overall, the equation ( f(f(x)) = x ) has solutions:( x = 0 ), ( x = -1 ), ( x = frac{-1 + sqrt{13}}{6} ), and ( x = frac{-1 - sqrt{13}}{6} ).But wait, the original question is to find ( x ) such that ( f(x) = f^{-1}(x) ). So, does that mean all these solutions are valid?I need to check if these solutions satisfy ( f(x) = f^{-1}(x) ). Because sometimes when we solve equations involving inverses, we might introduce extraneous solutions.Let me first check ( x = 0 ):Compute ( f(0) = 3(0)^2 + 2(0) - 1 = -1 ).Compute ( f^{-1}(0) ). Since ( f^{-1}(y) = x ) such that ( f(x) = y ). So, ( f(x) = 0 ) implies ( 3x^2 + 2x - 1 = 0 ).Solving ( 3x^2 + 2x - 1 = 0 ):Using quadratic formula:( x = frac{-2 pm sqrt{4 + 12}}{6} = frac{-2 pm sqrt{16}}{6} = frac{-2 pm 4}{6} ).So, ( x = frac{2}{6} = frac{1}{3} ) or ( x = frac{-6}{6} = -1 ).Therefore, ( f^{-1}(0) ) could be ( frac{1}{3} ) or ( -1 ). But since ( f ) is not one-to-one (it's a quadratic function), it doesn't have a true inverse unless we restrict the domain. So, perhaps ( f^{-1}(x) ) isn't uniquely defined here.This complicates things because ( f ) isn't invertible over all real numbers since it's not one-to-one. So, maybe the original approach is flawed.Wait, perhaps instead of setting ( f(f(x)) = x ), which gives all solutions where ( f ) composed with itself gives the identity, but ( f(x) = f^{-1}(x) ) would require that ( f ) is its own inverse, meaning ( f(f(x)) = x ). So, actually, the solutions to ( f(f(x)) = x ) are the solutions where ( f(x) = f^{-1}(x) ), but only if ( f ) is invertible.But since ( f ) is a quadratic function, it's not one-to-one over all real numbers, so it doesn't have an inverse unless we restrict its domain. Therefore, perhaps the problem assumes that ( f ) is invertible, maybe by restricting to a domain where it's increasing or decreasing.Alternatively, maybe the problem is expecting us to solve ( f(x) = f^{-1}(x) ) without worrying about the invertibility, just algebraically.Alternatively, perhaps another approach is better. Let me try to find ( f^{-1}(x) ) explicitly.Given ( f(x) = 3x^2 + 2x - 1 ), to find ( f^{-1}(x) ), we need to solve ( y = 3x^2 + 2x - 1 ) for ( x ) in terms of ( y ).So, ( 3x^2 + 2x - 1 - y = 0 ).This is a quadratic in ( x ):( 3x^2 + 2x - (1 + y) = 0 ).Using quadratic formula:( x = frac{-2 pm sqrt{4 + 12(1 + y)}}{6} = frac{-2 pm sqrt{4 + 12 + 12y}}{6} = frac{-2 pm sqrt{16 + 12y}}{6} ).Simplify:( x = frac{-2 pm 2sqrt{4 + 3y}}{6} = frac{-1 pm sqrt{4 + 3y}}{3} ).So, ( f^{-1}(x) = frac{-1 pm sqrt{4 + 3x}}{3} ).But since ( f ) is a quadratic, it's not one-to-one, so ( f^{-1}(x) ) isn't a function unless we restrict the domain. So, depending on the domain, we might take the positive or negative root.Assuming we take the principal (positive) root for the inverse, ( f^{-1}(x) = frac{-1 + sqrt{4 + 3x}}{3} ).Now, set ( f(x) = f^{-1}(x) ):( 3x^2 + 2x - 1 = frac{-1 + sqrt{4 + 3x}}{3} ).Multiply both sides by 3 to eliminate the denominator:( 9x^2 + 6x - 3 = -1 + sqrt{4 + 3x} ).Bring the -1 to the left side:( 9x^2 + 6x - 2 = sqrt{4 + 3x} ).Now, to eliminate the square root, square both sides:( (9x^2 + 6x - 2)^2 = 4 + 3x ).Compute the left side:Let me denote ( B = 9x^2 + 6x - 2 ), so ( B^2 = (9x^2 + 6x - 2)^2 ).Expanding:( (9x^2)^2 + (6x)^2 + (-2)^2 + 2(9x^2)(6x) + 2(9x^2)(-2) + 2(6x)(-2) ).Wait, that's a bit messy. Alternatively, use the formula ( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc ).So, ( (9x^2 + 6x - 2)^2 = (9x^2)^2 + (6x)^2 + (-2)^2 + 2(9x^2)(6x) + 2(9x^2)(-2) + 2(6x)(-2) ).Compute each term:- ( (9x^2)^2 = 81x^4 )- ( (6x)^2 = 36x^2 )- ( (-2)^2 = 4 )- ( 2(9x^2)(6x) = 108x^3 )- ( 2(9x^2)(-2) = -36x^2 )- ( 2(6x)(-2) = -24x )Now, add them all together:( 81x^4 + 36x^2 + 4 + 108x^3 - 36x^2 - 24x ).Combine like terms:- ( 81x^4 )- ( 108x^3 )- ( 36x^2 - 36x^2 = 0 )- ( -24x )- ( +4 )So, ( (9x^2 + 6x - 2)^2 = 81x^4 + 108x^3 - 24x + 4 ).Set equal to ( 4 + 3x ):( 81x^4 + 108x^3 - 24x + 4 = 4 + 3x ).Subtract ( 4 + 3x ) from both sides:( 81x^4 + 108x^3 - 24x + 4 - 4 - 3x = 0 ).Simplify:( 81x^4 + 108x^3 - 27x = 0 ).Factor out a common term, which is ( 9x ):( 9x(9x^3 + 12x^2 - 3) = 0 ).So, the equation is satisfied if either ( 9x = 0 ) or ( 9x^3 + 12x^2 - 3 = 0 ).First, ( 9x = 0 ) gives ( x = 0 ).Now, solve ( 9x^3 + 12x^2 - 3 = 0 ).Divide both sides by 3:( 3x^3 + 4x^2 - 1 = 0 ).Wait, this is the same cubic equation as before! So, the solutions are ( x = -1 ), ( x = frac{-1 + sqrt{13}}{6} ), and ( x = frac{-1 - sqrt{13}}{6} ).So, the solutions to ( f(x) = f^{-1}(x) ) are ( x = 0 ), ( x = -1 ), ( x = frac{-1 + sqrt{13}}{6} ), and ( x = frac{-1 - sqrt{13}}{6} ).But wait, earlier I was concerned about the invertibility of ( f ). Since ( f ) is a quadratic, it's not one-to-one over all real numbers, so ( f^{-1}(x) ) isn't a proper function unless we restrict the domain. So, perhaps only some of these solutions are valid.Let me check each solution in the original equation ( f(x) = f^{-1}(x) ).First, check ( x = 0 ):Compute ( f(0) = 3(0)^2 + 2(0) - 1 = -1 ).Compute ( f^{-1}(0) ). As before, solving ( f(x) = 0 ):( 3x^2 + 2x - 1 = 0 ).Solutions are ( x = frac{-2 pm sqrt{4 + 12}}{6} = frac{-2 pm 4}{6} ), so ( x = frac{1}{3} ) or ( x = -1 ).Therefore, ( f^{-1}(0) ) can be either ( frac{1}{3} ) or ( -1 ). But ( f(0) = -1 ), so if we take ( f^{-1}(0) = -1 ), then ( f(0) = f^{-1}(0) ) holds because both are -1. So, ( x = 0 ) is a valid solution.Next, check ( x = -1 ):Compute ( f(-1) = 3(-1)^2 + 2(-1) - 1 = 3 - 2 - 1 = 0 ).Compute ( f^{-1}(-1) ). Solve ( f(x) = -1 ):( 3x^2 + 2x - 1 = -1 ).Simplify:( 3x^2 + 2x = 0 ).Factor:( x(3x + 2) = 0 ).Solutions: ( x = 0 ) or ( x = -frac{2}{3} ).So, ( f^{-1}(-1) ) can be ( 0 ) or ( -frac{2}{3} ). But ( f(-1) = 0 ), so if we take ( f^{-1}(-1) = 0 ), then ( f(-1) = f^{-1}(-1) ) holds because both are 0. So, ( x = -1 ) is also a valid solution.Now, check ( x = frac{-1 + sqrt{13}}{6} ):Let me denote ( x = frac{-1 + sqrt{13}}{6} ).Compute ( f(x) ):( f(x) = 3x^2 + 2x - 1 ).Compute ( f^{-1}(x) ):Since ( f^{-1}(x) = frac{-1 + sqrt{4 + 3x}}{3} ) (assuming the principal root), let's compute both.First, compute ( f(x) ):Let me compute ( x = frac{-1 + sqrt{13}}{6} ).Compute ( x^2 ):( x^2 = left(frac{-1 + sqrt{13}}{6}right)^2 = frac{1 - 2sqrt{13} + 13}{36} = frac{14 - 2sqrt{13}}{36} = frac{7 - sqrt{13}}{18} ).So, ( 3x^2 = 3 times frac{7 - sqrt{13}}{18} = frac{7 - sqrt{13}}{6} ).Compute ( 2x = 2 times frac{-1 + sqrt{13}}{6} = frac{-2 + 2sqrt{13}}{6} = frac{-1 + sqrt{13}}{3} ).So, ( f(x) = 3x^2 + 2x - 1 = frac{7 - sqrt{13}}{6} + frac{-1 + sqrt{13}}{3} - 1 ).Convert all terms to sixths:( frac{7 - sqrt{13}}{6} + frac{-2 + 2sqrt{13}}{6} - frac{6}{6} ).Combine:( frac{7 - sqrt{13} - 2 + 2sqrt{13} - 6}{6} = frac{(7 - 2 - 6) + (-sqrt{13} + 2sqrt{13})}{6} = frac{-1 + sqrt{13}}{6} ).So, ( f(x) = frac{-1 + sqrt{13}}{6} ).Now, compute ( f^{-1}(x) ):( f^{-1}(x) = frac{-1 + sqrt{4 + 3x}}{3} ).Substitute ( x = frac{-1 + sqrt{13}}{6} ):Compute ( 4 + 3x = 4 + 3 times frac{-1 + sqrt{13}}{6} = 4 + frac{-3 + 3sqrt{13}}{6} = 4 - frac{1}{2} + frac{sqrt{13}}{2} = frac{8}{2} - frac{1}{2} + frac{sqrt{13}}{2} = frac{7 + sqrt{13}}{2} ).So, ( sqrt{4 + 3x} = sqrt{frac{7 + sqrt{13}}{2}} ).Hmm, this seems complicated. Let me see if this equals ( x ).Wait, ( f^{-1}(x) = frac{-1 + sqrt{4 + 3x}}{3} ).We have ( f(x) = frac{-1 + sqrt{13}}{6} ), and ( f^{-1}(x) ) is ( frac{-1 + sqrt{frac{7 + sqrt{13}}{2}}}{3} ).Wait, this seems messy. Maybe there's a better way to check.Alternatively, since ( f(x) = f^{-1}(x) ), then ( f(f(x)) = x ). We already know that ( x = frac{-1 + sqrt{13}}{6} ) is a solution to ( f(f(x)) = x ), so it should satisfy ( f(x) = f^{-1}(x) ).But let's verify numerically.Compute ( x = frac{-1 + sqrt{13}}{6} approx frac{-1 + 3.6055}{6} approx frac{2.6055}{6} approx 0.4342 ).Compute ( f(x) approx 3(0.4342)^2 + 2(0.4342) - 1 approx 3(0.1885) + 0.8684 - 1 approx 0.5655 + 0.8684 - 1 approx 0.4339 ).Compute ( f^{-1}(x) approx frac{-1 + sqrt{4 + 3(0.4342)}}{3} approx frac{-1 + sqrt{4 + 1.3026}}{3} approx frac{-1 + sqrt{5.3026}}{3} approx frac{-1 + 2.3027}{3} approx frac{1.3027}{3} approx 0.4342 ).So, ( f(x) approx 0.4339 ) and ( f^{-1}(x) approx 0.4342 ). These are approximately equal, considering rounding errors. So, ( x = frac{-1 + sqrt{13}}{6} ) is a valid solution.Now, check ( x = frac{-1 - sqrt{13}}{6} ):Compute ( x approx frac{-1 - 3.6055}{6} approx frac{-4.6055}{6} approx -0.7676 ).Compute ( f(x) = 3x^2 + 2x - 1 approx 3(0.5894) + 2(-0.7676) - 1 approx 1.7682 - 1.5352 - 1 approx -0.767 ).Compute ( f^{-1}(x) approx frac{-1 + sqrt{4 + 3(-0.7676)}}{3} approx frac{-1 + sqrt{4 - 2.3028}}{3} approx frac{-1 + sqrt{1.6972}}{3} approx frac{-1 + 1.3027}{3} approx frac{0.3027}{3} approx 0.1009 ).So, ( f(x) approx -0.767 ) and ( f^{-1}(x) approx 0.1009 ). These are not equal, so ( x = frac{-1 - sqrt{13}}{6} ) is not a valid solution.Wait, that's odd because earlier, when solving ( f(f(x)) = x ), we got this solution, but when checking ( f(x) = f^{-1}(x) ), it doesn't hold. So, this must be an extraneous solution introduced when squaring both sides.Therefore, the valid solutions are ( x = 0 ), ( x = -1 ), and ( x = frac{-1 + sqrt{13}}{6} ).But wait, earlier when checking ( x = -1 ), we saw that ( f(-1) = 0 ) and ( f^{-1}(-1) = 0 ) or ( -frac{2}{3} ). So, if we take ( f^{-1}(-1) = 0 ), then ( f(-1) = f^{-1}(-1) ) holds. But if we take ( f^{-1}(-1) = -frac{2}{3} ), then it doesn't hold. So, depending on the branch of the inverse function, ( x = -1 ) might or might not be a solution.This is getting complicated because ( f ) isn't one-to-one, so ( f^{-1}(x) ) isn't uniquely defined. Therefore, the solutions depend on how we define the inverse function.Given the problem statement, it's likely expecting the solutions where ( f(x) = f^{-1}(x) ) without worrying about the domain restrictions, so all solutions to ( f(f(x)) = x ) are valid, except those that don't satisfy ( f(x) = f^{-1}(x) ).From our checks, ( x = 0 ), ( x = -1 ), and ( x = frac{-1 + sqrt{13}}{6} ) satisfy ( f(x) = f^{-1}(x) ), while ( x = frac{-1 - sqrt{13}}{6} ) does not.Therefore, the values of ( x ) where ( f(x) = f^{-1}(x) ) are ( x = 0 ), ( x = -1 ), and ( x = frac{-1 + sqrt{13}}{6} ).But let me double-check ( x = -1 ):If ( x = -1 ), then ( f(-1) = 0 ), and ( f^{-1}(0) ) is either ( frac{1}{3} ) or ( -1 ). If we take ( f^{-1}(0) = -1 ), then ( f(-1) = f^{-1}(0) ), but we need ( f(-1) = f^{-1}(-1) ). Wait, no, the equation is ( f(x) = f^{-1}(x) ), so ( f(-1) = f^{-1}(-1) ).Compute ( f^{-1}(-1) ). As before, solving ( f(x) = -1 ):( 3x^2 + 2x - 1 = -1 ) leads to ( 3x^2 + 2x = 0 ), so ( x = 0 ) or ( x = -frac{2}{3} ).Therefore, ( f^{-1}(-1) ) can be ( 0 ) or ( -frac{2}{3} ). But ( f(-1) = 0 ), so if we take ( f^{-1}(-1) = 0 ), then ( f(-1) = f^{-1}(-1) ) holds. So, ( x = -1 ) is a valid solution.However, if we define ( f^{-1}(x) ) as the inverse function with a restricted domain where it's increasing, say ( x geq -frac{1}{3} ) (since the vertex of ( f(x) ) is at ( x = -frac{b}{2a} = -frac{2}{6} = -frac{1}{3} )), then ( f^{-1}(x) ) would be the branch where ( x geq -frac{1}{3} ), which would correspond to ( f^{-1}(x) = frac{-1 + sqrt{4 + 3x}}{3} ).In that case, ( f^{-1}(-1) ) would not be defined because ( -1 < -frac{1}{3} ). Therefore, ( x = -1 ) might not be a valid solution if we restrict the domain of ( f ) to make it invertible.This is a bit confusing. The problem doesn't specify the domain, so perhaps all solutions are acceptable. Therefore, the solutions are ( x = 0 ), ( x = -1 ), and ( x = frac{-1 + sqrt{13}}{6} ).But let me check the problem statement again: "For what value of ( x ) is ( f(x) ) equal to ( f^{-1}(x) )?" It says "value", singular, which might imply only one solution is expected. But we have multiple solutions.Looking back at the initial approach, when I set ( f(f(x)) = x ), I got four solutions, but upon checking, one was extraneous. So, the valid solutions are three: ( x = 0 ), ( x = -1 ), and ( x = frac{-1 + sqrt{13}}{6} ).However, considering the function's invertibility, perhaps only ( x = 0 ) and ( x = frac{-1 + sqrt{13}}{6} ) are valid because ( x = -1 ) might not be in the domain of the inverse function if we restrict it.Alternatively, the problem might accept all solutions regardless of the inverse function's domain.Given the ambiguity, but considering the problem likely expects the simplest solution, which is ( x = 0 ), as it's an integer and easy to verify.But to be thorough, I think the correct approach is to present all valid solutions, which are ( x = 0 ), ( x = -1 ), and ( x = frac{-1 + sqrt{13}}{6} ).However, upon re-examining the initial steps, when I set ( f(x) = f^{-1}(x) ), I ended up with the equation ( 81x^4 + 108x^3 - 27x = 0 ), which factors to ( 9x(9x^3 + 12x^2 - 3) = 0 ), leading to ( x = 0 ) and the roots of the cubic. The cubic had roots ( x = -1 ), ( x = frac{-1 + sqrt{13}}{6} ), and ( x = frac{-1 - sqrt{13}}{6} ). We saw that ( x = frac{-1 - sqrt{13}}{6} ) didn't satisfy ( f(x) = f^{-1}(x) ), so it's extraneous.Therefore, the valid solutions are ( x = 0 ), ( x = -1 ), and ( x = frac{-1 + sqrt{13}}{6} ).But since the problem asks for "the value" (singular), perhaps only one solution is expected. Given that ( x = 0 ) is the simplest and most straightforward, it's likely the intended answer.Alternatively, if multiple solutions are acceptable, all three should be listed.Given the initial problem didn't specify, but in the context of such problems, often only the principal solution is expected, which is ( x = 0 ).But to be precise, since we have three valid solutions, I should mention all of them.Wait, but in the initial problem, the user wrote:"Define the function ( f(x) = 3x^2 + 2x - 1 ). For what value of ( x ) is ( f(x) ) equal to ( f^{-1}(x) )?"They used "value" singular, but in reality, there are multiple solutions. So, perhaps the answer is all such ( x ).But in the initial assistant's response, they concluded ( x = 0 ) as the answer, but upon my detailed thinking, I found more solutions.So, perhaps the correct answer is ( x = 0 ), ( x = -1 ), and ( x = frac{-1 + sqrt{13}}{6} ).But to confirm, let me check ( x = frac{-1 + sqrt{13}}{6} ) again:Compute ( f(x) ) and ( f^{-1}(x) ):As before, ( f(x) = frac{-1 + sqrt{13}}{6} approx 0.4342 ).Compute ( f^{-1}(x) ):( f^{-1}(x) = frac{-1 + sqrt{4 + 3x}}{3} ).Substitute ( x approx 0.4342 ):( 4 + 3(0.4342) approx 4 + 1.3026 = 5.3026 ).( sqrt{5.3026} approx 2.3027 ).So, ( f^{-1}(x) approx frac{-1 + 2.3027}{3} approx frac{1.3027}{3} approx 0.4342 ).So, ( f(x) approx f^{-1}(x) approx 0.4342 ). Therefore, this is a valid solution.Similarly, ( x = -1 ):( f(-1) = 0 ), and ( f^{-1}(-1) = 0 ) (if we take the appropriate branch), so it's valid.Therefore, the solutions are ( x = 0 ), ( x = -1 ), and ( x = frac{-1 + sqrt{13}}{6} ).But the problem says "For what value of ( x )...", which is singular, but mathematically, there are multiple values. So, perhaps the answer is all such ( x ).However, in the initial assistant's response, they only concluded ( x = 0 ). So, maybe I should present all solutions.But to wrap up, the values of ( x ) where ( f(x) = f^{-1}(x) ) are ( x = 0 ), ( x = -1 ), and ( x = frac{-1 + sqrt{13}}{6} ).But let me check if ( x = frac{-1 + sqrt{13}}{6} ) is indeed a solution by substituting back into ( f(x) = f^{-1}(x) ):As computed earlier, ( f(x) approx 0.4342 ) and ( f^{-1}(x) approx 0.4342 ), so it holds.Therefore, the final answer is ( x = 0 ), ( x = -1 ), and ( x = frac{-1 + sqrt{13}}{6} ).But since the problem asks for "the value", perhaps listing all is appropriate.Alternatively, if the problem expects only one solution, ( x = 0 ) is the simplest.Given the ambiguity, but to be thorough, I think the correct approach is to present all valid solutions.

💡Okay, so I have this problem with three complex numbers: a + bi, c + di, and e + fi. I need to find the sum of d and f, given some conditions. Let me try to break this down step by step.First, I know that complex numbers have a real part and an imaginary part. So, each of these numbers can be written as (real part) + (imaginary part)i. For example, the first number is a + bi, where a is the real part and b is the coefficient of the imaginary part.The problem gives me a few pieces of information:1. b = 2. So, the imaginary part of the first complex number is 2i.2. e = -a - 2c. This relates the real part of the third complex number to the real parts of the first and second numbers.3. The sum of all three complex numbers is 2 - 3i.My goal is to find d + f, which are the coefficients of the imaginary parts of the second and third complex numbers.Let me write down the sum of the three complex numbers:(a + bi) + (c + di) + (e + fi) = 2 - 3i.I can substitute e with the expression given: e = -a - 2c. So, the third complex number becomes (-a - 2c) + fi.Now, let's plug that into the sum:(a + bi) + (c + di) + (-a - 2c + fi) = 2 - 3i.Let me simplify this expression by combining like terms. First, I'll handle the real parts:a (from the first number) + c (from the second number) + (-a - 2c) (from the third number).So, combining the real parts:a + c - a - 2c.The a and -a cancel each other out, leaving c - 2c, which simplifies to -c.Now, let's handle the imaginary parts:bi (from the first number) + di (from the second number) + fi (from the third number).Combining these:bi + di + fi.Since b is given as 2, this becomes 2i + di + fi.I can factor out the i:(2 + d + f)i.So, putting it all together, the sum of the three complex numbers is:(-c) + (2 + d + f)i = 2 - 3i.Now, for two complex numbers to be equal, their real parts must be equal, and their imaginary parts must be equal. So, I can set up two equations:1. Real parts: -c = 2.2. Imaginary parts: 2 + d + f = -3.Let me solve the first equation for c:-c = 2 Multiply both sides by -1: c = -2.Now, plug this value of c into the second equation:2 + d + f = -3.I need to solve for d + f. Let's subtract 2 from both sides:d + f = -3 - 2 d + f = -5.So, the sum of d and f is -5.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. I correctly substituted e = -a - 2c into the third complex number.2. When combining the real parts, I correctly simplified a + c - a - 2c to -c.3. For the imaginary parts, I correctly combined 2i + di + fi into (2 + d + f)i.4. Setting the real parts equal: -c = 2, so c = -2. That seems right.5. Then, setting the imaginary parts equal: 2 + d + f = -3. Subtracting 2 gives d + f = -5.Everything seems to check out. I don't see any errors in my calculations.So, the final answer is d + f = -5.

💡Okay, so I need to figure out how many more people attended the second concert compared to the first one. Let's see, the first concert had 65,899 people, and the second one had 66,018 people. I think I need to subtract the number of people at the first concert from the number at the second concert to find the difference. Alright, let's write that down: 66,018 minus 65,899. Hmm, doing this subtraction might be a bit tricky because of the large numbers, but I can break it down step by step. Maybe I'll start by subtracting the thousands first. So, 66,000 minus 65,000 is 1,000. That seems straightforward. Now, I have 18 left in the second concert number and 899 in the first concert number. Wait, that doesn't seem right. Maybe I should align the numbers properly to make sure I'm subtracting the correct place values.Let me write it out vertically:``` 66,018-65,899--------```Starting from the right, subtract the units place: 8 minus 9. Oh, I can't subtract 9 from 8, so I'll need to borrow from the tens place. The 1 in the tens place becomes 0, and the 8 in the units place becomes 18. Now, 18 minus 9 is 9.Next, the tens place: 0 minus 9. Again, I can't subtract 9 from 0, so I'll borrow from the hundreds place. The 0 in the hundreds place becomes 9, and the 0 in the tens place becomes 10. Now, 10 minus 9 is 1.Moving to the hundreds place: 9 minus 8 is 1.In the thousands place: 6 minus 5 is 1.Finally, in the ten-thousands place: 6 minus 6 is 0.Putting it all together, the difference is 119. So, 119 more people attended the second concert compared to the first one. That makes sense because 66,018 minus 65,899 equals 119. I think that's the right answer!