💡Okay, so George has some socks, and then he buys more, and his dad gives him some, and now he has 68 socks. I need to figure out how many socks he had at the beginning. Hmm, let's break this down step by step.First, let's think about what we know. George bought 36.0 new socks. That's a pretty big number, almost 40 socks! Then, his dad gave him 4.0 more. So, in total, George received 36 + 4 = 40 socks from buying and his dad. Now, he has 68 socks in total.Wait, so if he ended up with 68 socks after getting 40 more, then he must have had 68 - 40 = 28 socks initially. That seems straightforward, but let me double-check to make sure I'm not missing anything.Let me write this out as an equation to visualize it better. Let's let x be the number of socks George had initially. So, the equation would be:x (initial socks) + 36 (bought) + 4 (from dad) = 68 (total now)So, x + 36 + 4 = 68Combining the numbers, 36 + 4 = 40, so:x + 40 = 68To solve for x, I need to subtract 40 from both sides:x + 40 - 40 = 68 - 40Which simplifies to:x = 28Okay, so George had 28 socks initially. That makes sense because if he started with 28, bought 36, and got 4 more, that adds up to 28 + 36 + 4 = 68, which matches the total he has now.I think that's it. It seems pretty straightforward, but just to be thorough, let me check my calculations again.28 (initial) + 36 (bought) = 6464 + 4 (from dad) = 68Yep, that checks out. So, George must have had 28 socks to begin with.

💡First, I need to determine how much money Frank spent in total. He gave the cashier 20 and received 4 in change, so he spent 16.Next, I'll calculate the cost of the chips. Frank bought 2 bags of chips, and each bag costs 3. Therefore, the total cost for the chips is 2 multiplied by 3, which equals 6.Now, I'll find out how much money was spent on the chocolate bars by subtracting the cost of the chips from the total amount spent. So, 16 minus 6 equals 10 spent on chocolate bars.Since each chocolate bar costs 2, I'll divide the total amount spent on chocolate bars by the cost per bar to find out how many bars Frank bought. Therefore, 10 divided by 2 equals 5 chocolate bars.

💡Alright, I have this quadratic function given by ( y = (x + 1)(ax + 2a + 2) ), where ( a ) is a constant and ( a neq 0 ). There are three parts to this problem, and I need to tackle each one step by step. Let me start with part (1).**Problem (1): If ( a = 1 ), find the coordinates of the vertex of the graph of this function.**Okay, so if ( a = 1 ), I can substitute that into the equation. Let me do that:( y = (x + 1)(1 cdot x + 2 cdot 1 + 2) )Simplify the terms inside the second parenthesis:( 2 cdot 1 = 2 ), so ( 2 + 2 = 4 ). Therefore, the equation becomes:( y = (x + 1)(x + 4) )Now, I can expand this to get it into standard quadratic form. Let me multiply the two binomials:( y = x(x + 4) + 1(x + 4) )( y = x^2 + 4x + x + 4 )( y = x^2 + 5x + 4 )Alright, so now it's in the form ( y = ax^2 + bx + c ), where ( a = 1 ), ( b = 5 ), and ( c = 4 ). To find the vertex, I know the formula for the x-coordinate of the vertex is ( -frac{b}{2a} ). Let me compute that:( x = -frac{5}{2 cdot 1} = -frac{5}{2} )Now, to find the y-coordinate, I substitute this x-value back into the equation:( y = left(-frac{5}{2}right)^2 + 5 cdot left(-frac{5}{2}right) + 4 )Calculating each term:( left(-frac{5}{2}right)^2 = frac{25}{4} )( 5 cdot left(-frac{5}{2}right) = -frac{25}{2} )So,( y = frac{25}{4} - frac{25}{2} + 4 )To combine these, I need a common denominator, which is 4:( y = frac{25}{4} - frac{50}{4} + frac{16}{4} )( y = frac{25 - 50 + 16}{4} )( y = frac{-9}{4} )So, the vertex is at ( left(-frac{5}{2}, -frac{9}{4}right) ). That should be the answer for part (1).**Problem (2): If the graph of this quadratic function passes through one of the points ( (-1, 1) ), ( (-2, 3) ), or ( (0, -2) ), find the expression of this quadratic function.**Hmm, okay. So, the function passes through one of these three points. That means if I substitute the x and y values of each point into the equation, it should satisfy it. Since it's a quadratic function, and ( a ) is a constant, I can plug in each point and solve for ( a ). Let me try each point one by one.First, let's try the point ( (-1, 1) ):Substitute ( x = -1 ) and ( y = 1 ) into the equation:( 1 = (-1 + 1)(a cdot (-1) + 2a + 2) )Simplify the first parenthesis:( (-1 + 1) = 0 )So, the equation becomes:( 1 = 0 cdot (text{something}) )Which simplifies to:( 1 = 0 )That's not possible. So, the function doesn't pass through ( (-1, 1) ).Next, let's try the point ( (-2, 3) ):Substitute ( x = -2 ) and ( y = 3 ):( 3 = (-2 + 1)(a cdot (-2) + 2a + 2) )Simplify the first parenthesis:( (-2 + 1) = -1 )Now, simplify the second parenthesis:( a cdot (-2) = -2a )So, ( -2a + 2a + 2 = 0 + 2 = 2 )Therefore, the equation becomes:( 3 = (-1)(2) )( 3 = -2 )Again, that's not possible. So, the function doesn't pass through ( (-2, 3) ) either.Now, let's try the point ( (0, -2) ):Substitute ( x = 0 ) and ( y = -2 ):( -2 = (0 + 1)(a cdot 0 + 2a + 2) )Simplify the first parenthesis:( (0 + 1) = 1 )Simplify the second parenthesis:( a cdot 0 = 0 ), so ( 0 + 2a + 2 = 2a + 2 )Therefore, the equation becomes:( -2 = 1 cdot (2a + 2) )( -2 = 2a + 2 )Now, solve for ( a ):Subtract 2 from both sides:( -2 - 2 = 2a )( -4 = 2a )Divide both sides by 2:( a = -2 )Okay, so ( a = -2 ). Now, let's write the quadratic function with this value of ( a ):( y = (x + 1)(-2x + 2(-2) + 2) )Simplify inside the second parenthesis:( 2(-2) = -4 ), so ( -4 + 2 = -2 )Therefore, the function becomes:( y = (x + 1)(-2x - 2) )I can factor out a -2 from the second parenthesis:( y = (x + 1)(-2)(x + 1) )( y = -2(x + 1)^2 )So, the quadratic function is ( y = -2(x + 1)^2 ). That should be the answer for part (2).**Problem (3): If the graph of the quadratic function passes through the points ( (x_1, y_1) ) and ( (x_2, y_2) ), where ( x_1 + x_2 = 2 ), ( x_1 < x_2 ), and ( y_1 > y_2 ), prove that ( a < -frac{2}{5} ).**Alright, this seems more involved. Let me try to unpack this step by step.First, the quadratic function is ( y = (x + 1)(ax + 2a + 2) ). Let me expand this to standard form to make it easier to analyze.Expanding:( y = (x + 1)(ax + 2a + 2) )( y = x(ax + 2a + 2) + 1(ax + 2a + 2) )( y = ax^2 + 2a x + 2x + ax + 2a + 2 )Wait, hold on, let me do that multiplication correctly:First, multiply ( x ) by each term in the second parenthesis:( x cdot ax = a x^2 )( x cdot 2a = 2a x )( x cdot 2 = 2x )Then, multiply 1 by each term in the second parenthesis:( 1 cdot ax = a x )( 1 cdot 2a = 2a )( 1 cdot 2 = 2 )Now, combine all these terms:( y = a x^2 + 2a x + 2x + a x + 2a + 2 )Combine like terms:- ( a x^2 ) remains as is.- ( 2a x + a x = 3a x )- ( 2x ) remains as is.- ( 2a + 2 ) remains as is.So, the equation becomes:( y = a x^2 + (3a + 2) x + (2a + 2) )Alright, so in standard form, it's:( y = a x^2 + (3a + 2) x + (2a + 2) )Now, the graph passes through two points ( (x_1, y_1) ) and ( (x_2, y_2) ) with the conditions:1. ( x_1 + x_2 = 2 )2. ( x_1 < x_2 )3. ( y_1 > y_2 )We need to prove that ( a < -frac{2}{5} ).Let me think about how to approach this. Since the quadratic passes through these two points, their coordinates satisfy the equation. So, for each point, we can write an equation:For ( (x_1, y_1) ):( y_1 = a x_1^2 + (3a + 2) x_1 + (2a + 2) )For ( (x_2, y_2) ):( y_2 = a x_2^2 + (3a + 2) x_2 + (2a + 2) )Given that ( y_1 > y_2 ), subtracting the two equations gives:( y_1 - y_2 = a (x_1^2 - x_2^2) + (3a + 2)(x_1 - x_2) )Factor the differences of squares and linear terms:( y_1 - y_2 = a (x_1 - x_2)(x_1 + x_2) + (3a + 2)(x_1 - x_2) )Factor out ( (x_1 - x_2) ):( y_1 - y_2 = (x_1 - x_2) [a (x_1 + x_2) + (3a + 2)] )We know that ( x_1 + x_2 = 2 ), so substitute that in:( y_1 - y_2 = (x_1 - x_2) [a cdot 2 + 3a + 2] )( y_1 - y_2 = (x_1 - x_2) [2a + 3a + 2] )( y_1 - y_2 = (x_1 - x_2) [5a + 2] )Now, we know that ( y_1 > y_2 ), so ( y_1 - y_2 > 0 ). Also, ( x_1 < x_2 ), so ( x_1 - x_2 < 0 ). Therefore, the product ( (x_1 - x_2)(5a + 2) ) must be positive.Since ( x_1 - x_2 < 0 ), the term ( (5a + 2) ) must be negative to make the product positive (because negative times negative is positive). Therefore:( 5a + 2 < 0 )( 5a < -2 )( a < -frac{2}{5} )So, that's the result we needed to prove. Therefore, ( a < -frac{2}{5} ).Wait, let me double-check my steps to make sure I didn't make a mistake.1. Expanded the quadratic correctly to standard form.2. Plugged in the points into the equation, subtracted to get ( y_1 - y_2 ).3. Factored out ( x_1 - x_2 ) correctly.4. Substituted ( x_1 + x_2 = 2 ).5. Simplified the expression inside the brackets to ( 5a + 2 ).6. Noted that ( y_1 - y_2 > 0 ) and ( x_1 - x_2 < 0 ), so ( 5a + 2 < 0 ).7. Solved for ( a ) to get ( a < -frac{2}{5} ).Everything seems to check out. So, I think this is a valid proof.**Final Answer**(1) The vertex is at (boxed{left(-frac{5}{2}, -frac{9}{4}right)}).(2) The quadratic function is (boxed{y = -2(x + 1)^2}).(3) It is proven that (a < -frac{2}{5}), so the result is (boxed{a < -frac{2}{5}}).

💡Alright, so I've got this problem here with four different functions, and I need to figure out which one has a minimum value of 4. The options are labeled A through D, each with their own function. Let me take them one by one and see which one fits the bill.Starting with option A: ( y = x + frac{4}{x} ). Hmm, okay. I remember that for functions like this, where you have a variable and its reciprocal multiplied by a constant, you can use calculus to find the minimum or maximum. Alternatively, I think the AM-GM inequality might apply here. Let me recall: the AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. So, maybe I can apply that here.But wait, in this case, ( x ) can be positive or negative because it's just ( x ) without any restrictions. If ( x ) is positive, then both ( x ) and ( frac{4}{x} ) are positive, so AM-GM would work. But if ( x ) is negative, then ( y ) becomes negative as well because ( x ) is negative and ( frac{4}{x} ) is also negative. So, as ( x ) approaches negative infinity, ( y ) would approach negative infinity. That means the function doesn't have a lower bound; it can go to negative infinity. So, it can't have a minimum value of 4 because it can go much lower. So, option A is probably not the answer.Moving on to option B: ( y = sin x + frac{4}{sin x} ) with ( 0 < x < pi ). Okay, so ( sin x ) in this interval ranges from 0 to 1 because sine of 0 is 0 and sine of ( pi/2 ) is 1, and then it decreases back to 0 at ( pi ). So, ( sin x ) is positive in this interval. Therefore, both ( sin x ) and ( frac{4}{sin x} ) are positive. Maybe I can apply AM-GM here as well.Let me set ( a = sin x ) and ( b = frac{4}{sin x} ). Then, the AM-GM inequality tells us that ( frac{a + b}{2} geq sqrt{ab} ). Plugging in the values, we get ( frac{sin x + frac{4}{sin x}}{2} geq sqrt{sin x cdot frac{4}{sin x}} ). Simplifying the right side, the ( sin x ) cancels out, so we have ( sqrt{4} = 2 ). Multiplying both sides by 2, we get ( sin x + frac{4}{sin x} geq 4 ). So, the minimum value is 4, right?But wait, the AM-GM inequality requires that ( a = b ) for equality. So, when does ( sin x = frac{4}{sin x} )? Let's solve that equation: ( sin^2 x = 4 ), which implies ( sin x = 2 ) or ( sin x = -2 ). But ( sin x ) can't be 2 or -2 because its range is only between -1 and 1. So, equality can never be achieved here. That means the minimum value is actually greater than 4. So, option B doesn't have a minimum value of 4 either.Okay, moving on to option C: ( y = 4 log_{3}x + log_{x}3 ). Hmm, logarithms. Let me see. First, I know that ( log_{x}3 ) can be rewritten using the change of base formula. Specifically, ( log_{x}3 = frac{1}{log_{3}x} ). So, substituting that in, the function becomes ( y = 4 log_{3}x + frac{1}{log_{3}x} ).Let me set ( t = log_{3}x ). Then, the function becomes ( y = 4t + frac{1}{t} ). Now, this looks similar to the functions in options A and B. Let's see if I can apply AM-GM here. The AM-GM inequality says that ( frac{4t + frac{1}{t}}{2} geq sqrt{4t cdot frac{1}{t}} ). Simplifying the right side, the ( t ) cancels out, so we get ( sqrt{4} = 2 ). Multiplying both sides by 2, we get ( 4t + frac{1}{t} geq 4 ). So, the minimum value is 4.But again, equality holds when ( 4t = frac{1}{t} ), which implies ( 4t^2 = 1 ), so ( t^2 = frac{1}{4} ), hence ( t = frac{1}{2} ) or ( t = -frac{1}{2} ). But ( t = log_{3}x ), and since ( x ) must be positive (because logarithm is only defined for positive numbers), ( t ) can be positive or negative depending on whether ( x ) is greater than 1 or between 0 and 1.If ( t = frac{1}{2} ), then ( log_{3}x = frac{1}{2} ), so ( x = 3^{1/2} = sqrt{3} ). That's valid. If ( t = -frac{1}{2} ), then ( log_{3}x = -frac{1}{2} ), so ( x = 3^{-1/2} = frac{1}{sqrt{3}} ). That's also valid. So, both these points give ( y = 4 ). So, does that mean the minimum value is 4?Wait, but earlier with option B, the equality couldn't be achieved because ( sin x ) couldn't be 2. But here, ( t ) can be ( frac{1}{2} ) or ( -frac{1}{2} ), which correspond to valid ( x ) values. So, does that mean this function actually attains the minimum value of 4? Hmm, that seems conflicting with my initial thought.Wait, let me double-check. If ( t = frac{1}{2} ), then ( y = 4 * frac{1}{2} + frac{1}{frac{1}{2}} = 2 + 2 = 4 ). Similarly, if ( t = -frac{1}{2} ), then ( y = 4 * (-frac{1}{2}) + frac{1}{-frac{1}{2}} = -2 - 2 = -4 ). Oh, wait a second! So, when ( t = -frac{1}{2} ), ( y = -4 ). That's actually a lower value than 4. So, does that mean the function can go as low as -4?But hold on, the problem is asking for the minimum value of 4. So, if the function can attain -4, which is lower than 4, then 4 isn't the minimum. So, maybe I made a mistake here. Let me think again.The function ( y = 4 log_{3}x + log_{x}3 ) can indeed take negative values because ( log_{3}x ) can be negative if ( x ) is between 0 and 1. So, the function doesn't have a minimum value of 4; it can go lower. Therefore, option C isn't the correct answer either.Alright, moving on to option D: ( y = 4e^{x} + e^{-x} ). Hmm, exponentials. I know that ( e^{x} ) is always positive, so both terms here are positive for all real ( x ). That means the function is always positive, so it can't go to negative infinity or anything like that. Maybe I can apply AM-GM here as well.Let me set ( a = 4e^{x} ) and ( b = e^{-x} ). Then, the AM-GM inequality tells us that ( frac{a + b}{2} geq sqrt{ab} ). Plugging in the values, we get ( frac{4e^{x} + e^{-x}}{2} geq sqrt{4e^{x} cdot e^{-x}} ). Simplifying the right side, ( e^{x} cdot e^{-x} = e^{0} = 1 ), so ( sqrt{4 * 1} = 2 ). Multiplying both sides by 2, we get ( 4e^{x} + e^{-x} geq 4 ). So, the minimum value is 4.But when does equality hold? It holds when ( a = b ), so ( 4e^{x} = e^{-x} ). Let's solve for ( x ). Dividing both sides by ( e^{x} ), we get ( 4 = e^{-2x} ). Taking the natural logarithm of both sides, ( ln 4 = -2x ), so ( x = -frac{ln 4}{2} = -ln 2 ). That's a valid real number, so equality is achieved at ( x = -ln 2 ). Therefore, the function does attain the minimum value of 4 at that point.So, putting it all together:- Option A: Can go to negative infinity, so minimum isn't 4.- Option B: Theoretical minimum is 4, but it's unattainable because ( sin x ) can't be 2.- Option C: The function can actually go lower than 4, so 4 isn't the minimum.- Option D: The function has a minimum value of 4, which is attainable.Therefore, the correct answer must be option D.**Final Answer**boxed{D}

💡Okay, so I need to figure out the range of the real number ( m ) such that the function ( f(x) = frac{1}{3}x^3 - frac{1}{2}x^2 + m ) is a "mean value function" on the interval ([0, m]). First, let me make sure I understand the definition correctly. A function ( f(x) ) is called a "mean value function" on ([a, b]) if there exist two points ( x_1 ) and ( x_2 ) within the open interval ( (a, b) ) such that the derivative at both points equals the average rate of change of the function over ([a, b]). In mathematical terms, that means:[f'(x_1) = f'(x_2) = frac{f(b) - f(a)}{b - a}]So, in this problem, ( a = 0 ) and ( b = m ). Therefore, we need to find ( m ) such that there are two distinct points ( x_1 ) and ( x_2 ) in ( (0, m) ) where the derivative of ( f(x) ) equals the average rate of change over ([0, m]).Let me start by computing the average rate of change of ( f(x) ) over ([0, m]). That would be:[frac{f(m) - f(0)}{m - 0} = frac{f(m) - f(0)}{m}]So, I need to compute ( f(m) ) and ( f(0) ).Calculating ( f(m) ):[f(m) = frac{1}{3}m^3 - frac{1}{2}m^2 + m]Calculating ( f(0) ):[f(0) = frac{1}{3}(0)^3 - frac{1}{2}(0)^2 + m = m]So, the average rate of change is:[frac{f(m) - f(0)}{m} = frac{left(frac{1}{3}m^3 - frac{1}{2}m^2 + mright) - m}{m} = frac{frac{1}{3}m^3 - frac{1}{2}m^2}{m}]Simplifying this:[frac{frac{1}{3}m^3 - frac{1}{2}m^2}{m} = frac{1}{3}m^2 - frac{1}{2}m]So, the average rate of change is ( frac{1}{3}m^2 - frac{1}{2}m ).Next, I need to find the derivative of ( f(x) ) because we're looking for points where the derivative equals this average rate of change. Let's compute ( f'(x) ):[f'(x) = frac{d}{dx}left(frac{1}{3}x^3 - frac{1}{2}x^2 + mright) = x^2 - x]So, ( f'(x) = x^2 - x ).We need to find two distinct points ( x_1 ) and ( x_2 ) in ( (0, m) ) such that:[f'(x_1) = f'(x_2) = frac{1}{3}m^2 - frac{1}{2}m]Which translates to solving the equation:[x^2 - x = frac{1}{3}m^2 - frac{1}{2}m]Let me denote the right-hand side as a constant ( C ) for simplicity:[C = frac{1}{3}m^2 - frac{1}{2}m]So, the equation becomes:[x^2 - x - C = 0]Substituting back ( C ):[x^2 - x - left(frac{1}{3}m^2 - frac{1}{2}mright) = 0]Simplify:[x^2 - x - frac{1}{3}m^2 + frac{1}{2}m = 0]Let me write this as:[x^2 - x + left(-frac{1}{3}m^2 + frac{1}{2}mright) = 0]This is a quadratic equation in terms of ( x ). For this quadratic equation to have two distinct real solutions in the interval ( (0, m) ), certain conditions must be satisfied.First, the quadratic equation must have two distinct real roots. This requires the discriminant ( D ) to be positive. The discriminant of a quadratic equation ( ax^2 + bx + c = 0 ) is ( D = b^2 - 4ac ).In our case, ( a = 1 ), ( b = -1 ), and ( c = -frac{1}{3}m^2 + frac{1}{2}m ). So, the discriminant is:[D = (-1)^2 - 4 times 1 times left(-frac{1}{3}m^2 + frac{1}{2}mright) = 1 - 4left(-frac{1}{3}m^2 + frac{1}{2}mright)]Simplify this:[D = 1 + frac{4}{3}m^2 - 2m]For two distinct real roots, we require ( D > 0 ):[1 + frac{4}{3}m^2 - 2m > 0]Multiply both sides by 3 to eliminate the fraction:[3 + 4m^2 - 6m > 0]Rearranged:[4m^2 - 6m + 3 > 0]Let me compute the discriminant of this quadratic in ( m ) to see if it ever equals zero or becomes negative. The discriminant ( D' ) is:[D' = (-6)^2 - 4 times 4 times 3 = 36 - 48 = -12]Since ( D' < 0 ), the quadratic ( 4m^2 - 6m + 3 ) is always positive for all real ( m ). Therefore, the discriminant ( D > 0 ) is always satisfied, meaning the quadratic in ( x ) always has two distinct real roots, regardless of ( m ).However, we also need these roots to lie within the interval ( (0, m) ). So, the next step is to ensure that both roots ( x_1 ) and ( x_2 ) satisfy ( 0 < x_1 < x_2 < m ).To find the roots of the quadratic equation ( x^2 - x - frac{1}{3}m^2 + frac{1}{2}m = 0 ), we can use the quadratic formula:[x = frac{1 pm sqrt{1 + frac{4}{3}m^2 - 2m}}{2}]Simplify the expression under the square root:[sqrt{1 + frac{4}{3}m^2 - 2m} = sqrt{frac{4}{3}m^2 - 2m + 1}]Let me denote this as ( sqrt{A} ), where ( A = frac{4}{3}m^2 - 2m + 1 ).So, the roots are:[x = frac{1 pm sqrt{frac{4}{3}m^2 - 2m + 1}}{2}]We need both roots to be greater than 0 and less than ( m ).First, let's analyze the smaller root:[x_1 = frac{1 - sqrt{frac{4}{3}m^2 - 2m + 1}}{2}]And the larger root:[x_2 = frac{1 + sqrt{frac{4}{3}m^2 - 2m + 1}}{2}]We need both ( x_1 > 0 ) and ( x_2 < m ).Let's first ensure ( x_1 > 0 ):[frac{1 - sqrt{frac{4}{3}m^2 - 2m + 1}}{2} > 0]Multiply both sides by 2:[1 - sqrt{frac{4}{3}m^2 - 2m + 1} > 0]Which simplifies to:[sqrt{frac{4}{3}m^2 - 2m + 1} < 1]Square both sides (since both sides are positive):[frac{4}{3}m^2 - 2m + 1 < 1]Subtract 1 from both sides:[frac{4}{3}m^2 - 2m < 0]Factor out ( m ):[mleft(frac{4}{3}m - 2right) < 0]This inequality holds when ( m ) is between 0 and ( frac{3}{2} ) because:- For ( m > 0 ), the term ( frac{4}{3}m - 2 ) is negative when ( m < frac{3}{2} ).- So, the product is negative when ( 0 < m < frac{3}{2} ).Therefore, ( x_1 > 0 ) implies ( 0 < m < frac{3}{2} ).Next, let's ensure ( x_2 < m ):[frac{1 + sqrt{frac{4}{3}m^2 - 2m + 1}}{2} < m]Multiply both sides by 2:[1 + sqrt{frac{4}{3}m^2 - 2m + 1} < 2m]Subtract 1 from both sides:[sqrt{frac{4}{3}m^2 - 2m + 1} < 2m - 1]Now, for the square root to be real, the expression inside must be non-negative:[frac{4}{3}m^2 - 2m + 1 geq 0]But earlier, we saw that this quadratic is always positive because its discriminant is negative, so this condition is always satisfied.Now, back to the inequality:[sqrt{frac{4}{3}m^2 - 2m + 1} < 2m - 1]First, the right-hand side must be positive because the square root is always non-negative:[2m - 1 > 0 implies m > frac{1}{2}]So, ( m > frac{1}{2} ).Now, square both sides (since both sides are positive):[frac{4}{3}m^2 - 2m + 1 < (2m - 1)^2]Compute the right-hand side:[(2m - 1)^2 = 4m^2 - 4m + 1]So, the inequality becomes:[frac{4}{3}m^2 - 2m + 1 < 4m^2 - 4m + 1]Subtract ( frac{4}{3}m^2 - 2m + 1 ) from both sides:[0 < 4m^2 - 4m + 1 - frac{4}{3}m^2 + 2m - 1]Simplify:[0 < left(4m^2 - frac{4}{3}m^2right) + (-4m + 2m) + (1 - 1)]Calculate each term:- ( 4m^2 - frac{4}{3}m^2 = frac{12}{3}m^2 - frac{4}{3}m^2 = frac{8}{3}m^2 )- ( -4m + 2m = -2m )- ( 1 - 1 = 0 )So, the inequality becomes:[0 < frac{8}{3}m^2 - 2m]Multiply both sides by 3 to eliminate the fraction:[0 < 8m^2 - 6m]Factor out ( 2m ):[0 < 2m(4m - 3)]So, the inequality ( 2m(4m - 3) > 0 ) holds when:- Both factors are positive: ( m > 0 ) and ( 4m - 3 > 0 implies m > frac{3}{4} )- Or both factors are negative: ( m < 0 ) and ( 4m - 3 < 0 implies m < frac{3}{4} ). But since ( m > frac{1}{2} ) from earlier, this case is irrelevant.Therefore, ( 2m(4m - 3) > 0 ) implies ( m > frac{3}{4} ).Combining this with the earlier condition ( m > frac{1}{2} ), we have ( m > frac{3}{4} ).Now, let's summarize the conditions we have so far:1. From ( x_1 > 0 ): ( 0 < m < frac{3}{2} )2. From ( x_2 < m ): ( m > frac{3}{4} )Therefore, combining these two, we get:[frac{3}{4} < m < frac{3}{2}]But we also need to ensure that ( m > 0 ), which is already satisfied in this interval.Wait, let me double-check. When we considered ( x_1 > 0 ), we found ( 0 < m < frac{3}{2} ). Then, for ( x_2 < m ), we found ( m > frac{3}{4} ). So, intersecting these two intervals gives ( frac{3}{4} < m < frac{3}{2} ).Is there anything else I need to consider? Maybe checking the behavior of the function or ensuring that the roots are distinct and lie within the interval.Wait, another thing: the quadratic equation ( x^2 - x - frac{1}{3}m^2 + frac{1}{2}m = 0 ) must have both roots in ( (0, m) ). So, besides the conditions on the roots, we should also ensure that the quadratic is positive or negative appropriately at the endpoints.Wait, actually, since we already ensured that both roots are within ( (0, m) ) by checking ( x_1 > 0 ) and ( x_2 < m ), maybe we don't need additional conditions. But just to be thorough, let me check the value of the quadratic at ( x = 0 ) and ( x = m ).Compute ( g(0) ):[g(0) = 0^2 - 0 - frac{1}{3}m^2 + frac{1}{2}m = -frac{1}{3}m^2 + frac{1}{2}m]Compute ( g(m) ):[g(m) = m^2 - m - frac{1}{3}m^2 + frac{1}{2}m = left(m^2 - frac{1}{3}m^2right) + left(-m + frac{1}{2}mright) = frac{2}{3}m^2 - frac{1}{2}m]So, ( g(0) = -frac{1}{3}m^2 + frac{1}{2}m ) and ( g(m) = frac{2}{3}m^2 - frac{1}{2}m ).For the quadratic to cross the x-axis twice between 0 and ( m ), the function should be negative at ( x = 0 ) and positive at ( x = m ), or vice versa. But since the leading coefficient is positive (1), the parabola opens upwards. Therefore, if ( g(0) < 0 ) and ( g(m) > 0 ), the quadratic will cross the x-axis twice between 0 and ( m ).So, let's check these conditions:1. ( g(0) < 0 ):[-frac{1}{3}m^2 + frac{1}{2}m < 0]Multiply both sides by 6 to eliminate denominators:[-2m^2 + 3m < 0]Factor:[m(-2m + 3) < 0]This inequality holds when ( m ) is between 0 and ( frac{3}{2} ):- For ( m > 0 ), the term ( -2m + 3 ) is positive when ( m < frac{3}{2} ).- So, the product is negative when ( 0 < m < frac{3}{2} ).2. ( g(m) > 0 ):[frac{2}{3}m^2 - frac{1}{2}m > 0]Multiply both sides by 6:[4m^2 - 3m > 0]Factor:[m(4m - 3) > 0]This inequality holds when:- ( m > 0 ) and ( 4m - 3 > 0 implies m > frac{3}{4} )- Or ( m < 0 ) and ( 4m - 3 < 0 implies m < frac{3}{4} ). But since ( m > 0 ), we only consider ( m > frac{3}{4} ).So, combining these two conditions:- ( g(0) < 0 ) gives ( 0 < m < frac{3}{2} )- ( g(m) > 0 ) gives ( m > frac{3}{4} )Thus, the intersection is ( frac{3}{4} < m < frac{3}{2} ), which is consistent with our earlier conclusion.Therefore, the range of ( m ) is ( left( frac{3}{4}, frac{3}{2} right) ).Just to recap:1. Calculated the average rate of change.2. Found the derivative of ( f(x) ).3. Set up the quadratic equation for the derivative equal to the average rate of change.4. Ensured the quadratic has two distinct real roots by checking the discriminant (which was always positive).5. Ensured both roots lie within ( (0, m) ) by analyzing the conditions on ( x_1 ) and ( x_2 ).6. Verified by checking the signs of the quadratic at the endpoints ( x = 0 ) and ( x = m ).Everything seems to check out, so I'm confident the range is ( frac{3}{4} < m < frac{3}{2} ).**Final Answer**The range of the real number ( m ) is boxed{left( dfrac{3}{4}, dfrac{3}{2} right)}.

💡First, I'll denote the number of red mushrooms Bill gathered as ( R ).Ted gathered 6 blue mushrooms, and half of them have white spots. So, the number of white-spotted blue mushrooms is:[frac{6}{2} = 3]Bill gathered 6 brown mushrooms, and all of them have white spots. Therefore, there are 6 white-spotted brown mushrooms.Two-thirds of the red mushrooms have white spots. Thus, the number of white-spotted red mushrooms is:[frac{2}{3}R]The total number of white-spotted mushrooms is 17. This gives us the equation:[frac{2}{3}R + 6 + 3 = 17]Simplifying the equation:[frac{2}{3}R + 9 = 17][frac{2}{3}R = 8]To solve for ( R ), multiply both sides by the reciprocal of ( frac{2}{3} ), which is ( frac{3}{2} ):[R = 8 times frac{3}{2} = 12]Therefore, Bill gathered 12 red mushrooms.