answer:We want to find the values of lambda such that there exists an x in 0,1 for which f(x) neq x but f(f(x)) = x. We are given that f(x) = lambda x(1-x). First, let's find the fixed points of f(x), i.e., the values of x for which f(x) = x: lambda x(1-x) = x implies lambda x - lambda x^2 = x implies lambda x - x = lambda x^2 implies x(lambda - 1) = lambda x^2 implies x = 0, frac{lambda-1}{lambda} qquad (lambda neq 0) Thus, the fixed points of f(x) are 0 and (lambda-1)/lambda. Next, we need to determine for which values of lambda do we have f(f(x)) = x when f(x) neq x. Let's compute f(f(x)) first: begin{align*} f(f(x)) &= lambda (lambda x(1-x)) (1 - lambda x(1-x)) &= lambda^2 x (1-x) (1 - lambda(1-2x+lambda x^2)) &= lambda^2 x (1-x) (1 - lambda + 2lambda x - lambda^2 x^2) &= lambda^2 x (1 - lambda - lambda x + 2lambda x - lambda^2 x^2 + lambda^2 x^2 - lambda^3 x^3) &= lambda^2 x (1 - lambda + 2lambda x - lambda^2 x^2 - lambda^3 x^3) end{align*} Now, we want to solve the equation f(f(x)) = x subject to the constraint f(x) neq x, or equivalently, x notin {0, (lambda-1)/lambda}. If x neq 0, then we can divide both sides of the equation f(f(x)) = x by x to obtain: lambda^2 (1 - lambda + 2lambda x - lambda^2 x^2 - lambda^3 x^3) = 1 Expanding this equation, we get begin{align*} lambda^2 - lambda^3 + 2lambda^3 x - lambda^4 x^2 - lambda^5 x^3 &= 1 lambda^2(1-lambda + 2lambda x - lambda^2x^2 - lambda^3 x^3) - 1 &= 0 lambda^2(lambda-1)(2x-1)-1 &= 0 end{align*} If lambda = 1, then f(x) = x, so we must have lambda neq 1. Then, we can rearrange the equation to obtain x = frac{1}{2} + frac{1}{2lambda(lambda-1)} This expression for x must lie in the interval (0,1), so we need to consider the sign of the denominator in the above expression and check the boundary cases. Note that the denominator is positive if 0 < lambda < 1 or lambda > 2. Thus, we have three cases: 1. lambda in (0,1): In this case, the denominator is positive, and the numerator is always less than 1, so x lies in (0,1). 2. lambda in (1,2): In this case, the denominator is still positive, but the numerator is now greater than 1, so this interval of lambda should be discarded. 3. lambda > 2: In this case, the denominator is positive, and the numerator is always less than 1, so x lies in (0,1). Therefore, the values of lambda for which there exists an x in 0,1 such that f(x) neq x but f(f(x)) = x are 0 < lambda < 1 and lambda > 2.,Let's first find the fixed points of the function f: Those are the values x for which f(x) = x. So we need to solve the equation: x = lambda x(1-x) There are three fixed points for this function: 1. x = 0 2. x = 1 3. x = frac{1}{2} if lambda = 2 Now, we'll use the fact that f(f(x)) = x holds. Note that f(x) can achieve any value within the interval 0, lambda. Case 1: x in 0, frac{1}{2}. So, f(x) in 0, lambda frac{1}{2}. Now suppose that f(x) = y, then y = lambda y (1-y) implies y(1-frac{lambda}{2}) = 0. If y = 0, then f(x) = x, this doesn't help us. But if y neq 0, then we have (1-frac{lambda}{2}) = 1 which gives us lambda = 2. Case 2: x in frac{1}{2},1 f(x) in frac{lambda}{2}, lambda. Similarly, we have (1-frac{lambda}{2}) = 0, this is only possible if lambda = 2. If lambda = 2, then there will be an x in frac{1}{2},1 such that f(x) = frac{1}{2} neq x, but f(f(x)) = x. However, note that this happens because frac{1}{2} is a fixed point of f. Therefore, the value lambda = 2 is the only one that satisfies the given condition within the interval 0,4. However, since there is no restriction on the an x such that f(x) neq,We have that f(f(x)) = f(lambda x(1 - x)) = lambda cdot lambda x(1 - x) (1 - lambda x(1 - x)),so we want to solve lambda cdot lambda x(1 - x) (1 - lambda x(1 - x)) = x. Note that if f(x) = x, then f(f(x)) = f(x) = x, so any roots of lambda x(1 - x) = x will also be roots of lambda cdot lambda x(1 - x) (1 - lambda x(1 - x)) = x. Thus, we should expect lambda x(1 - x) - x to be a factor of lambda cdot lambda x(1 - x) (1 - lambda x(1 - x)) - x. Indeed, lambda cdot lambda x(1 - x) (1 - lambda x(1 - x)) - x = (lambda x(1 - x) - x)(lambda^2 x^2 - (lambda^2 + lambda) x + lambda + 1).The discriminant of lambda^2 x^2 - (lambda^2 + lambda) x + lambda + 1 is (lambda^2 + lambda)^2 - 4 lambda^2 (lambda + 1) = lambda^4 - 2 lambda^3 - 3 lambda^2 = lambda^2 (lambda + 1)(lambda - 3).This is nonnegative when lambda = 0 or 3 le lambda le 4. If lambda = 0, then f(x) = 0 for all x in 0,1. If lambda = 3, then the equation f(f(x)) = x becomes (3x(1 - x) - x)(9x^2 - 12x + 4) = 0.The roots of 9x^2 - 12x + 4 = 0 are both frac{2}{3}, which satisfy f(x) = x. On the other hand, for lambda > 3, the roots of lambda x(1 - x) = x are x = 0 and x = frac{lambda - 1}{lambda}. Clearly x = 0 is not a root of lambda^2 x^2 - (lambda^2 + lambda) x + lambda + 1 = 0. Also, if x = frac{lambda - 1}{lambda}, then lambda^2 x^2 - (lambda^2 + lambda) x + lambda + 1 = lambda^2 left( frac{lambda - 1}{lambda} right)^2 - (lambda^2 + lambda) cdot frac{lambda - 1}{lambda} + lambda + 1 = 3 - lambda neq 0.Furthermore, the product of the roots is frac{lambda + 1}{lambda^2}, which is positive, so either both roots are positive or both roots are negative. Since the sum of the roots is frac{lambda^2 + lambda}{lambda^2} > 0, both roots are positive. Also, frac{lambda^2 + lambda}{lambda} = 1 + frac{1}{lambda} < frac{4}{3},so at least one root must be less than 1. Therefore, the set of lambda that satisfy the given condition is lambda in (3,4.

answer:We can write the expression as begin{align*} 2x^2 + 2xy + 4y + 5y^2 - x &= (x^2 + 2xy + y^2) + left( x^2 - x + frac{1}{4} right) + (4y^2 + 4y + 1) - frac{1}{4} - 1 &= (x + y)^2 + left( x - frac{1}{2} right)^2 + (2y + 1)^2 - frac{5}{4}. end{align*}We see that the minimum value is -frac{5}{4}}, which occurs at x = frac{1}{2} and y = -frac{1}{2.,To find the minimum value of the given expression, we can complete the square for both x and y terms. Let's start by moving the constant term to the other side of the equation: -x - 5y = -left(2x^2+2xy+5y^2+5y-xright) Now, let's separate the terms with x and y: begin{aligned} -x &= -left(2x^2+2xy-xright) -5y &= -left(5y^2+2xy-5yright) end{aligned} Next, we'll complete the square for both x and y terms: For x-terms: -x = -left((x+1)^2-1right) = (x+1)^2-1-x-1 For y-terms: -5y = -left((y+5/2)^2-frac{25}{4}right) = (y+5/2)^2-frac{25}{4}+10y Now, let's add and subtract the x^2 and y^2 terms from both sides: begin{aligned} -x &= -(x+1)^2+1-(x-1)^2 -5y &= -(y+5/2)^2+frac{25}{4}-(y-5/2)^2 end{aligned} Now, we can rewrite the original equation as: left((x^2+x+1right))^2+left((y^2+y+5/2)^2(frac{25}{4}+10yright) In order to find the minimum value of the expression, we need to find the minimum values of (x^2+x+1) and (y^2+y+5/2). For (x^2+x+1), we can complete the square again: (x^2+x+1) = (x+1/2)^2-frac{1}{4} , To find the minimum value of the given expression, we can complete the square for both x and y terms. First, let's rewrite the expression as follows: 2x^2 + 2xy + frac{1}{2}y^2 + frac{9}{2}y^2 - x. Now, complete the square for the x and frac{1}{2}y terms: 2left(x^2 + xcdot y + frac{1}{4}y^2right) + frac{9}{2}y^2 - frac{1}{2}y^2 - x. Notice that the term inside the first parentheses is a perfect square trinomial: 2left(left(x + frac{1}{2}yright)^2 - frac{1}{4}y^2right) + frac{4}{2}y^2 - x. Now, distribute the 2: 2(x + frac{1}{2}y)^2 - y^2 - x + 2y^2. Combine like terms: 2(x + frac{1}{2}y)^2 + 2y^2 - x. Since x and y are real numbers, the minimum value of (x + frac{1}{2}y)^2 is 0, which occurs when x = -frac{1}{2}y. Plugging this back in, we get 2(0) + 2y^2 - left(-frac{1}{2}yright) = boxed{frac{5}{4}y^2}. Since y is a real number, the minimum value of frac{5}{4}y^2 is 0, which occurs when y = 0. Thus, the minimum value of the given expression is achieved when x = -frac{1}{2}y = 0: boxed{0}.

answer:To find the maximum value of the given expression, we use the Cauchy-Schwarz inequality. For any nonnegative real numbers a, b, c, and d, we have the inequality: (a^2 + b^2 + c^2 + d^2) cdot (0 + 0 + 0 + 1) le (a + b + c + d)^2 Substituting the given condition a + b + c + d = 1, we get: (a^2 + b^2 + c^2 + d^2) cdot 1 le 1^2 Now, we divide both sides by 1, which gives us: a^2 + b^2 + c^2 + d^2 le 1 Since the inequality holds with equality when a = b = c = d = 1/4, the maximum value of the expression is: a^2 + b^2 + c^2 + d^2 = 1, We have a^2 + b^2 + c^2 + d^2 = (a+b+c+d)^2 - 2(ab+ac+ad+bc+bd+cd) by expanding and using the fact that a+b+c+d=1. We also know that ab+ac+ad+bc+bd+cd ge frac{9}{4}(a+b+c+d)^2 = frac{9}{4}, which is a result of AM-GM Inequality. Now plug back in: begin{align*} a^2 + b^2 + c^2 + d^2 &=1 - 2(ab+ac+ad+bc+bd+cd) &le 1- 2 cdot frac{9}{4} = boxed{frac{1}{4}}. end{align*}Ok, we got frac{1}{4} and want to show that there is a way to get this value, let's take frac{1}{2} for a,b,c,d, they are nonnegative and sum up to 1, and clearly: a^2 + b^2 + c^2 + d^2 = left(frac{1}{2}right)^2 + left(frac{1}{2}right)^2 + left(frac{1}{2}right)^2 + left(frac{1}{2}right)^2 = boxed{frac{1}{4}}.So we are done. square,I notice that this problem involves finding the maximum value of a quadratic expression subject to a linear constraint. This reminds me of a technique called Lagrange multipliers, which can be used to find the extrema of a function subject to a constraint. To apply this technique, I need to define a function f(a,b,c,d) that I want to maximize, and a function g(a,b,c,d) that represents the constraint. In this case, I can let f(a,b,c,d) = a^2 + b^2 + c^2 + d^2 and g(a,b,c,d) = a + b + c + d - 1. Then, I need to find the values of a, b, c, d, and a constant lambda that satisfy the system of equations nabla f = lambda nabla g, quad g = 0. The gradient of f is the vector of partial derivatives (2a, 2b, 2c, 2d), and the gradient of g is the constant vector (1,1,1,1). So, the system of equations becomes begin{cases} 2a = lambda 2b = lambda 2c = lambda 2d = lambda a + b + c + d = 1. end{cases} I notice that this system implies that a = b = c = d = lambda/2. Substituting this into the last equation, I get lambda = 1/2. Therefore, the only possible critical point is (a,b,c,d) = (1/4, 1/4, 1/4, 1/4). To check if this is a maximum or a minimum, I can use the second derivative test. The Hessian matrix of f is the matrix of second partial derivatives, which in this case is just 2I, where I is the identity matrix. The Hessian matrix is positive definite, which means that f is convex and the critical point is a minimum. Therefore, the minimum value of f is f(1/4, 1/4, 1/4, 1/4) = 4 cdot (1/4)^2 = 1/4. Since f is bounded above by 1, the maximum value of f must be 1, and it is attained when one of the variables is 1 and the rest are 0.

answer:To find the value of log_{9} 4, we can use the following property of logarithms: log_b(a)=frac{log_c(a)}{log_c(b)} Let's replace the variables in the property with the given values: log_9(4)=frac{log_3(4)}{log_3(9)} Now we have to calculate each individual logarithm: 1. log_3(4): Since log_3(6) approx 1.63, we can use the properties of logarithms to find log_3(4). We know that 4 < 6 and the logarithm function is increasing, so log_3(4) < log_3(6) approx 1.63. Using a calculator, we find that log_3(4) approx 1.6094. 2. log_3(9): Using the same reasoning as in the previous step, we know that log_3(9) > log_3(6) approx 1.63. Using a calculator, we find that log_3(9) approx 2. Now, we can put the values back into the equation: log_9(4)=frac{1.6094}{2} approx 0.8047 So, the value of log_{9} 4, accurate to two decimal places, is approximately 0.80.,I know that log_{9} 4 is the same as frac{log_{3} 4}{log_{3} 9}, by the change of base formula. To find log_{3} 4, I can use the fact that 4 = 2^2, so log_{3} 4 = 2 log_{3} 2. I also know that log_{3} 9 = 2, since 9 = 3^2. Therefore, log_{9} 4 = frac{2 log_{3} 2}{2} = log_{3} 2. Now, how can I find log_{3} 2? I notice that 6 = 2 cdot 3, so log_{3} 6 = log_{3} 2 + log_{3} 3. Since log_{3} 3 = 1, I can solve for log_{3} 2 by subtracting 1 from both sides: log_{3} 2 = log_{3} 6 - 1. The problem gives me an approximate value for log_{3} 6, which is 1.63. Therefore, log_{3} 2 approx 1.63 - 1 = 0.63. This is the value of log_{9} 4, accurate to two decimal places., We can rewrite log_{9} 4 as dfrac{log_{3} 4}{log_{3} 9}. Now we need to calculate the values of log_{3} 4 and log_{3} 9. log_{3} 4 = log_{3} (2^2) = 2log_{3} 2. Since log_{3} 6 approx 1.63, we have log_{3} 2 approx 1.63/2 approx 0.815. Therefore, log_{3} 4 approx 0.815 times 2 approx 1.63. log_{3} 9 = log_{3} (3^2) = 2log_{3} 3 = 2. So, dfrac{log_{3} 4}{log_{3} 9} = dfrac{1.63}{2} approx 0.815. However, this result is not accurate enough since we need the answer up to two decimal places. We can use the original formula and calculate log_{3} 4 and log_{3} 9 more precisely using a calculator: log_{3} 4 approx 1.2619 log_{3} 9 = log_{3} (3^2) = 2log_{3} 3 = 2 Therefore, log_{9} 4 = dfrac{log_{3} 4}{log_{3} 9} approx dfrac{1.2619}{2} approx 0.63095. Thus, the value of log_{9} 4, accurate to two decimal places, is approximately 0.63.