answer：Okay, so I need to prove that the composition of two homotheties is either a homothety or a parallel translation using complex numbers. Hmm, homotheties... I remember they are transformations that scale points about a center. So, if I have two homotheties, one after another, their composition should either be another homothety or a translation. First, let me recall what a homothety is in complex numbers. A homothety with center ( A ) and ratio ( k ) can be represented as ( H_A^k(z) = k(z - A) + A ). So, it's like scaling the point ( z ) by ( k ) relative to the center ( A ), and then translating back by ( A ).Alright, so if I have two homotheties, ( H_{A_1}^{k_1} ) and ( H_{A_2}^{k_2} ), their composition would be ( H_{A_2}^{k_2} circ H_{A_1}^{k_1} ). That means I first apply ( H_{A_1}^{k_1} ) to a point ( z ), and then apply ( H_{A_2}^{k_2} ) to the result.Let me write down the expressions step by step. First, applying ( H_{A_1}^{k_1} ) to ( z ):[ H_{A_1}^{k_1}(z) = k_1(z - A_1) + A_1 ]Then, applying ( H_{A_2}^{k_2} ) to this result:[ H_{A_2}^{k_2}(H_{A_1}^{k_1}(z)) = k_2 left( k_1(z - A_1) + A_1 - A_2 right) + A_2 ]Let me simplify this expression step by step. First, distribute ( k_2 ) inside:[ k_2 cdot k_1(z - A_1) + k_2(A_1 - A_2) + A_2 ]Which is:[ k_1 k_2(z - A_1) + k_2 A_1 - k_2 A_2 + A_2 ]Now, let's expand ( k_1 k_2(z - A_1) ):[ k_1 k_2 z - k_1 k_2 A_1 + k_2 A_1 - k_2 A_2 + A_2 ]Combine like terms. The terms involving ( A_1 ) are ( -k_1 k_2 A_1 + k_2 A_1 ), which can be factored as:[ k_2 A_1 (1 - k_1) ]Similarly, the terms involving ( A_2 ) are ( -k_2 A_2 + A_2 ), which can be factored as:[ A_2 (1 - k_2) ]So, putting it all together, the composition is:[ k_1 k_2 z + k_2 (1 - k_1) A_1 + (1 - k_2) A_2 ]Now, this expression is of the form ( w = k z + c ), where ( k = k_1 k_2 ) and ( c = k_2 (1 - k_1) A_1 + (1 - k_2) A_2 ). In complex analysis, a transformation of the form ( w = k z + c ) is either a homothety (if ( k neq 1 )) or a translation (if ( k = 1 )). So, let's analyze the two cases:1. **Case 1: ( k_1 k_2 = 1 )** If ( k_1 k_2 = 1 ), then the scaling factor ( k = 1 ), so the transformation becomes: [ w = z + c ] where ( c = k_2 (1 - k_1) A_1 + (1 - k_2) A_2 ). Since ( k_1 k_2 = 1 ), we can express ( k_2 = frac{1}{k_1} ). Substituting this into the expression for ( c ): [ c = frac{1}{k_1} (1 - k_1) A_1 + left(1 - frac{1}{k_1}right) A_2 ] Simplifying: [ c = left( frac{1}{k_1} - 1 right) A_1 + left( 1 - frac{1}{k_1} right) A_2 ] Factor out ( left( frac{1}{k_1} - 1 right) ): [ c = left( frac{1}{k_1} - 1 right) (A_1 - A_2) ] This shows that ( c ) is a scalar multiple of ( A_1 - A_2 ), meaning the translation vector ( c ) is parallel to the line ( A_1 A_2 ). Therefore, the composition is a translation ( T_a ) where ( a ) is parallel to ( A_1 A_2 ).2. **Case 2: ( k_1 k_2 neq 1 )** If ( k_1 k_2 neq 1 ), then the transformation ( w = k z + c ) is a homothety with ratio ( k = k_1 k_2 ). To find the center ( A ) of this homothety, we set up the equation: [ k z + c = k(z - A) + A ] Expanding the right-hand side: [ k z - k A + A ] Equate this to the left-hand side: [ k z + c = k z - k A + A ] Subtract ( k z ) from both sides: [ c = -k A + A ] Factor out ( A ): [ c = A (1 - k) ] Solving for ( A ): [ A = frac{c}{1 - k} ] Substitute ( c = k_2 (1 - k_1) A_1 + (1 - k_2) A_2 ) and ( k = k_1 k_2 ): [ A = frac{k_2 (1 - k_1) A_1 + (1 - k_2) A_2}{1 - k_1 k_2} ] This shows that the center ( A ) lies on the line ( A_1 A_2 ) because it is a weighted average of ( A_1 ) and ( A_2 ). Therefore, the composition is a homothety ( H_A^k ) with center ( A ) on ( A_1 A_2 ) and ratio ( k = k_1 k_2 ).So, putting it all together, the composition of two homotheties is either a translation (if ( k_1 k_2 = 1 )) or another homothety (if ( k_1 k_2 neq 1 )), with the center lying on the line through the original centers and the ratio being the product of the original ratios.**Final Answer**The composition of two homotheties is either a translation or another homothety, specifically:[H_{A_{2}}^{k_{2}} circ H_{A_{1}}^{k_{1}} = begin{cases}T_{a}, & text{if } k_{1}k_{2} = 1, H_{A}^{k}, & text{if } k_{1}k_{2} neq 1,end{cases}]where ( a ) is parallel to ( A_1A_2 ) and ( A ) lies on ( A_1A_2 ) with ( k = k_1k_2 ). Thus, the final result is boxed{H_{A_{2}}^{k_{2}} circ H_{A_{1}}^{k_{1}} = begin{cases} T_{a}, & k_{1}k_{2} = 1, H_{A}^{k}, & k_{1}k_{2} neq 1. end{cases}}.

answer：Okay, so I have this function f(x) = a ln x - x + 2, where a is a real number. There are three parts to this problem. Let me try to tackle each part step by step.Starting with part (I): Find the monotonic intervals of the function f(x).Hmm, monotonic intervals mean where the function is increasing or decreasing. To find that, I remember I need to take the derivative of f(x) and analyze its sign.So, f(x) = a ln x - x + 2.The derivative f’(x) would be the derivative of a ln x, which is a/x, minus the derivative of x, which is 1, and the derivative of 2 is 0. So, f’(x) = a/x - 1.To find where the function is increasing or decreasing, I need to find where f’(x) is positive or negative.Let me set f’(x) equal to zero to find critical points:a/x - 1 = 0Multiply both sides by x: a - x = 0 => x = a.So, the critical point is at x = a.Now, I need to consider the sign of f’(x) around x = a.Case 1: If a ≤ 0.Wait, a is a real number, but ln x is only defined for x > 0, so the domain of f(x) is x > 0. So, a can be any real number, but x is always positive.If a is less than or equal to 0, then a/x is non-positive because a is non-positive and x is positive. So, f’(x) = a/x - 1. Since a/x ≤ 0, then f’(x) = (something ≤ 0) - 1, which is ≤ -1, which is always negative. So, f(x) is decreasing on (0, ∞) when a ≤ 0.Case 2: If a > 0.Then, x = a is a critical point. Let's analyze the intervals around x = a.For x < a: Let's pick a number less than a, say x = a/2.f’(a/2) = a/(a/2) - 1 = 2 - 1 = 1 > 0. So, f’(x) is positive when x < a.For x > a: Let's pick x = 2a.f’(2a) = a/(2a) - 1 = 1/2 - 1 = -1/2 < 0. So, f’(x) is negative when x > a.Therefore, when a > 0, f(x) is increasing on (0, a) and decreasing on (a, ∞).So, summarizing part (I):- If a ≤ 0, f(x) is decreasing on (0, ∞).- If a > 0, f(x) is increasing on (0, a) and decreasing on (a, ∞).Alright, that seems solid. Let me move on to part (II).Part (II): Prove that when a = 3, f(x) ≤ x² always holds.So, f(x) = 3 ln x - x + 2. We need to show that 3 ln x - x + 2 ≤ x² for all x > 0.Let me define a new function g(x) = f(x) - x² = 3 ln x - x + 2 - x².If I can show that g(x) ≤ 0 for all x > 0, then f(x) ≤ x² holds.To analyze g(x), let's find its maximum value. If the maximum of g(x) is ≤ 0, then we're done.First, compute the derivative g’(x):g’(x) = derivative of 3 ln x is 3/x, derivative of -x is -1, derivative of 2 is 0, derivative of -x² is -2x.So, g’(x) = 3/x - 1 - 2x.Let me write that as g’(x) = (3/x) - 1 - 2x.To find critical points, set g’(x) = 0:3/x - 1 - 2x = 0Multiply both sides by x to eliminate the denominator:3 - x - 2x² = 0Rearranged: 2x² + x - 3 = 0Wait, let me check that:3 - x - 2x² = 0Multiply both sides by -1: 2x² + x - 3 = 0Wait, is that correct? Let me double-check:Original equation after multiplying by x: 3 - x - 2x² = 0So, 2x² + x - 3 = 0 is incorrect because 3 - x - 2x² = 0 is equivalent to -2x² - x + 3 = 0, which is 2x² + x - 3 = 0 when multiplied by -1.So, 2x² + x - 3 = 0.Let me solve this quadratic equation.Using quadratic formula: x = [-b ± sqrt(b² - 4ac)] / (2a)Here, a = 2, b = 1, c = -3.Discriminant D = 1² - 4*2*(-3) = 1 + 24 = 25.So, sqrt(D) = 5.Thus, x = [-1 ± 5]/4.So, two solutions:x = (-1 + 5)/4 = 4/4 = 1x = (-1 - 5)/4 = -6/4 = -1.5But x must be positive, so x = 1 is the critical point.So, the only critical point is at x = 1.Now, let's analyze the behavior of g’(x) around x = 1.For x < 1, say x = 0.5:g’(0.5) = 3/(0.5) - 1 - 2*(0.5) = 6 - 1 - 1 = 4 > 0So, g’(x) is positive when x < 1.For x > 1, say x = 2:g’(2) = 3/2 - 1 - 4 = 1.5 - 1 - 4 = -3.5 < 0So, g’(x) is negative when x > 1.Therefore, g(x) is increasing on (0, 1) and decreasing on (1, ∞). So, the maximum of g(x) occurs at x = 1.Compute g(1):g(1) = 3 ln 1 - 1 + 2 - (1)^2 = 0 - 1 + 2 - 1 = 0.So, the maximum value of g(x) is 0 at x = 1.Therefore, g(x) ≤ 0 for all x > 0, which implies f(x) ≤ x² for all x > 0 when a = 3.That proves part (II). Good.Now, part (III): If for any x₁ ∈ [1, e], there always exists x₂ ∈ [1, e], such that f(x₁) + f(x₂) = 4, find the value of the real number a.Hmm, this seems a bit more involved. Let me try to parse this.So, for any x₁ in [1, e], there exists an x₂ in [1, e] such that f(x₁) + f(x₂) = 4.So, for every x₁, there is some x₂ (which may depend on x₁) such that their f values add up to 4.I need to find the value of a such that this condition holds.Let me think about what this implies.First, let's note that [1, e] is a closed interval, so f(x) is continuous on [1, e], so it attains its maximum and minimum there.Given that, for any x₁, f(x₁) can be paired with some f(x₂) such that their sum is 4.This suggests that the function f(x) is symmetric in some way around 2, or that the function's range is such that for every value y in the range, 4 - y is also in the range.Wait, that might be a key insight.If for every y = f(x₁), there exists an x₂ such that f(x₂) = 4 - y, then the range of f must be symmetric around 2, because if y is in the range, then 4 - y is also in the range.So, the range of f on [1, e] must be symmetric around 2.Alternatively, the function f(x) must satisfy that for every x in [1, e], there exists another x' in [1, e] such that f(x) + f(x') = 4.This is similar to saying that the function is symmetric with respect to the point (c, 2) for some c, but I'm not sure.Alternatively, perhaps the function is such that f(x) + f(k/x) = 4 or something, but I don't know.Alternatively, maybe the function is linear or something.Wait, but f(x) is a logarithmic function, so it's not linear.Alternatively, perhaps the function is such that f(x) + f(e/x) = 4, but I need to check.Wait, let me test this idea.Suppose that f(x) + f(e/x) = 4.Let me compute f(x) + f(e/x):f(x) = a ln x - x + 2f(e/x) = a ln(e/x) - (e/x) + 2 = a (1 - ln x) - e/x + 2So, f(x) + f(e/x) = a ln x - x + 2 + a (1 - ln x) - e/x + 2Simplify:a ln x - x + 2 + a - a ln x - e/x + 2The a ln x and -a ln x cancel out.So, we have (-x + a - e/x + 4)So, f(x) + f(e/x) = -x + a - e/x + 4We want this to equal 4 for all x.So, set -x + a - e/x + 4 = 4Simplify: -x + a - e/x = 0So, a = x + e/xBut a is a constant, so x + e/x must be equal to a for all x in [1, e].But x + e/x is a function that varies with x.Compute x + e/x on [1, e].At x = 1: 1 + e/1 = 1 + eAt x = e: e + e/e = e + 1So, x + e/x is equal to 1 + e at both ends, but in between, it's different.Wait, let me compute its derivative to see its behavior.Let h(x) = x + e/xh’(x) = 1 - e/x²Set h’(x) = 0: 1 - e/x² = 0 => x² = e => x = sqrt(e)So, h(x) has a critical point at x = sqrt(e).Compute h(sqrt(e)) = sqrt(e) + e / sqrt(e) = sqrt(e) + sqrt(e) = 2 sqrt(e)Which is approximately 2 * 1.6487 ≈ 3.2974Compare to h(1) = 1 + e ≈ 1 + 2.718 ≈ 3.718h(e) = e + 1 ≈ 2.718 + 1 ≈ 3.718So, h(x) has a minimum at x = sqrt(e) of approximately 3.2974, and maximum at the endpoints of approximately 3.718.So, h(x) ranges from about 3.2974 to 3.718 on [1, e].Therefore, if a = x + e/x, a would have to vary between approximately 3.2974 and 3.718, but a is a constant. So, this approach might not work.Alternatively, maybe my initial assumption is wrong.Perhaps instead of f(x) + f(e/x) = 4, it's something else.Wait, maybe f(x) is symmetric around x = something, but I don't know.Alternatively, perhaps the function f(x) is such that its minimum plus its maximum equals 4, so that for any x₁, you can find an x₂ such that f(x₁) + f(x₂) = 4.Wait, that might be the case.Let me think: If the function f(x) on [1, e] has a minimum m and a maximum M, then for any y in [m, M], there exists z in [m, M] such that y + z = 4.This would require that the range of f(x) is symmetric around 2, so that M = 4 - m.So, if M + m = 4, then for any y, z = 4 - y is also in the range.Therefore, the condition is that the maximum of f(x) plus the minimum of f(x) equals 4.So, M + m = 4.So, if I can compute M and m in terms of a, set M + m = 4, and solve for a.That seems promising.So, let me try that.First, find the maximum and minimum of f(x) on [1, e].From part (I), we know that f(x) is increasing on (0, a) and decreasing on (a, ∞) when a > 0.So, on [1, e], depending on where a is, the maximum could be at x = a if a is inside [1, e], or at the endpoints.So, let's consider different cases for a.Case 1: a ≤ 1.Then, on [1, e], f(x) is decreasing because a ≤ 1, so x ≥ a, and f’(x) = a/x - 1 ≤ 1/x - 1. At x = 1, f’(1) = a - 1 ≤ 0. So, f(x) is decreasing on [1, e].Therefore, maximum at x = 1: f(1) = a ln 1 - 1 + 2 = 0 - 1 + 2 = 1.Minimum at x = e: f(e) = a ln e - e + 2 = a*1 - e + 2 = a - e + 2.So, M = 1, m = a - e + 2.Then, M + m = 1 + (a - e + 2) = a - e + 3.Set this equal to 4: a - e + 3 = 4 => a = e + 1.But wait, in this case, a ≤ 1, but e + 1 ≈ 3.718, which is greater than 1. So, this is a contradiction. Therefore, no solution in this case.Case 2: 1 < a < e.Then, on [1, e], f(x) is increasing on [1, a] and decreasing on [a, e]. So, the maximum is at x = a: f(a) = a ln a - a + 2.The minimum is at one of the endpoints. Let's compute f(1) and f(e):f(1) = 1, as before.f(e) = a - e + 2.So, which is smaller, f(1) = 1 or f(e) = a - e + 2?Compute f(e) - f(1) = (a - e + 2) - 1 = a - e + 1.Since 1 < a < e, a - e + 1 < e - e + 1 = 1.But a - e + 1 is greater than 1 - e + 1 = 2 - e ≈ 2 - 2.718 ≈ -0.718.So, f(e) - f(1) can be negative or positive depending on a.Wait, let's solve when f(e) = f(1):a - e + 2 = 1 => a = e - 1 ≈ 1.718.So, if a < e - 1, then f(e) < f(1). If a > e - 1, then f(e) > f(1).But since 1 < a < e, and e - 1 ≈ 1.718, so:- If 1 < a < e - 1, then f(e) < f(1), so minimum is f(e).- If e - 1 < a < e, then f(e) > f(1), so minimum is f(1).Wait, let me verify:If a = 2, which is greater than e - 1 ≈ 1.718, then f(e) = 2 - e + 2 ≈ 4 - 2.718 ≈ 1.282, which is greater than f(1) = 1.If a = 1.5, which is less than e - 1, then f(e) = 1.5 - e + 2 ≈ 3.5 - 2.718 ≈ 0.782, which is less than f(1) = 1.So, yes, that seems correct.Therefore, in this case:- If 1 < a < e - 1, then m = f(e) = a - e + 2, M = f(a) = a ln a - a + 2.- If e - 1 < a < e, then m = f(1) = 1, M = f(a) = a ln a - a + 2.So, let's compute M + m in both subcases.Subcase 2a: 1 < a < e - 1.M + m = [a ln a - a + 2] + [a - e + 2] = a ln a - a + 2 + a - e + 2 = a ln a + 4 - e.Set this equal to 4: a ln a + 4 - e = 4 => a ln a = e.So, a ln a = e.We need to solve for a in (1, e - 1).But let's see: a ln a = e.Let me consider the function h(a) = a ln a.At a = e: h(e) = e * 1 = e.But in this subcase, a < e - 1 ≈ 1.718, which is less than e.Compute h(1) = 1 * 0 = 0.h(e - 1) = (e - 1) ln(e - 1) ≈ (2.718 - 1) * ln(1.718) ≈ 1.718 * 0.542 ≈ 0.931.So, h(a) increases from 0 to approximately 0.931 as a goes from 1 to e - 1.But we need h(a) = e ≈ 2.718, which is way higher. So, no solution in this subcase.Subcase 2b: e - 1 < a < e.M + m = [a ln a - a + 2] + 1 = a ln a - a + 3.Set this equal to 4: a ln a - a + 3 = 4 => a ln a - a = 1 => a (ln a - 1) = 1.So, we have a (ln a - 1) = 1.We need to solve for a in (e - 1, e).Let me define k(a) = a (ln a - 1).Compute k(e - 1):k(e - 1) = (e - 1)(ln(e - 1) - 1) ≈ (1.718)(0.542 - 1) ≈ 1.718*(-0.458) ≈ -0.786.k(e) = e (1 - 1) = 0.So, k(a) goes from approximately -0.786 to 0 as a goes from e - 1 to e.But we need k(a) = 1, which is positive. Since k(a) is negative in this interval, there is no solution here either.Therefore, in case 2, no solution.Case 3: a ≥ e.Then, on [1, e], since a ≥ e, and from part (I), f(x) is increasing on (0, a). Since a ≥ e, f(x) is increasing on [1, e].Therefore, maximum at x = e: f(e) = a - e + 2.Minimum at x = 1: f(1) = 1.So, M = a - e + 2, m = 1.Thus, M + m = (a - e + 2) + 1 = a - e + 3.Set this equal to 4: a - e + 3 = 4 => a = e + 1.Now, check if a = e + 1 satisfies a ≥ e, which it does, since e + 1 ≈ 3.718 > e ≈ 2.718.Therefore, a = e + 1 is a solution.But let me verify if this works.So, if a = e + 1, then f(x) = (e + 1) ln x - x + 2.On [1, e], since a = e + 1 > e, f(x) is increasing on [1, e].Therefore, f(1) = (e + 1)*0 - 1 + 2 = 1.f(e) = (e + 1)*1 - e + 2 = e + 1 - e + 2 = 3.So, the range of f(x) on [1, e] is [1, 3].Now, for any x₁ ∈ [1, e], f(x₁) ∈ [1, 3].We need to find x₂ ∈ [1, e] such that f(x₁) + f(x₂) = 4.So, for any y ∈ [1, 3], we need to find z ∈ [1, 3] such that y + z = 4.Which means z = 4 - y.Since y ∈ [1, 3], z = 4 - y ∈ [1, 3].Therefore, for any y, z is also in [1, 3], so such x₂ exists because f is continuous and surjective onto [1, 3].Wait, but f is increasing, so it's bijective from [1, e] to [1, 3].Therefore, for any y ∈ [1, 3], there exists a unique x ∈ [1, e] such that f(x) = y.Therefore, for any x₁, f(x₁) = y, then x₂ is the unique x such that f(x) = 4 - y.Since f is increasing, this x₂ exists in [1, e].Therefore, a = e + 1 satisfies the condition.Is this the only solution?From the previous cases, in case 1 and 2, we didn't find any solutions, only in case 3, we got a = e + 1.Therefore, the value of a is e + 1.So, the answer is a = e + 1.**Final Answer**The value of the real number ( a ) is (boxed{e + 1}).

answer：First, determine the amount of meat required per taco. Carmen uses 4 pounds of meat for 10 tacos, so each taco requires 4/10 = 0.4 pounds of meat.Next, calculate the total amount of meat needed for 35 tacos by multiplying the meat per taco by the number of tacos: 0.4 pounds/taco * 35 tacos = 14 pounds.Therefore, Carmen needs 14 pounds of meat to make 35 tacos.

answer：First, I need to determine the total distance between the markings on the number line. If there are seven equally spaced markings from 0 to 35, I can calculate the distance between each marking by dividing the total distance by the number of intervals. Next, I'll use the calculated distance per interval to find the value of ( z ). If ( z ) is four intervals away from a starting point, I'll multiply the distance per interval by four and add it to the starting value to find ( z ).Finally, I'll verify the calculation to ensure accuracy and select the correct answer from the given options.

answer：To find the y-intercept of the line ( x - 2y - 3 = 0 ), I need to determine the value of ( y ) when ( x = 0 ).First, substitute ( x = 0 ) into the equation:[0 - 2y - 3 = 0]Simplify the equation:[-2y - 3 = 0]Add 3 to both sides:[-2y = 3]Divide both sides by -2:[y = -frac{3}{2}]Therefore, the y-intercept of the line is (-frac{3}{2}).

answer：Okay, so I have this inequality to solve: (2^{2x - 7} < 2^{x - 3}). Hmm, I remember that when you have exponents with the same base, you can compare the exponents directly if the base is greater than 1. Since 2 is greater than 1, that should work here. Alright, so if (2^{2x - 7}) is less than (2^{x - 3}), then the exponent on the left side must be less than the exponent on the right side. That means (2x - 7 < x - 3). Okay, that makes sense. Now I just need to solve this linear inequality for x.Let me subtract x from both sides to get the x terms on one side. So, (2x - x - 7 < -3), which simplifies to (x - 7 < -3). Now, I need to isolate x, so I'll add 7 to both sides. That gives me (x < -3 + 7), which simplifies to (x < 4).Wait, let me double-check that. Starting with (2x - 7 < x - 3), subtracting x from both sides gives (x - 7 < -3), and then adding 7 to both sides gives (x < 4). Yeah, that seems right. So, the solution is all real numbers less than 4.I think I got it. Since the base is the same and greater than 1, comparing the exponents directly was the key step. Then solving the resulting linear inequality was straightforward. I don't think I made any mistakes here, but maybe I should plug in a number less than 4 to verify.Let's try x = 3. Plugging into the original inequality: (2^{2(3) - 7} = 2^{6 - 7} = 2^{-1} = 0.5), and (2^{3 - 3} = 2^{0} = 1). So, 0.5 < 1, which is true. What about x = 4? (2^{2(4) - 7} = 2^{8 - 7} = 2^{1} = 2), and (2^{4 - 3} = 2^{1} = 2). So, 2 < 2 is not true, which means x cannot be equal to 4. That confirms that x must be less than 4.Maybe I should also try a number greater than 4 to see if it doesn't satisfy the inequality. Let's take x = 5. (2^{2(5) - 7} = 2^{10 - 7} = 2^{3} = 8), and (2^{5 - 3} = 2^{2} = 4). So, 8 < 4 is false, which is consistent with our solution that x must be less than 4.I think I've covered all the bases here. I compared the exponents correctly, solved the inequality step by step, and even tested values to make sure my solution makes sense. Yeah, I'm confident that the range of x is all real numbers less than 4.