💡Okay, so I need to find the equation of the tangent line that passes through the point (0, √3) and is tangent to the circle C: (x - 1)² + y² = 4. Hmm, let me think about how to approach this.First, I remember that a tangent line to a circle is perpendicular to the radius at the point of tangency. So, if I can find the point where the tangent touches the circle, I can then find the equation of the tangent line.But wait, I don't know the point of tangency yet. Maybe I can use the fact that the tangent line passes through (0, √3). So, perhaps I can set up an equation for the tangent line and then use the condition that it touches the circle at exactly one point.Let me recall the general equation of a line. It can be written as y = mx + b, where m is the slope and b is the y-intercept. Since the line passes through (0, √3), when x = 0, y = √3. So, substituting that into the equation, I get √3 = m*0 + b, which means b = √3. So, the equation of the tangent line is y = mx + √3.Now, I need to find the value of m such that this line is tangent to the circle (x - 1)² + y² = 4. For a line to be tangent to a circle, the distance from the center of the circle to the line must be equal to the radius of the circle.The center of the circle is at (1, 0), and the radius is 2. So, the distance from (1, 0) to the line y = mx + √3 should be equal to 2.The formula for the distance from a point (x₀, y₀) to the line Ax + By + C = 0 is |Ax₀ + By₀ + C| / √(A² + B²). Let me rewrite the equation of the tangent line in standard form. Starting from y = mx + √3, I can subtract y to get mx - y + √3 = 0. So, A = m, B = -1, and C = √3.Now, plugging the center (1, 0) into the distance formula:Distance = |m*1 + (-1)*0 + √3| / √(m² + (-1)²) = |m + √3| / √(m² + 1)This distance should equal the radius, which is 2. So, I set up the equation:|m + √3| / √(m² + 1) = 2To solve for m, I can square both sides to eliminate the absolute value and the square root:(m + √3)² / (m² + 1) = 4Expanding the numerator:(m² + 2m√3 + 3) / (m² + 1) = 4Multiply both sides by (m² + 1):m² + 2m√3 + 3 = 4(m² + 1)Expanding the right side:m² + 2m√3 + 3 = 4m² + 4Bring all terms to one side:m² + 2m√3 + 3 - 4m² - 4 = 0Combine like terms:-3m² + 2m√3 - 1 = 0Multiply both sides by -1 to make it easier:3m² - 2m√3 + 1 = 0Now, this is a quadratic equation in terms of m. Let me write it as:3m² - 2√3 m + 1 = 0I can solve this using the quadratic formula. For an equation ax² + bx + c = 0, the solutions are:m = [-b ± √(b² - 4ac)] / (2a)Here, a = 3, b = -2√3, and c = 1.Plugging in the values:m = [2√3 ± √( ( -2√3 )² - 4*3*1 )] / (2*3)Calculate the discriminant:( -2√3 )² = 4*3 = 124*3*1 = 12So, discriminant = 12 - 12 = 0Wait, discriminant is zero, which means there's only one solution. That makes sense because the line is tangent, so there should be only one slope that satisfies the condition.So, m = [2√3 ± 0] / 6 = (2√3)/6 = √3/3So, the slope m is √3/3.Therefore, the equation of the tangent line is y = (√3/3)x + √3.Let me just verify this. If I plug in x = 0, y should be √3, which it is. Now, let's check if this line is indeed tangent to the circle.To confirm, I can solve the system of equations:1. (x - 1)² + y² = 42. y = (√3/3)x + √3Substitute equation 2 into equation 1:(x - 1)² + [(√3/3)x + √3]² = 4Let me expand this:First, (x - 1)² = x² - 2x + 1Second, [(√3/3)x + √3]²:Let me factor out √3: √3[(1/3)x + 1]So, squared, it's 3[(1/3)x + 1]² = 3[(1/9)x² + (2/3)x + 1] = (1/3)x² + 2x + 3So, putting it all together:(x² - 2x + 1) + ( (1/3)x² + 2x + 3 ) = 4Combine like terms:x² + (1/3)x² = (4/3)x²-2x + 2x = 01 + 3 = 4So, the equation becomes:(4/3)x² + 4 = 4Subtract 4 from both sides:(4/3)x² = 0Multiply both sides by 3/4:x² = 0So, x = 0Therefore, the only solution is x = 0, which means y = (√3/3)*0 + √3 = √3So, the point of tangency is (0, √3). Wait, but that's the point we started with. Hmm, does that mean that (0, √3) is the point where the tangent touches the circle?Let me check if (0, √3) is on the circle:(x - 1)² + y² = (0 - 1)² + (√3)² = 1 + 3 = 4, which equals the radius squared. So yes, (0, √3) is on the circle.So, the tangent line at (0, √3) is y = (√3/3)x + √3.But wait, isn't the tangent line at a point on the circle supposed to be perpendicular to the radius at that point? Let me check the slope.The radius from the center (1, 0) to the point (0, √3) has a slope of (√3 - 0)/(0 - 1) = √3 / (-1) = -√3.The slope of the tangent line is √3/3. Are these slopes negative reciprocals? Let's see: (-√3) * (√3/3) = (-3)/3 = -1. Yes, they multiply to -1, so they are perpendicular. That checks out.So, everything seems consistent. The tangent line passes through (0, √3) and is tangent to the circle at that point, with the correct slope.I think that's the solution. The equation of the tangent line is y = (√3/3)x + √3.**Final Answer**The equation of the tangent line is boxed{y = frac{sqrt{3}}{3}x + sqrt{3}}.

💡Okay, so I have this problem where I'm given two equations: (x + y = 8) and (x^2 - y^2 = 24). I need to find the value of (x - y). Hmm, let me think about how to approach this.First, I remember that (x^2 - y^2) is a difference of squares, which can be factored into ((x + y)(x - y)). That seems useful because I already know (x + y) is 8. So, if I substitute that into the factored form, I can set up an equation to solve for (x - y).Let me write that down:(x^2 - y^2 = (x + y)(x - y))Given that (x + y = 8), I can substitute that in:(24 = 8 times (x - y))Now, I need to solve for (x - y). To do that, I can divide both sides of the equation by 8:(x - y = frac{24}{8})Calculating that, I get:(x - y = 3)Wait, does that make sense? Let me double-check my steps. I used the difference of squares correctly, substituted the known sum, and then solved for the difference. It seems straightforward. Maybe I should verify by plugging numbers back in.If (x - y = 3), and (x + y = 8), I can set up a system of equations:1. (x + y = 8)2. (x - y = 3)Adding these two equations together, I get:(2x = 11)So, (x = frac{11}{2}) or 5.5. Then, substituting back into (x + y = 8):(5.5 + y = 8)So, (y = 8 - 5.5 = 2.5). Now, let's check if (x^2 - y^2) equals 24:(x^2 = (5.5)^2 = 30.25)(y^2 = (2.5)^2 = 6.25)Subtracting these: (30.25 - 6.25 = 24). Yep, that checks out. So, (x - y = 3) is correct.I think that's solid. I didn't make any calculation errors, and the verification confirms the answer. So, I'm confident that (x - y) is indeed 3.

💡First, I need to find the square root of 300 using the given approximation of √3 ≈ 1.732.I can express 300 as 100 multiplied by 3.Then, I'll take the square root of both parts: √300 = √(100 × 3) = √100 × √3.Since √100 is 10, the equation becomes √300 = 10 × √3.Substituting the given value of √3 ≈ 1.732, I get √300 ≈ 10 × 1.732.Performing the multiplication, 10 × 1.732 equals 17.32.Therefore, the square root of 300 is approximately 17.32.

💡Okay, so I have this problem about a 17x17 table where some squares are colored black. The goal is to find the minimal number of initially black squares, n, such that by repeatedly painting entire lines (rows, columns, or diagonals) black whenever at least 6 squares in that line are already black, I can eventually turn the entire table black. First, I need to understand the rules clearly. A line is any row, column, or the two main diagonals. In one step, if a line has at least 6 black squares, I can paint the entire line black. So, the key is to figure out the smallest number of initially black squares that can trigger a chain reaction, turning the whole table black.I think the strategy should involve placing the initial black squares in such a way that they can activate as many lines as possible, each activation potentially leading to more activations. Since each activation can turn an entire line black, which can then help in activating other lines.Let me consider the structure of the table. It's a 17x17 grid, so there are 17 rows, 17 columns, and 2 main diagonals. Each line has 17 squares. To activate a line, I need at least 6 black squares in it. So, if I can arrange the initial black squares so that multiple lines each have 6 black squares, I can start painting them black one by one.But I need to minimize the number of initial black squares. So, I should look for overlaps where a single black square can contribute to multiple lines. For example, a square at the intersection of a row and a column can help in both activating the row and the column.Maybe I can start by focusing on the diagonals. If I can get both main diagonals to each have 6 black squares, I can paint them black, which would then help in activating rows and columns. But wait, the main diagonals have 17 squares each, so to activate them, I need 6 black squares on each. But if I place 6 black squares on each diagonal, that's 12 squares, but some might overlap at the center if they intersect. In a 17x17 grid, the main diagonals intersect at the center square (9,9). So, if I place 6 black squares on each diagonal, I might have one overlapping square, meaning I need 11 squares in total.But 11 squares might not be enough because after painting the diagonals, I still need to activate rows and columns. Each diagonal has 17 squares, so painting them would add 17 black squares each, but since they intersect, the total would be 33 squares. But wait, initially, I only have 11 black squares. After painting the diagonals, I would have 33 black squares, which is a significant increase.But I need to check if 11 initial black squares are enough. Let's see. If I place 6 black squares on the main diagonal and 6 on the other main diagonal, overlapping at the center, that's 11 squares. Then, in the first step, I can paint both diagonals black, turning 33 squares black. Now, each row and column that intersects with the diagonals will have at least 2 black squares (from the diagonals). But 2 is less than 6, so I can't activate any rows or columns yet.Hmm, that's a problem. So, painting the diagonals gives me more black squares, but not enough to activate any rows or columns. I need a way to have some rows or columns already have 6 black squares after the initial step.Maybe I should also place some black squares in specific rows or columns to get them to 6. For example, if I have a row with 6 black squares, I can paint it black, which would then contribute to the columns it intersects.Alternatively, perhaps a combination of diagonals and some strategically placed black squares in rows and columns can help. Let me think about how to cover multiple lines with minimal squares.Another approach is to consider that each black square can be part of a row, a column, and possibly a diagonal. So, if I place black squares at intersections where multiple lines meet, each square can contribute to multiple lines.Wait, but in a 17x17 grid, each square is part of one row, one column, and at most two diagonals (the main diagonals if it's on them). So, to maximize the coverage, I should place black squares where they can contribute to as many lines as possible.Maybe focusing on the center area where multiple lines intersect. For example, the center square is part of both main diagonals, its row, and its column. So, placing a black square there would help in four lines. But I need 6 in each line to activate it.Alternatively, maybe arranging the initial black squares in a way that forms a sort of "cross" in the center, covering multiple rows and columns, which can then be activated and spread outwards.But I'm not sure if that's the most efficient. Let me try to think in terms of the minimal number. If I can find a configuration where each initial black square contributes to multiple lines, I can minimize n.Suppose I place 6 black squares in a single row. That would allow me to paint that entire row black, adding 17 black squares. Then, those 17 black squares would each contribute to their respective columns. If each column now has at least 1 black square (from the painted row), but that's not enough to activate them. I need 6 in each column.So, maybe I need to have multiple rows and columns each with 6 black squares. But that would require more initial squares.Wait, perhaps a better strategy is to have overlapping lines. For example, if I have a set of rows and columns each with 6 black squares, but arranged so that the black squares overlap as much as possible.Let me consider that each black square can be part of one row and one column. So, if I have r rows each with 6 black squares, and c columns each with 6 black squares, the total number of initial black squares would be r*6 + c*6 - overlaps. But overlaps can't exceed the number of intersections between the rows and columns.This seems complicated. Maybe I should look for known results or similar problems. I recall something about the "minimum number of clues" in Sudoku, but this is different. Or maybe it's similar to the set cover problem, where I need to cover all lines with the minimal number of squares.Alternatively, think about it as a graph problem, where each line is a node, and each square is an edge connecting a row, column, and possibly diagonals. Then, the problem becomes covering all nodes with the minimal number of edges such that each node has at least 6 edges connected to it.But I'm not sure if that's the right way to model it. Maybe another approach is to think about how many lines I need to activate, and how each initial black square can help in activating multiple lines.There are 17 rows, 17 columns, and 2 diagonals, totaling 36 lines. Each line needs at least 6 black squares to be activated. So, in total, I need to cover 36 lines with 6 black squares each, but each black square can cover up to 4 lines (row, column, and two diagonals if it's on them). However, most squares are only on one row, one column, and possibly one or two diagonals.But this is getting too abstract. Maybe I should try a smaller grid to get some intuition. Let's say a 3x3 grid, and see what the minimal n would be there, with a lower threshold, say 2 instead of 6. Maybe that can give me some insight.In a 3x3 grid, to paint all squares black, starting with n black squares, where if a line has at least 2 black squares, you can paint the entire line. What's the minimal n?I think in that case, you could place 2 black squares on a diagonal, then paint the diagonal, which would give you 3 black squares on that diagonal. Then, each row and column intersecting the diagonal would have 1 black square, which isn't enough. So, maybe you need to place 2 black squares on a row and 2 on a column, overlapping at one square. So, total of 3 black squares. Then, paint the row and column, which would give you 3 + 3 -1 = 5 black squares. Then, the other rows and columns would have 2 black squares each, allowing you to paint them.Wait, but in the 3x3 case, maybe n=3 is sufficient. So, scaling up, perhaps in the 17x17 case, the minimal n is related to 6 times something, but adjusted for overlaps.But I'm not sure. Maybe I should think about the problem in terms of linear algebra. Each line is an equation, and each square is a variable. We need to set enough variables to 1 (black) such that each equation (line) has at least 6 ones. But this is a bit too abstract.Alternatively, think about it as a covering problem. Each initial black square can cover multiple lines. So, to cover all 36 lines with each line needing at least 6 covers, and each square can cover up to 4 lines, the minimal n would be at least (36*6)/4 = 54. But this is a lower bound, and the actual minimal n might be higher due to overlaps and the structure of the grid.But 54 seems too high because the total number of squares is 289, and we're looking for a much smaller n. Wait, maybe I made a mistake in the calculation. Each line needs 6 black squares, but each black square can contribute to multiple lines. So, the total number of "line contributions" needed is 36*6=216. Each black square can contribute to up to 4 lines, so the minimal n is at least 216/4=54. So, 54 is a lower bound.But in reality, it's probably higher because not all squares can contribute to 4 lines. Only the ones on the diagonals can contribute to 4 lines, others contribute to 2 or 3. So, maybe the lower bound is higher.Wait, but in the problem, after the initial n black squares, you can perform multiple steps. So, maybe the initial n doesn't need to cover all lines, but just enough to start a chain reaction.So, perhaps the minimal n is less than 54. Maybe around 26 or something. Wait, I think I remember a similar problem where the minimal number was 26 for a 17x17 grid, but I'm not sure.Wait, let me think again. If I can activate lines step by step, each activation can help in activating more lines. So, maybe I don't need all lines to have 6 black squares initially, but just enough to start a chain.For example, if I have 6 black squares in a row, I can paint the entire row, which then gives me 17 black squares in that row, which can help in activating the columns they are in. Each column now has at least 1 black square, but that's not enough. However, if I have multiple rows painted, each column would have more black squares.So, maybe if I paint several rows, each column would accumulate enough black squares to be painted. Similarly, painting columns can help in painting rows.But how to minimize the initial n? Maybe place 6 black squares in each of 5 rows, but that would be 30 squares, which seems high. Alternatively, arrange the initial black squares so that they cover multiple lines.Wait, perhaps using the diagonals. If I place 6 black squares on each diagonal, overlapping at the center, that's 11 squares. Then, painting both diagonals would give me 33 black squares. Then, each row and column would have 2 black squares (from the diagonals). Not enough, but maybe if I have some additional black squares in specific rows or columns, I can get them to 6.Alternatively, maybe place 6 black squares in a single row and a single column, overlapping at one square. So, 6 + 6 -1 = 11 squares. Then, paint the row and column, which would give me 17 + 17 -1 = 33 black squares. Now, each row and column would have at least 1 black square, but not enough. However, the intersection of the painted row and column would have 2 black squares, but still not enough.Hmm, this seems tricky. Maybe I need to have multiple rows and columns each with 6 black squares, arranged in such a way that their intersections allow for a chain reaction.Wait, perhaps using a grid where the initial black squares form a sort of "frame" around the grid, covering the outer rows and columns, which can then be painted and help in painting the inner rows and columns.But I'm not sure. Maybe I should look for a pattern or a known solution. I think in similar problems, the minimal number is often related to the size of the grid and the threshold for activation. For a 17x17 grid with a threshold of 6, the minimal n is likely 26 or 27.Wait, let me think about it differently. Suppose I have n initial black squares. Each time I paint a line, I add 17 - k black squares, where k is the number already black in that line. So, to maximize the number of new black squares per step, I should paint lines with as few black squares as possible, i.e., exactly 6.But to do that, I need to have lines with exactly 6 black squares. So, maybe arrange the initial black squares so that multiple lines have exactly 6, allowing me to paint them one by one, each time increasing the number of black squares and potentially creating more lines with 6 black squares.But how to arrange that? Maybe place 6 black squares in several lines, ensuring that their intersections don't overlap too much, so that each line has exactly 6.Wait, but in a 17x17 grid, if I place 6 black squares in each of 5 rows, that's 30 squares, but maybe some can overlap in columns, reducing the total.Alternatively, maybe place 6 black squares in each of 5 rows and 5 columns, overlapping at certain points. But this is getting too vague.Wait, maybe the minimal n is 26. Because 26 is the minimal number such that in a 17x17 grid, you can arrange them to cover enough lines to start a chain reaction.But I'm not sure. Maybe I should try to calculate it more carefully.Let me consider that each initial black square can contribute to a row, a column, and possibly a diagonal. So, to cover all 36 lines with each needing 6 black squares, and each square contributing to up to 4 lines, the minimal n is at least (36*6)/4 = 54. But since we can perform multiple steps, maybe the initial n can be less.Wait, but in each step, we can paint a line, which adds 17 - k black squares, where k is the number already black in that line. So, the more lines we can paint early on, the more black squares we can add, potentially allowing us to paint more lines.So, maybe the minimal n is around 26 because 26 is the minimal number such that, with careful arrangement, you can paint lines step by step until the entire grid is black.But I'm not entirely sure. Maybe I should look for a pattern or a known solution. I think in similar problems, the minimal number is often related to the size of the grid and the threshold for activation. For a 17x17 grid with a threshold of 6, the minimal n is likely 26 or 27.Wait, I think I remember a similar problem where the minimal number was 26 for a 17x17 grid, but I'm not sure. Alternatively, maybe it's 27.Wait, let me think about it again. If I have 26 black squares, can I arrange them so that I can paint lines step by step until the entire grid is black? Maybe, but I'm not sure. Alternatively, 27 might be the minimal number.I think I'll go with 27 as the minimal n. So, the answer is 27.

💡First, identify the original number of cards, which is 9.Next, determine the total number of cards after some were added, which is 13.To find out how many cards were added, subtract the original number of cards from the total number of cards.So, 13 minus 9 equals 4.Therefore, 4 cards were added.

💡Okay, so I need to find the minimum value of this function: f(x) = 2^(sqrt(x² - 4x + 4)) + sqrt(x² - 2x). Hmm, that looks a bit complicated, but let's break it down step by step.First, I should figure out the domain of the function because square roots require the expressions inside them to be non-negative. Let's look at each square root separately.Starting with sqrt(x² - 4x + 4). I notice that x² - 4x + 4 is a quadratic expression. Maybe it can be factored? Let me try: x² - 4x + 4. Hmm, that looks like a perfect square. Yes, it factors to (x - 2)². So sqrt((x - 2)²) is just |x - 2|, the absolute value of x - 2. Since absolute values are always non-negative, this part is defined for all real numbers. So no restrictions from this term.Next, the other square root: sqrt(x² - 2x). Again, let's see if this can be factored or simplified. x² - 2x can be written as x(x - 2). So sqrt(x(x - 2)). For this to be real, the expression inside must be greater than or equal to zero. So x(x - 2) ≥ 0.To solve x(x - 2) ≥ 0, I can find the critical points where the expression equals zero: x = 0 and x = 2. Now, I'll test intervals around these points to see where the product is non-negative.- For x < 0: Let's pick x = -1. Then (-1)(-1 - 2) = (-1)(-3) = 3, which is positive.- For 0 < x < 2: Let's pick x = 1. Then (1)(1 - 2) = (1)(-1) = -1, which is negative.- For x > 2: Let's pick x = 3. Then (3)(3 - 2) = (3)(1) = 3, which is positive.So the expression x(x - 2) is non-negative when x ≤ 0 or x ≥ 2. Therefore, the domain of f(x) is (-∞, 0] ∪ [2, ∞).Alright, so f(x) is defined for x ≤ 0 and x ≥ 2. Now, I need to find the minimum value of f(x) in this domain.Let me analyze the function f(x) = 2^(sqrt(x² - 4x + 4)) + sqrt(x² - 2x). Let's rewrite sqrt(x² - 4x + 4) as |x - 2|, as I did earlier. So f(x) becomes 2^{|x - 2|} + sqrt(x² - 2x).Now, let's consider the two intervals separately: x ≤ 0 and x ≥ 2.First, let's look at x ≥ 2.For x ≥ 2, |x - 2| = x - 2, so f(x) = 2^{x - 2} + sqrt(x² - 2x).Let me see if I can simplify sqrt(x² - 2x). Maybe complete the square?x² - 2x = (x - 1)² - 1. So sqrt(x² - 2x) = sqrt((x - 1)² - 1). Hmm, that might not help much. Maybe I can just consider the behavior of the function as x increases.As x increases, 2^{x - 2} grows exponentially, and sqrt(x² - 2x) grows like x for large x. So f(x) is increasing for x ≥ 2. Therefore, the minimum on this interval should be at x = 2.Let me compute f(2):f(2) = 2^{|2 - 2|} + sqrt(2² - 2*2) = 2^0 + sqrt(4 - 4) = 1 + 0 = 1.Okay, so at x = 2, f(x) = 1.Now, let's look at x ≤ 0.For x ≤ 0, |x - 2| = 2 - x, so f(x) = 2^{2 - x} + sqrt(x² - 2x).Again, let's see if I can analyze the behavior as x decreases.As x becomes more negative, 2^{2 - x} becomes 2^{2 - x} = 2^{2} * 2^{-x} = 4 * (1/2)^x. Since x is negative, -x is positive, so (1/2)^x grows exponentially as x becomes more negative. Therefore, 2^{2 - x} grows exponentially as x decreases.On the other hand, sqrt(x² - 2x) for x ≤ 0. Let's see: x² is positive, and -2x is positive since x is negative. So x² - 2x is positive, and sqrt(x² - 2x) is real. As x becomes more negative, x² grows quadratically, and -2x grows linearly, so overall, sqrt(x² - 2x) behaves like |x| for large |x|, which is linear.So, as x approaches negative infinity, 2^{2 - x} dominates and grows exponentially, while sqrt(x² - 2x) grows linearly. Therefore, f(x) tends to infinity as x approaches negative infinity.Now, what about near x = 0? Let's compute f(0):f(0) = 2^{|0 - 2|} + sqrt(0² - 2*0) = 2^2 + sqrt(0) = 4 + 0 = 4.So at x = 0, f(x) = 4.Since for x ≤ 0, f(x) is decreasing as x increases towards 0, but f(x) increases as x moves away from 0 into more negative values. Wait, that might not be correct. Let me think.Wait, for x ≤ 0, as x increases towards 0, 2^{2 - x} decreases because as x increases, 2 - x decreases, so 2^{2 - x} decreases. Similarly, sqrt(x² - 2x) also decreases as x increases towards 0 because x² decreases and -2x decreases (since x is negative). So both terms are decreasing as x increases towards 0. Therefore, f(x) is decreasing on (-∞, 0].Therefore, the minimum on x ≤ 0 would be at x = 0, which is 4.Comparing the two intervals, the minimum value on x ≥ 2 is 1 at x = 2, and the minimum on x ≤ 0 is 4 at x = 0. Therefore, the overall minimum value of f(x) is 1.Wait, but let me double-check. Is there any possibility that somewhere between x = 2 and x = infinity, the function might dip below 1? But I thought f(x) is increasing for x ≥ 2, so the minimum is at x = 2.Similarly, for x ≤ 0, f(x) is decreasing towards x = 0, so the minimum is at x = 0, which is 4.Therefore, the overall minimum is 1 at x = 2.But just to be thorough, let me check another point in x ≥ 2, say x = 3.f(3) = 2^{|3 - 2|} + sqrt(3² - 2*3) = 2^1 + sqrt(9 - 6) = 2 + sqrt(3) ≈ 2 + 1.732 ≈ 3.732, which is greater than 1.And at x = 4, f(4) = 2^2 + sqrt(16 - 8) = 4 + sqrt(8) ≈ 4 + 2.828 ≈ 6.828, which is even larger.So yes, it's increasing for x ≥ 2, so x = 2 is indeed the minimum.Similarly, for x ≤ 0, let's check x = -1.f(-1) = 2^{|-1 - 2|} + sqrt{(-1)^2 - 2*(-1)} = 2^{3} + sqrt{1 + 2} = 8 + sqrt{3} ≈ 8 + 1.732 ≈ 9.732, which is much larger than 4.And at x = -2, f(-2) = 2^{|-2 - 2|} + sqrt{4 - (-4)} = 2^4 + sqrt{8} = 16 + 2.828 ≈ 18.828, which is even larger.So yes, the function is increasing as x becomes more negative, so the minimum on x ≤ 0 is at x = 0, which is 4.Therefore, the overall minimum value of f(x) is 1 at x = 2.