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âLet function h and f have properties such that h(x) = 4f^{-1}(x) and f(x) = frac{40}{x + 5}. For what value of x does h(x) = 20?
đĄOkay, so I have this problem where I'm given two functions, h and f. The problem says that h(x) is equal to 4 times the inverse of f evaluated at x, which is written as h(x) = 4f^{-1}(x). Also, I'm given that f(x) is equal to 40 divided by (x + 5), so f(x) = 40/(x + 5). The question is asking me to find the value of x for which h(x) equals 20.Alright, let me break this down step by step. First, I need to understand what h(x) = 4f^{-1}(x) means. Since h(x) is given in terms of the inverse function of f, I might need to find f inverse first. But before that, maybe I can work directly with the equation h(x) = 20.So, if h(x) = 20, then according to the given equation, 4f^{-1}(x) = 20. To solve for f^{-1}(x), I can divide both sides by 4. That would give me f^{-1}(x) = 20 / 4, which simplifies to f^{-1}(x) = 5.Now, since f^{-1}(x) = 5, this means that the inverse function of f evaluated at x is 5. Remembering that if f^{-1}(x) = a, then f(a) = x. So, in this case, if f^{-1}(x) = 5, then f(5) must equal x.Okay, so now I need to find f(5). I'm given that f(x) = 40/(x + 5). So, substituting 5 into the function f, I get f(5) = 40/(5 + 5). Let me compute that: 5 + 5 is 10, so f(5) = 40/10, which is 4.Therefore, x must be 4 because f(5) = 4, and since f^{-1}(4) = 5, h(4) = 4 * 5 = 20.Wait, let me double-check that. If x is 4, then h(4) = 4f^{-1}(4). I need to make sure that f^{-1}(4) is indeed 5. To find f^{-1}(4), I can set f(y) = 4 and solve for y. So, f(y) = 40/(y + 5) = 4. Multiplying both sides by (y + 5) gives 40 = 4(y + 5). Dividing both sides by 4, we get 10 = y + 5. Subtracting 5 from both sides, y = 5. So, f^{-1}(4) is indeed 5, which means h(4) = 4 * 5 = 20. That checks out.Hmm, is there another way to approach this problem? Maybe by finding the inverse function f^{-1}(x) explicitly and then plugging it into h(x). Let me try that method to confirm.To find f^{-1}(x), I need to solve the equation y = 40/(x + 5) for x in terms of y. So, starting with y = 40/(x + 5), I can multiply both sides by (x + 5) to get y(x + 5) = 40. Then, distribute y: yx + 5y = 40. Now, isolate the term with x: yx = 40 - 5y. Finally, solve for x: x = (40 - 5y)/y.So, the inverse function f^{-1}(y) is (40 - 5y)/y. Therefore, f^{-1}(x) = (40 - 5x)/x.Now, let's plug this into h(x): h(x) = 4f^{-1}(x) = 4 * [(40 - 5x)/x]. Simplify that: h(x) = (160 - 20x)/x.We need to find x such that h(x) = 20. So, set up the equation: (160 - 20x)/x = 20.Multiply both sides by x to eliminate the denominator: 160 - 20x = 20x.Now, add 20x to both sides: 160 = 40x.Divide both sides by 40: x = 160 / 40 = 4.So, again, we get x = 4. That's consistent with the previous method. Therefore, I'm confident that the value of x for which h(x) = 20 is indeed 4.Just to recap, the key steps were:1. Understanding that h(x) is defined in terms of the inverse function of f.2. Setting h(x) equal to 20 and solving for f^{-1}(x).3. Recognizing that f^{-1}(x) = 5 implies f(5) = x.4. Calculating f(5) using the given function f(x).5. Verifying the result by finding the inverse function explicitly and plugging it back into h(x).Both approaches led me to the same answer, so I feel pretty certain that x = 4 is correct. I don't think I made any calculation errors, but it's always good to check your work, especially when dealing with inverse functions, which can sometimes be tricky.Another thing I could consider is graphing the functions to visualize what's happening. If I graph f(x) = 40/(x + 5), it's a hyperbola with a vertical asymptote at x = -5. The inverse function f^{-1}(x) would be a reflection over the line y = x, so it would also be a hyperbola but with a horizontal asymptote. The function h(x) = 4f^{-1}(x) would then be a vertical scaling of the inverse function by a factor of 4.If I were to graph h(x), I could look for the point where it intersects y = 20. That intersection should occur at x = 4, which matches our algebraic solution. Visualizing the graphs might help reinforce the concept of inverse functions and how scaling affects the graph.Additionally, thinking about the relationship between f and f^{-1}, it's important to remember that they are reflections over the line y = x. So, if f maps 5 to 4, then f^{-1} maps 4 back to 5. This bijective relationship is crucial when working with inverse functions, ensuring that each input has a unique output and vice versa.In summary, by carefully applying the properties of inverse functions and solving step by step, I arrived at the solution x = 4. It's always reassuring to approach the problem from multiple angles, whether it's directly solving for the inverse or finding the inverse function explicitly, to confirm that the answer is consistent.
âFind the equation of the line that passes through the point P(1, e) on the curve y = e^x and is perpendicular to the tangent line of the curve at that point.
đĄOkay, so I need to find the equation of a line that passes through the point P(1, e) on the curve y = e^x, and this line has to be perpendicular to the tangent line of the curve at that point. Hmm, let me break this down step by step.First, I remember that to find a line perpendicular to another line, I need to know the slope of the original line. Since the line we're looking for is perpendicular to the tangent line at P, I should start by finding the slope of the tangent line at that point.The curve is given by y = e^x. To find the slope of the tangent line at any point on this curve, I need to compute the derivative of y with respect to x. The derivative of e^x is still e^x, right? So, dy/dx = e^x.Now, I need to evaluate this derivative at the point x = 1 because that's where our point P(1, e) is located. Plugging in x = 1, the slope of the tangent line at P is e^1, which is just e. So, the slope of the tangent line is e.Since the line we want is perpendicular to this tangent line, its slope will be the negative reciprocal of e. The negative reciprocal of e is -1/e. So, the slope of our desired line is -1/e.Now that I have the slope of the line and a point it passes through, I can use the point-slope form of a line's equation to find the equation. The point-slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point.Plugging in the values, we have y - e = (-1/e)(x - 1). Let me write that out:y - e = (-1/e)(x - 1)Now, I can simplify this equation. First, distribute the (-1/e) on the right side:y - e = (-1/e)x + (1/e)Next, I'll add e to both sides to solve for y:y = (-1/e)x + (1/e) + eTo combine the constants, I need to express e in terms of e. Since e is the same as e^1, and 1/e is e^(-1), I can write the equation as:y = (-1/e)x + e + (1/e)But to make this equation look neater, maybe I can combine the constants. Let's see:e + (1/e) can be written as (e^2 + 1)/e. Let me check that:e + (1/e) = (e^2)/e + 1/e = (e^2 + 1)/e. Yes, that works.So, substituting back into the equation:y = (-1/e)x + (e^2 + 1)/eNow, to write this in standard form (Ax + By + C = 0), I'll multiply both sides by e to eliminate the denominators:e*y = -x + e^2 + 1Then, bring all terms to one side:x + e*y - e^2 - 1 = 0So, the equation of the line is x + e*y - e^2 - 1 = 0.Let me double-check my steps to make sure I didn't make any mistakes.1. Found the derivative of y = e^x, which is correct: dy/dx = e^x.2. Evaluated at x = 1: slope of tangent is e. That seems right.3. The slope of the perpendicular line is the negative reciprocal, so -1/e. That makes sense.4. Used point-slope form with point (1, e): y - e = (-1/e)(x - 1). Correct.5. Distributed and simplified: y = (-1/e)x + (1/e) + e. Then combined constants to (e^2 + 1)/e. That looks good.6. Multiplied both sides by e: e*y = -x + e^2 + 1. Then rearranged to x + e*y - e^2 - 1 = 0. Perfect.I think that's all correct. So, the equation of the line is x + e*y - e^2 - 1 = 0.
âGiven a parallelogram ABCD, the coordinates of three vertices are A(-1, 2), B(3, 4), and C(4, -2). If a point (x, y) lies within the parallelogram ABCD (including the boundary), determine the range of values for z = 2x - 5y.
đĄOkay, so I have this problem about a parallelogram ABCD with three given vertices: A(-1, 2), B(3, 4), and C(4, -2). I need to find the range of values for z = 2x - 5y where (x, y) is any point inside or on the boundary of the parallelogram. Hmm, okay, let's break this down step by step.First, I remember that in a parallelogram, the opposite sides are equal and parallel. Also, the diagonals bisect each other. That might help me find the missing vertex D. So, if I can find D, I'll have all four vertices of the parallelogram. Once I have all the vertices, I can evaluate z at each of them because z is a linear function, and its maximum and minimum values on a convex polygon (like a parallelogram) occur at the vertices.Alright, let's find vertex D. Since the diagonals bisect each other, the midpoint of diagonal AC should be the same as the midpoint of diagonal BD. The midpoint formula is ((x1 + x2)/2, (y1 + y2)/2). Let's calculate the midpoint of AC first.Midpoint of AC:x-coordinate: (-1 + 4)/2 = 3/2 = 1.5y-coordinate: (2 + (-2))/2 = 0/2 = 0So, the midpoint is (1.5, 0).Now, this should be the same as the midpoint of BD. Let's denote D as (x, y). Then, the midpoint of BD would be:x-coordinate: (3 + x)/2y-coordinate: (4 + y)/2Setting these equal to the midpoint of AC:(3 + x)/2 = 1.5(4 + y)/2 = 0Solving for x and y:From the x-coordinate:(3 + x)/2 = 1.5Multiply both sides by 2: 3 + x = 3Subtract 3: x = 0From the y-coordinate:(4 + y)/2 = 0Multiply both sides by 2: 4 + y = 0Subtract 4: y = -4Wait, that can't be right. If D is (0, -4), let's check if ABCD is a parallelogram. Let me plot these points mentally: A(-1, 2), B(3, 4), C(4, -2), D(0, -4). Hmm, does this form a parallelogram?Let me check the vectors. Vector AB should be equal to vector DC, and vector AD should be equal to vector BC.Vector AB = B - A = (3 - (-1), 4 - 2) = (4, 2)Vector DC = C - D = (4 - 0, -2 - (-4)) = (4, 2)Okay, that matches.Vector AD = D - A = (0 - (-1), -4 - 2) = (1, -6)Vector BC = C - B = (4 - 3, -2 - 4) = (1, -6)That also matches.So, D is indeed (0, -4). Wait, but in the initial problem statement, the user had a different D. Hmm, maybe I made a mistake.Wait, no, the user didn't provide D, so I need to find it. But according to my calculation, D is (0, -4). Let me double-check the midpoint calculation.Midpoint of AC: (-1 + 4)/2 = 1.5, (2 + (-2))/2 = 0. Correct.Midpoint of BD: (3 + x)/2 = 1.5, so x = 0. (4 + y)/2 = 0, so y = -4. Yes, that's correct.So, D is (0, -4). Okay, so now I have all four vertices: A(-1, 2), B(3, 4), C(4, -2), D(0, -4).Now, I need to find the range of z = 2x - 5y for any point (x, y) inside or on the parallelogram ABCD.Since z is a linear function, its maximum and minimum values over the parallelogram will occur at the vertices. So, I can evaluate z at each vertex and find the maximum and minimum values.Let's calculate z at each vertex:At A(-1, 2):z = 2*(-1) - 5*(2) = -2 - 10 = -12At B(3, 4):z = 2*(3) - 5*(4) = 6 - 20 = -14At C(4, -2):z = 2*(4) - 5*(-2) = 8 + 10 = 18At D(0, -4):z = 2*(0) - 5*(-4) = 0 + 20 = 20Wait, so z at D is 20, which is higher than at C. But earlier, I thought the maximum was at C. Hmm, so I need to check my calculations again.Wait, let's recalculate z at D:z = 2x - 5yx = 0, y = -4z = 2*0 - 5*(-4) = 0 + 20 = 20Yes, that's correct. So, z at D is 20, which is higher than at C. So, the maximum z is 20, and the minimum z is -14.Wait, but in the initial problem, the user had D at (4, 0), which gave z at D as 8. But according to my calculation, D is (0, -4), giving z as 20. So, there's a discrepancy here.Wait, maybe I made a mistake in finding D. Let me double-check.Given A(-1, 2), B(3, 4), C(4, -2). To find D, since ABCD is a parallelogram, vector AB should equal vector DC, and vector AD should equal vector BC.Vector AB = B - A = (3 - (-1), 4 - 2) = (4, 2)Vector DC = C - D = (4 - x, -2 - y) = (4 - x, -2 - y)Since vector AB = vector DC:4 = 4 - x => x = 02 = -2 - y => y = -4So, D is (0, -4). That seems correct.Alternatively, using the midpoint method:Midpoint of AC is ((-1 + 4)/2, (2 + (-2))/2) = (1.5, 0)Midpoint of BD is ((3 + x)/2, (4 + y)/2) = (1.5, 0)So, (3 + x)/2 = 1.5 => x = 0(4 + y)/2 = 0 => y = -4Yes, D is (0, -4). So, z at D is 20.But in the initial problem, the user had D at (4, 0), which is different. Maybe the user made a mistake in their calculation. Let me check their calculation.In the initial problem, the user wrote:"For the parallelogram ABCD with vertices A(-1, 2), B(3, 4), and C(4, -2), we can find the coordinates of D using the fact that the diagonals bisect each other. Let D have coordinates (x_D, y_D). Since AC and BD are diagonals that bisect each other, we have:[x_D + (-1) = frac{3 + 4}{2} quad text{and} quad y_D + 2 = frac{4 + (-2)}{2}.]Solving these, we get:[x_D = 4 quad text{and} quad y_D = 0.]"Wait, that seems incorrect. The midpoint of AC is ((-1 + 4)/2, (2 + (-2))/2) = (1.5, 0). The midpoint of BD should also be (1.5, 0). So, (3 + x_D)/2 = 1.5 => x_D = 0, and (4 + y_D)/2 = 0 => y_D = -4. So, the user incorrectly set up the equations. They wrote x_D + (-1) = (3 + 4)/2, which is not correct. The correct equation is (x_D + 3)/2 = 1.5, not x_D + (-1) = 1.5.So, the user incorrectly calculated D as (4, 0), but the correct D is (0, -4). Therefore, their z at D was 8, but the correct z at D is 20.So, in my calculation, z at D is 20, which is higher than at C, which is 18. Therefore, the maximum z is 20, and the minimum z is -14.Wait, but let me confirm this. Maybe I should also check the edges of the parallelogram to ensure that there are no higher or lower values of z on the edges.Since z is linear, its extrema occur at the vertices, so checking the vertices is sufficient. However, just to be thorough, let's consider the edges.Let's parameterize each edge and see if z can attain higher or lower values.First, edge AB from A(-1, 2) to B(3, 4). Let's parameterize this as:x = -1 + t*(3 - (-1)) = -1 + 4ty = 2 + t*(4 - 2) = 2 + 2twhere t â [0, 1]Then, z = 2x - 5y = 2*(-1 + 4t) - 5*(2 + 2t) = -2 + 8t -10 -10t = (-2 -10) + (8t -10t) = -12 - 2tSince t â [0, 1], z ranges from -12 -0 = -12 to -12 -2 = -14. So, on edge AB, z ranges from -14 to -12.Edge BC from B(3, 4) to C(4, -2). Parameterize as:x = 3 + t*(4 - 3) = 3 + ty = 4 + t*(-2 - 4) = 4 -6tt â [0, 1]z = 2x -5y = 2*(3 + t) -5*(4 -6t) = 6 + 2t -20 +30t = (6 -20) + (2t +30t) = -14 +32tt â [0,1], so z ranges from -14 +0 = -14 to -14 +32 = 18.Edge CD from C(4, -2) to D(0, -4). Parameterize as:x = 4 + t*(0 -4) = 4 -4ty = -2 + t*(-4 - (-2)) = -2 -2tt â [0,1]z = 2x -5y = 2*(4 -4t) -5*(-2 -2t) = 8 -8t +10 +10t = (8 +10) + (-8t +10t) = 18 +2tt â [0,1], so z ranges from 18 +0 =18 to 18 +2=20.Edge DA from D(0, -4) to A(-1, 2). Parameterize as:x =0 + t*(-1 -0) = -ty = -4 + t*(2 - (-4)) = -4 +6tt â [0,1]z = 2x -5y = 2*(-t) -5*(-4 +6t) = -2t +20 -30t = 20 -32tt â [0,1], so z ranges from 20 -0=20 to 20 -32= -12.So, on edge DA, z ranges from -12 to 20.Wait, but on edge DA, when t=0, z=20, and when t=1, z= -12.So, compiling all the ranges:- Edge AB: -14 to -12- Edge BC: -14 to 18- Edge CD: 18 to 20- Edge DA: -12 to 20So, the overall minimum z is -14, and the overall maximum z is 20.Therefore, the range of z is from -14 to 20.But wait, in the initial problem, the user had D at (4,0), which gave z at D as 8, and their maximum z was 18. But with the correct D at (0, -4), z at D is 20, which is higher.So, the correct range should be [-14, 20].But let me double-check my calculations for z at D.z = 2x -5yx=0, y=-4z= 0 -5*(-4)=0 +20=20. Yes, correct.Similarly, at C(4, -2):z=2*4 -5*(-2)=8 +10=18.At B(3,4):z=6 -20=-14.At A(-1,2):z=-2 -10=-12.So, the vertices give z as -12, -14, 18, 20.Therefore, the minimum z is -14, and the maximum z is 20.Hence, the range of z is [-14, 20].But wait, in the initial problem, the user had D at (4,0), which gave z at D as 8, and their maximum z was 18. So, their answer was [-14,18], but that was based on an incorrect D.So, the correct answer should be [-14,20].Therefore, I think the user made a mistake in finding D, and consequently, their maximum z was incorrect.So, to summarize, the correct range of z is from -14 to 20.
âGiven the function ( f(x) = frac{ln (1+x)}{x} ):1. Prove that ( f(x) > frac{2}{x+2} ) when ( x > 0 ).2. Determine the value of the real number ( k ) such that the inequality ( f(x) < frac{1 + kx}{1 + x} ) holds for ( x > -1 ) and ( x neq 0 ).
đĄAlright, so I have this function ( f(x) = frac{ln(1+x)}{x} ), and I need to tackle two parts. Let me start with the first one.**1. Prove that ( f(x) > frac{2}{x+2} ) when ( x > 0 ).**Hmm, okay. So I need to show that ( frac{ln(1+x)}{x} > frac{2}{x+2} ) for all positive ( x ). Maybe I can rearrange this inequality to make it easier to handle. Let's see:Starting with:[ frac{ln(1+x)}{x} > frac{2}{x+2} ]If I multiply both sides by ( x ) (since ( x > 0 ), the inequality direction remains the same), I get:[ ln(1+x) > frac{2x}{x+2} ]Alright, so now I need to show that ( ln(1+x) ) is greater than ( frac{2x}{x+2} ) for all ( x > 0 ). Maybe I can define a function that represents the difference between these two expressions and analyze its behavior.Let me define:[ h(x) = ln(1+x) - frac{2x}{x+2} ]If I can show that ( h(x) > 0 ) for all ( x > 0 ), then the original inequality holds. To do this, I can look at the derivative of ( h(x) ) to see if it's increasing or decreasing.Calculating the derivative:[ h'(x) = frac{d}{dx}left(ln(1+x) - frac{2x}{x+2}right) ][ h'(x) = frac{1}{1+x} - frac{2(x+2) - 2x}{(x+2)^2} ][ h'(x) = frac{1}{1+x} - frac{4}{(x+2)^2} ]Hmm, that looks a bit complicated. Maybe I can combine these terms into a single fraction to simplify:[ h'(x) = frac{(x+2)^2 - 4(1+x)}{(1+x)(x+2)^2} ]Expanding the numerator:[ (x+2)^2 = x^2 + 4x + 4 ][ 4(1+x) = 4x + 4 ]So subtracting:[ x^2 + 4x + 4 - 4x - 4 = x^2 ]Therefore:[ h'(x) = frac{x^2}{(1+x)(x+2)^2} ]Since ( x > 0 ), the numerator ( x^2 ) is positive, and the denominator ( (1+x)(x+2)^2 ) is also positive. So ( h'(x) > 0 ) for all ( x > 0 ). This means that ( h(x) ) is increasing on the interval ( (0, infty) ).Now, let's check the value of ( h(x) ) at ( x = 0 ):[ h(0) = ln(1+0) - frac{2 cdot 0}{0+2} = 0 - 0 = 0 ]Since ( h(x) ) is increasing and ( h(0) = 0 ), it follows that ( h(x) > 0 ) for all ( x > 0 ). Therefore:[ ln(1+x) > frac{2x}{x+2} ]Dividing both sides by ( x ) (which is positive), we get:[ frac{ln(1+x)}{x} > frac{2}{x+2} ]So the first part is proved.**2. Determine the value of the real number ( k ) such that the inequality ( f(x) < frac{1 + kx}{1 + x} ) holds for ( x > -1 ) and ( x neq 0 ).**Alright, now I need to find ( k ) such that:[ frac{ln(1+x)}{x} < frac{1 + kx}{1 + x} ]for all ( x > -1 ) and ( x neq 0 ).Let me rewrite the inequality:[ frac{ln(1+x)}{x} < frac{1 + kx}{1 + x} ]Multiplying both sides by ( x(1+x) ) (I need to be careful about the sign of ( x ) here, but since ( x > -1 ) and ( x neq 0 ), ( 1+x > 0 ), so the inequality direction remains the same if ( x > 0 ), but if ( -1 < x < 0 ), multiplying by ( x ) would reverse the inequality. Hmm, maybe I should consider both cases separately.)Alternatively, let's bring everything to one side:[ frac{ln(1+x)}{x} - frac{1 + kx}{1 + x} < 0 ]Combine the fractions:[ frac{(1+x)ln(1+x) - x(1 + kx)}{x(1+x)} < 0 ]So the numerator is:[ (1+x)ln(1+x) - x - kx^2 ]Let me define:[ g(x) = (1+x)ln(1+x) - x - kx^2 ]So the inequality becomes:[ frac{g(x)}{x(1+x)} < 0 ]Now, since ( x(1+x) > 0 ) when ( x > 0 ) (because both ( x ) and ( 1+x ) are positive), and ( x(1+x) < 0 ) when ( -1 < x < 0 ) (because ( x ) is negative and ( 1+x ) is positive, making their product negative). Therefore, the sign of the entire expression depends on the sign of ( g(x) ).For ( x > 0 ):[ frac{g(x)}{x(1+x)} < 0 Rightarrow g(x) < 0 ]For ( -1 < x < 0 ):[ frac{g(x)}{x(1+x)} < 0 Rightarrow g(x) > 0 ]So, ( g(x) ) must be negative for ( x > 0 ) and positive for ( -1 < x < 0 ). To find ( k ), I need to analyze the behavior of ( g(x) ).First, let's compute the first derivative of ( g(x) ):[ g'(x) = frac{d}{dx}left[(1+x)ln(1+x) - x - kx^2right] ][ g'(x) = ln(1+x) + frac{1}{1+x} - 1 - 2kx ]Simplify:[ g'(x) = ln(1+x) - 2kx ]Now, the second derivative:[ g''(x) = frac{d}{dx}left[ln(1+x) - 2kxright] ][ g''(x) = frac{1}{1+x} - 2k ]Let's analyze ( g''(x) ). For ( x > 0 ):[ frac{1}{1+x} < 1 ]So if ( 2k geq 1 ), then ( g''(x) leq 0 ). This would mean ( g'(x) ) is decreasing for ( x > 0 ).Similarly, for ( -1 < x < 0 ):[ frac{1}{1+x} > 1 ]So if ( 2k leq 1 ), then ( g''(x) geq 0 ). This would mean ( g'(x) ) is increasing for ( -1 < x < 0 ).Let's consider ( k = frac{1}{2} ). Then:[ g''(x) = frac{1}{1+x} - 1 ]For ( x > 0 ):[ g''(x) = frac{1}{1+x} - 1 < 0 ]So ( g'(x) ) is decreasing.For ( -1 < x < 0 ):[ g''(x) = frac{1}{1+x} - 1 > 0 ]So ( g'(x) ) is increasing.Now, let's evaluate ( g'(x) ) at ( x = 0 ):[ g'(0) = ln(1+0) - 2k cdot 0 = 0 ]Since ( g'(x) ) is decreasing for ( x > 0 ) and ( g'(0) = 0 ), it means ( g'(x) < 0 ) for ( x > 0 ). Therefore, ( g(x) ) is decreasing for ( x > 0 ), and since ( g(0) = 0 ), it follows that ( g(x) < 0 ) for ( x > 0 ).For ( -1 < x < 0 ), ( g'(x) ) is increasing and ( g'(0) = 0 ). Since ( g'(x) ) is increasing and starts from some value at ( x = -1 ) (but we don't need to compute it exactly), it means that ( g'(x) < 0 ) for ( -1 < x < 0 ) because it's increasing towards 0. Therefore, ( g(x) ) is increasing for ( -1 < x < 0 ), and since ( g(0) = 0 ), it follows that ( g(x) > 0 ) for ( -1 < x < 0 ).Thus, with ( k = frac{1}{2} ), the inequality ( f(x) < frac{1 + kx}{1 + x} ) holds for all ( x > -1 ) and ( x neq 0 ).I think this makes sense. Let me double-check by plugging in ( k = frac{1}{2} ) into the original inequality and testing a value, say ( x = 1 ):[ f(1) = frac{ln 2}{1} approx 0.6931 ][ frac{1 + frac{1}{2} cdot 1}{1 + 1} = frac{1.5}{2} = 0.75 ]Indeed, ( 0.6931 < 0.75 ), so it holds for ( x = 1 ).Testing ( x = -0.5 ):[ f(-0.5) = frac{ln(0.5)}{-0.5} = frac{-0.6931}{-0.5} approx 1.3862 ][ frac{1 + frac{1}{2} cdot (-0.5)}{1 + (-0.5)} = frac{1 - 0.25}{0.5} = frac{0.75}{0.5} = 1.5 ]Here, ( 1.3862 < 1.5 ), so it also holds for ( x = -0.5 ).Therefore, ( k = frac{1}{2} ) seems to be the correct value.
âGiven the set M={x mid (x-1)^2 < 4, x in mathbb{R}}, and N={-1, 0, 1, 2, 3}, then the intersection M cap N is ( ).A: {0, 1, 2}B: {-1, 0, 1, 2}C: {-1, 0, 2, 3}D: {0, 1, 2, 3}
đĄFirst, I need to determine the elements of set ( M ) by solving the inequality ( (x-1)^2 < 4 ).I start by taking the square root of both sides:[sqrt{(x-1)^2} < sqrt{4}]This simplifies to:[|x - 1| < 2]Next, I split the absolute value inequality into two separate inequalities:1. ( x - 1 < 2 )2. ( -(x - 1) < 2 ) which simplifies to ( x - 1 > -2 )Solving the first inequality:[x - 1 < 2 x < 3]Solving the second inequality:[x - 1 > -2 x > -1]Combining these results, I find that:[-1 < x < 3]So, set ( M ) is ( M = {x mid -1 < x < 3} ).Now, I look at set ( N = {-1, 0, 1, 2, 3} ) and find the intersection ( M cap N ). The elements in ( N ) that are also in ( M ) are ( 0, 1, ) and ( 2 ).Therefore, the intersection ( M cap N ) is ( {0, 1, 2} ).
âIf the function f(x) is an odd function, and for x < 0, f(x) = ln(-x) + 3x, then the equation of the tangent line to the curve at the point (1, 3) is (ãã)A: y + 3 = -2(x - 1)B: y - 3 = 2(x - 1)C: y + 3 = 4(x - 1)D: y - 3 = 4(x + 1)
đĄOkay, so I have this problem here about an odd function. Hmm, I remember that odd functions have the property where f(-x) = -f(x). That means if I know the function for negative x, I can find it for positive x by just flipping the sign. The function given is for x < 0, which is f(x) = ln(-x) + 3x. Alright, so I need to find the equation of the tangent line at the point (1, 3). Tangent lines require the derivative, right? But before I can find the derivative, I need to make sure I know what the function looks like for positive x since the point (1, 3) is in the positive x region.Since the function is odd, f(-x) = -f(x). Let me plug in a positive x into this. If x is positive, then -x is negative, so f(-x) would be ln(-(-x)) + 3*(-x) which simplifies to ln(x) - 3x. But because it's an odd function, f(-x) is also equal to -f(x). So, setting them equal: -f(x) = ln(x) - 3x. Therefore, f(x) = -ln(x) + 3x. Got it, so for x > 0, f(x) is -ln(x) + 3x.Now, to find the tangent line at (1, 3), I need the derivative f'(x). Let me compute that. The derivative of -ln(x) is -1/x, and the derivative of 3x is 3. So, f'(x) = -1/x + 3.Next, I need to evaluate this derivative at x = 1 to get the slope of the tangent line. Plugging in x = 1: f'(1) = -1/1 + 3 = -1 + 3 = 2. So the slope is 2.Now, with the point (1, 3) and the slope 2, I can write the equation of the tangent line using point-slope form, which is y - y1 = m(x - x1). Plugging in the values: y - 3 = 2(x - 1). Looking at the options, that's option B: y - 3 = 2(x - 1). Let me double-check my steps to make sure I didn't make a mistake. I used the property of odd functions correctly to find f(x) for positive x, computed the derivative properly, evaluated it at x = 1, and then wrote the equation of the tangent line. Everything seems to add up. So I think option B is the correct answer.