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Question 3 a) Given the function: f(x) = 3(x-2)^2 + 2 , determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4. b) The points of intersection between the lines tangent to g(x) when x=2 and the line tangent to f(x) when x=5 are the centers of three circles with a radius of 5. Determine the equations of the circles where g(x)= squareroot[x] .
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Part A a) Given the function: f(x) = 3(x-2)^2 + 2 , determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4. <ul><li>First of all, what is a derivative? </li></ul><ul><ul><li>In mathematics, the derivative of a function is the graph of the function's slope or its rate of change, at any given point. </li></ul></ul><ul><ul><li>In general, a derivative is something has been derived or come from something else </li></ul></ul><ul><li>The process of finding a derivative is called differentiation. </li></ul>This question involves fundamental knowledge on derivatives.
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Intro. a) Given the function: f(x) = 3(x-2)^2 + 2 , determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4. The derivative of any constant is zero. So the derivative of 5, 10, 11 is “zero.” To understand this, graph y=5, y=10, and y=11. They all have no slope, so the derivative is zero. The derivative of any variable to the exponent “1” is “1”. To understand this, graph y=x. X is the variable. You should see through rise over run that the slope is “1.” Y=x is also just another form of y=mx+b where m=1. Common Notation This is a common form of Leibniz’s Notation. This is Lagrange’s notation, also known as prime notation.
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Intro. It is hard to explain this stuff since I’m fairly new to this stuff, so I won’t go into very much detail. Power Rule The derivative of a power of x is equal to the product of the exponent times x with an exponent reduced by 1 For example: Let’s find the derivative of x^6 You can see that n=6. The derivative of x^6 is 6x^5. Now let’s proceed to my question.
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Part A Continued a) Given the function: f(x) = 3(x-2)^2 + 2 , determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4. First, you put the equation into general form. f(x) = 3(x-2)^2 + 2 f(x) = 3(x^2 -4x +4) + 2 f(x) = 3x^2 -12x + 12 + 2 f(x) = 3x^2 -12x + 14 Now, we can find the derivative of this equation. f 1 (x) = 3(2)(x) – (12x^0) - 0 = 6x – 12 The exponent on each reduces by one and you multiply the base by the original exponent. The derivative of any constant is zero like mentioned earlier.
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Part A Continued a) Given the function: f(x) = 3(x-2)^2 + 2 , determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4. So this is the equation to obtain the slope of a line tangent to any point on f(x). f 1 (x) = 3(2)(x) – (12x^0) - 0 = 6x – 12 We want the line tangent when x=4, so we plug in 4 for x. f 1 (4) = 6(4) – 12 = 24 – 12 = 12 This is just the slope, but we need to find the equation. If you think about it, we can use the point-slope formula
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Part A Continued a) Given the function: f(x) = 3(x-2)^2 + 2 , determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4. This is the point-slope formula. Y-Y1 = M(X-X1) The point when x=4 is found on both graphs obviously. However, we need the y-coordinate of the point. What we do is plug 4 into the original equation, not the derivative one because that equation determines slopes. f(x) = 3x^2 -12x + 14 f(4) = 3(4^2) -12(4) + 14 f(4) = 48 – 48 +14 f(4) = 14 Therefore, the point is (4,14). Now let’s plug that in.
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Part A Continued a) Given the function: f(x) = 3(x-2)^2 + 2 , determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4. This is the point-slope formula. Y – 14 = 12(X-4) Y – 14 = 12x – 48 Y = 12x – 48 + 14 Y = 12x – 34 This is the equation of the line tangent to f(x) when x=4. Let’s take a look at a graph to see this.
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Part A End a) Given the function: f(x) = 3(x-2)^2 + 2 , determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4. You can see that at (4,14), the line Y = 12x – 34 touches f(x) = 3x^2 -12x + 14
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Part B b) The points of intersection between the lines tangent to g(x) when x=2 and the line tangent to f(x) when x=5 are the centers of three circles with a radius of 5. Determine the equations of the circles where g(x)= squareroot[x] . We now have two functions: g(x) = squareroot[x] f(x) = 3(x-2)^2 + 2 Let’s start by finding the line tangent to g(x) when x=4.
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Part B b) The points of intersection between the lines tangent to g(x) when x=2 and the line tangent to f(x) when x=5 are the centers of three circles with a radius of 5. Determine the equations of the circles where g(x)= squareroot[x] . g(x) = squareroot[x] We have x already, which is 4, so let’s find y. g(4) = squareroot[4] g(4) = +2 and -2 Oh snap, that’s two values. This is because for every x value on the graph, there are 2 y values, excluding the vertex. Now let’s obtain the slope
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Part B b) The points of intersection between the lines tangent to g(x) when x=2 and the line tangent to f(x) when x=5 are the centers of three circles with a radius of 5. Determine the equations of the circles where g(x)= squareroot[x] . We must find the derivative of g(x) g(x) = squareroot[x] g 1 (x) = (1/2)x^(-1/2) The squareroot of x is equal to x^(1/2). We multiply that exponent by x and subtract 1 from ½ which gives us -1/2. Now we plug in x to obtain the slope. g 1 (x) = (1/2)x^(-1/2) g 1 (4) = (1/2)4^(-1/2) *exponents first* *bedmas* g 1 (4) = (1/2)(1/-2) or (1/2)(1/2) g 1 (4) = (1/-4) or (1/4) Okay, so these are our slopes. Now we find the equations.
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Part B What a great slope we have ! Slope = +1/4 and -1/4 Now let’s use the point-slope formula to solve for the equation. Earlier, we obtained 2 y values, +2 and -2. We have four possibilities for the equation now. Y-Y1 = M(X-X1) Let’s use 2 as our y value with the 2 slopes Y – 2 = ¼(x-4) Y = (x/4) + 1 Or Y – 2 = -¼(x-4) Y = (-x/4) + 3 Let’s use -2 as our y value with the 2 slopes Y + 2 = ¼(x-4) Y = (x/4) - 3 Or Y + 2 = -¼(x-4) Y = (-x/4) - 1 Amazingly, the equations in the black are the correct ones. The next slide shows a graph.
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Part B We know the derivative of f(x) from earlier, except not the y-value when x=16. f 1 (x) = 6x – 12 f(x) = 3(x-2)^2 + 2 Let’s find the y-value f(16) = 3(16-2)^2 + 2 f(16) = 3(14)^2 + 2 f(16) = 3(16-2)^2 + 2 f(16) = 590 Therefore, the point is (16, 590). Now we need to obtain the slope. b) The points of intersection between the lines tangent to g(x) when x=2 and the line tangent to f(x) when x=5 are the centers of three circles with a radius of 5. Determine the equations of the circles where g(x)= squareroot[x] .
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Part B We use the derivative of f(x) to find the slope. f 1 (x) = 6x – 12 Plug in 16 for X f 1 (16) = 6(16) – 12 f 1 (16) = 96 – 12 f 1 (16) = 84 Now that we have the slope and 2 points, we can find the equation of the line. Y-Y1 = M(X-X1) Y-590 = 84(x-16) Y = 84x – 1344 + 590 Y = 84x – 754 Great. This is the equation of the line tangent to f(x) when x=16. Let’s see a graph to verify this. b) The points of intersection between the lines tangent to g(x) when x=2 and the line tangent to f(x) when x=5 are the centers of three circles with a radius of 5. Determine the equations of the circles where g(x)= squareroot[x] .
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Part B The graphs are very ugly, but you can see that the equations are correct. Now we must find the points of intersection between the tangent lines!
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Part B Here are the three equations of tangent lines we have. To find the points of intersection, we make two equations equal other because at that specific point, they share the same coordinate. For example, making the y’s equal other shows which values of y are shared amongst the graphs. From g(x), we have: 1) Y = (x/4) + 1 2) Y = (-x/4) - 1 From f(x), we have: 3) Y = 84x – 754 There are going to be three intersections because each of these graphs are lines that never end. I assigned each of the equations numbers to show which intersections I’ll be finding.
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Part B 1,2 1) Y = (x/4) + 1 2) Y = (-x/4) – 1 (x/4) + 1 = (-x/4) – 1 (x/4) – (-x/4) = – 1 -1 2x/4 = -2 X = -4 Now we must find the y-value to find the coordinate. We can plug x into anyone of the equations above because they both contain a point when x=-4 Y = (-4/4) + 1 Y = 0 Our first point of intersection is (-4,0)
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Part B 1,3 1) Y = (x/4) + 1 3) Y = 84x - 754 (x/4) + 1 = 84x – 754 (x/4) – (84x) = – 754 – 1 (x – 336x)/4 = -755 x(1 – 336) = -3020 x(-335) = -3020 x = 604/67 This is approximately 9.0149 Now we must find the y-value to find the coordinate. We can plug x into anyone of the equations above because they both contain a point when x=604/67 Y = ((604/67)/4) + 1 Y = 151/67 + 67/67 Y = 218/67 This approximately equals 3.2537 Our second point of intersection is ( 604/67 , 218/67)
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Part B 2,3 2) Y = (-x/4) - 1 3) Y = 84x - 754 (-x/4) - 1 = 84x – 754 (-x/4) – (84x) = – 754 + 1 (-x – 336x)/4 = -753 -337X = -3012 x = 3012/337 This approximately equals 8.9377 Now we must find the y-value to find the coordinate. We can plug x into anyone of the equations above because they both contain a point when x=3012/337 Y = (-(3012/337)/4) + 1 Y = -753/337 - 337/337 Y = - 1090/337 This approximately equals -3.2344 Our third point of intersection is ( 3012/337 , 1090/337 )
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Part B Alright, so our final step is to find the equations of the circles where their origins are the intersections we found. Each circle has a radius of 5. This is perhaps the HARDEST part of the problem because it requires so much work to get this far. Our points of intersection are: ( -4 , 0 ) ( 604/67 , 218/67 ) ( 3012/337 , 1090/337 ) The radius is 5. Now we just plug them into the equation: (x-h)^2 + (y-k)^2 = r^2 (x+4)^2 + (y-0)^2 = 25 (x-604/67)^2 + (y-218/67)^2 = 25 (x-3012/337)^2 + (y-1090/337)^2 = 25
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