Three function questions...?
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1. when b = 0\n2. only if it has zero slope\n3. it must have a y-intercept. it can have no x-intercept (for example y=x²+1)
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1) Yes.\n2) No.\n3) No.
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Use "..."\n\nI don't really get it, but there's some documentation here: http://courses.cs.vt.edu/~cs1704/notes/A02.AdvancedFunctionParameters.pdf
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1) No\n2) Yes\n3) No\n4) Yes
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The last link was the most informative.\nThe function is as follows:\n(x' y')T = [(1 1)T (1 -1)T] (x y)T\n\nIf you multiply the basis vectors by this matrix you get:\n(0 2)T for the (1 -1)T vector and\n(-1 -5)T for the (-3 2)T vector.\n\n(0, 2) = c1 * (1, -1) + c2 * (-3, 2)\nso:\n0 = c1 - 3 * c2\n2 = - c1 + 2 * c2\nc1 = -6\nc2 = -2\n\nlikewise:\n(-1, -5) = c3 * (1, -1) + c4 * (-3, 2)\nc3 = 17 and c4 = 6\n\nM = [(-6 -2)T (17 6)T]
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1. p'q\n\n2. p'\n\n3. pq' + p'\n\nI think i got the correct ones !
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1) 21.69\n2) 14.10\n3) 7.59
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http://www.functionx.com/vb/index.htm\nvisit this site, i hope it can help you. good luck
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Take a look at this site.\nhttp://www.functionx.com/visualc/index.htm
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1. c\n2. c\n3. c\n4. c\n5. c
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Instantaneous rate of change of a function (7 words),\n\nOR,\n\nSlope of the tangent line (5 words). :P
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Because of the way you would call the function \nsomefunct(param1, param2, param3); // canonical\nsomefunct(param1, param2); // compiler knows you mean to use the default param for 3 because it´s the third param you don´t provide.
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1. Undefined (No Slope)\n2. Undefined again.
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1 -> b\n2 -> e\n3 -> c\n4 -> d\n5 -> a\n6 -> f
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First one: d= - 1\n\nEDIT: the rest I'm too lazy to do. Do your own homework! =D
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Well, I see you've used the correct order of operations this time. That's good - it shows that you're improving.\n\n#1: Apply the product and chain rules to find f'(x), then evaluate at 0:\n\nf'(x) = d(√(x³+5x+121))/dx (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * d(x³+5x+121)/dx * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nf'(x) = 1/(2√(x³+5x+121)) * (3x²+5) * (x²+x+11) + (2x+1)√(x³+5x+121)\n\nNow evaluating :\n\nf'(0) = 1/(2√121)*5*11 + √121 = 5/2 + 11 = 27/2\n\n#2: First, evaluate the function at y=1 to get the value of x:\n\nxy^2 -2y + 4y^3 = 6\nx - 1 + 4 = 6\nx = 3\n\nNow, implicitly differentiate both sides:\n\nd(xy^2 -2y + 4y^3)/dx = d(6)/dx\ny² + 2xy dy/dx - 2 dy/dx + 12y² dy/dx = 0\n\nSolve for dy/dx:\n\ny² + (2xy - 2 + 12y²) dy/dx = 0\n(2xy - 2 + 12y²) dy/dx = -y²\ndy/dx = -y²/(2xy - 2 + 12y²)\n\nNow evaluate at (3, 1):\n\ndy/dx = -1/(6 - 2 + 12) = -1/16\n\n#3: First, find the derivative:\n\nf'(x) = (x²+x-2 - (2x+1)x)/(x²+x-2)² (using the quotient rule)\n\nSimplifying:\n\nf'(x) = (x²+x-2 - 2x² - x)/(x²+x-2)²\nf'(x) = (-x²-2)/(x²+x-2)²\n\nNow we must find where f'(x) < 0\n\n(-x²-2)/(x²+x-2)² < 0\n-x²-2 < 0 (we know (x²+x-2)² is nonnegative, and if the expression on the left is defined, it is also nonzero. Thus it is positive and we can multiply by it without changing the inequality)\n-x² < 2\n\nNow, since x is a real number, this expression will always be true. Thus, if f'(x) is even defined, x is decreasing. Now f'(x) is defined provied that (x²+x-2)²≠0. So:\n\n(x²+x-2)²≠0\n(x-1)²(x+2)²≠0\nx≠1 and x≠-2\n\nWhich makes sense, since those are vertical asymptotes of the original function. Thus, the function is decreasing on:\n\n(-∞, -2) ∪ (-2, 1) ∪ (1, ∞)
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bool function(int a, int b, int c)\n{\nreturn a || b || c;\n}