Let f(X) be a function in D and g(X) a function on bdy D.?
607760
Well, when you maximize or minimize with constraint, you use Lagrange multipliers. So set that up. Now write the Euler-Lagrange equations for your system. Be sure to consider the boundary carefully!
791940
In order to a od function the function have to obay:\nf(x) = - f (-x) <=>\n-3x^9 + 4x^5 = -(-3(-x)^9 + 4(-x)^5) <=>\n-3x^9 + 4x^5 = 3(-x)^9 - 4(-x)^5 <=>\n-3x^9 + 4x^5 = -3x^9 + 4x^5 ( True)\nthe function is od.
1034892
What is the function? I assume it is \n\n(dy/dx is equal to 1 only if y=x+C whre C is a constant)\n\nTo prove use the definition of the derivative which is dy/dx as a limit of delta_x appraching of (delta_y/delta_x)\nwe have \ny2-y1=x2+C-x1-C then\ndelta_y=x2-X1\n\ndelta_y/delta_x=(x2-x1)/(x2-x1)=1 QED\n(the is no need to take a limit, however in must be used. I hav eommited it for lack of better representation)\n\nBest
754435
I | AUG is not correct.  You cannot differentiate power functions by the exponent rule.  (Even doing that, she made an error.)  The implicit differentiation done by Greg G is correct, or you can use the functional power rule:\n\nd(ƒ^g)/dx = (ƒ^g)*[(dƒ/dx)*(g/ƒ) + (dg/dx)*lnƒ]
1082751
1)   f(x) = x+1/x =(x^2+1)/x  \n      g(x) = x-2/x =(x^2-2)/x\n(f+g)(x) = (x^2+x^2+1-2)/x = (2*x^2-1)/x = 2*x -1/x\n\n2)  f(x) = -2/x\n     g(x) = (x+1) / (x-3)\n(f+g)(x) = -2/x + (x+1)/(x-3) = {-2*(x-3)+x*(x+1)}/x*(x-3)\n              = (-2x+6+x^2-x)/(x^2-3x) =(x^2-x+6)/x^2-3x
1166733
dy/dx = {(3x^2-2)(4x-3)^4 - (x^3-2x)(4)(4x-3)^3(4)}/(4x-3)^8\n= {(3x^2-2)(4x-3)^4 - 16(x^3-2x)(4x-3)^3}/(4x-3)^8\n\nUsing the quotient rule.  d/dx f(x)/g(x) = {f'(x)g(x) - g'(x)f(x)}/g(x)^2
837082
Hint: Let C=X&#92;O and let f(x)=dist(x,C). The other way is trivial.
939187
First, evaluate these functions to understand what they mean. Here is a listing of n and f(n)\n\n0 1 2 3 4 5 6 ...\n0 2 4 6 8 10 12 ...\n\nHere is a listing of n and g(n):\n\n0 1 2 3 4 5 6 ...\n0 0 1 1 2 2 3 ...\n\nEvaluate g • f using g • f (n) = g(f(n)) to get its listing:\n\n0 1 2 3 4 5 6 ...\n0 1 2 3 4 5 6 ...\n\nwhich provides evidence for question (a). g(f(n)) = g(2n) = n, because 2n is even.\n\nThe listing of f • g is\n\n0 1 2 3 4 5 6...\n0 0 2 2 4 4 3 ...\n\nAnalyze f(g(n)) = 2 [n/2]. What happens when n is even, and when n is odd?\n\nA function, F, is injective if different values never map to the same value. Or, the listing does not have duplicates. To prove that F is not injective is easy: just find a counterexample, that is a and b such that F(a) = F(b), but a ≠ b. To prove that F is injective you must show that in general if F(x) = F(y) then x = y. \n\nLook at the listings. If a listing has duplicates then the function is not injective. A proof is just providing a counterexample. Example: f • g has duplicates; the simplest are 0 and 1: f • g(0) = f • g(1), and 0 ≠ 1.The listing of f has no duplicates. To prove that f is injective, prove that if 2x = 2y then x = y. That should be easy!\n\nA function is surjective if the listing contains every number in the domain, that is all the numbers: 0, 1, 2, 3,...\n\nThe listing of f does not contain 1. So f is not surjective. A proof is to show that there is no n such that f(n) = 1. That's easy because f(n) is even!\n\nThe listing of g appears to contain all the numbers. Does it contain 2006? Can you find n such that g(n) = 2006? Such that n/2 = 2006? Yes, n = 4012. Similarly, for any number (instead of 2006).\n\nSo, I have provided some strategies and examples to let you now make your own proofs.
74348
If you look at the standard epsilon/delta proof that G(x) = x^2 is continuous 0, it all works for F(x) too. \n\nThat is, given epsilon (e), you can find delta (d) such that /G(x) - G(0)/ < e when /x - 0/ < d.\n(Since for -1 < x < 1,  x^2 < |x|, just pick d = min(e, 1))\n\nNow looking at F(x), given d, pick that same e you would use with G(x). If x is rational, F(x) = G(x)\nso /G(x) - G(0)/ = /F(x) - F(0)/ < e.  If x is irrational, /G(x) - G(0)/ = /0 - 0/ < e.
1032260
1. if x=f(t) and y=f(t) then\n\nx^4+y^4=6y\n\nd/dt { x^4+y^4=6y}\n\n4x^3 (dx/dt) + 4y^3(dy/dt) =6(dy/dt)\n\nand you have to have the x=f(t) & y=f(t) in order to substitute for the derivatives.\n\n2. if y=f(x) and x=f(t) then\n\nx^4+y^4=6y\n\nd/dt { x^4+y^4=6y}\n\n4x^3 (dx/dt) + 4y^3(dy/dx)(dx/dt) =6(dy/dt)(dx/dt)\n\nthen substitute for the derivatives of the functions y=f(x) & x=f(t)
576554
Hint: If an analytic function is 0 on a set with its accumulation point, then the function is identically 0.\n\nNow, either f or g has to have this property since at each point one or the other is 0.
873165
f(g(x)) = (g(x)+4)^2 = (8x-8+4)^2 = (8x-4)^2 = 64x^2-64x+16\n\nThat's your composite function.\n\nHope this helps.
865303
1) If h(x) = f(g(x)), then g(x) = 4+3x and f(x) = x^5\n2) Assuming your function is 3sqrt(x^2 -8), then\ng(x) = x^2 - 8 and f(x) = 3sqrt(x)
824963
Given:  BX/D = CE/F\n\nFind:  "X"\n\nSolution:\nBX/D = CE/F\n(B/D)(X) = CE/F\n(D/B)(B/D)(X) = (D/B)(CE/F)\n(1)(X) = CDE/BF\n\nAnswer:\nX = CDE over BF  or  X = CDE/BF
457089
G=uys\nF=***\nE=verthing
940999
Differentiate the function as you would normally except that when the derivative of y is taken make sure to replace y with y'. So this would be,\n(1/2)(xy)^(-2/3)(xy') + y = 0 - y'\nThen, solve for y' by moving all the y' over to one side and factoring it out.\ny' + (1/2)(xy)^(-2/3)(xy') + y = 0\n\ny'((1/2)(xy)^(-2/3)x + 1) = -y\n\ny' = -y / ((1/2(xy)^(-2/3)x + 1)\n\nThe second derivative is (quotient rule),\ny'' = (.5(xy)^(-2/3) + 1)(-y') - (-y)(1/3)(xy)^(-5/3)xy(xy')(y)\n\ny'' = .5(xy)^(-2/3)y' - y' + (1/3)x^2*y^3*y'(xy)^(-5/3)\nFactor,\ny'' = y'{(xy)^(-5/3)[.5 - 1 + (1/3)x^2*y^3(xy)]}\n\ny'' = y'{(xy)^(-5/3)[- .5 + (1/3)x^2*y^3(xy)]}\n\n(check my work)
977406
Here is an outline:\n\nf is an ONTO function. So for EVERY element b in B there is an\nelement a in A so that f(a) = b.  That is what onto means.  All the\nelements in B have some element in A that f maps it into.\n\nSince g is not one to one there must be two elements  in B that get mapped into the same element in C.  Call them b1 and b2. Show that b1 and b2 are part of the codomain of f.  f(x) maps into all B, so there must be a a1 and a2 that get mapped into b1 and b2.  So then we have g(f(a1)) and g(f(a2) that map into the same element, and hence g(f(x)) is not one to one.