Use MVT to prove  l sin(a)-sin(b)l is less than or equal to la-bl for all a,b?
817419
Proof.\n\nsin : [a, b] → ℝ is a continuous differentiable function, so by the mean value theorem, ∃ c ∈ (a, b) such that\n\nsin'(c) = (sin(b) - sin(a)) / (b - a), i.e.\ncos(c) * (b - a) = sin(a) - sin(b).\n\nNow since ∀ x ∈ ℝ, -1 ≤ cos(x) ≤ 1, it follows that\n\n| sin(a) - sin(b) |\n= | sin(b) - sin(a) |\n= | cos(c) * (b - a) |\n= | cos(c) | * | b - a |\n≤ 1 * | b - a |\n= | a - b |. ∎
850884
using chain rule and product rule\n\ny'= 3x^(2)*tan(4x)\n         +4x^(3)*sec^(2)(4x)\n  = x^2(3tan(4x)+4x*sec^(2)(4x) \n\ni hope that this helps
795252
Let a and b be any two real numbers. Consider the number x defined by\n\nx = ab + (-a)(b) + (-a)(-b).\n\nWe can write\nx = ab + (-a)[ (b) + (-b) ]       (factor out -a)\n  = ab + (-a)(0)\n  = ab + 0\n  = ab.\n\nAlso,\nx = [ a + (-a) ]b + (-a)(-b)      (factor out b)\n  = 0 * b + (-a)(-b)\n  = 0 + (-a)(-b)\n  = (-a)(-b).\n\nSo we have\n      x = ab\nand\n      x = (-a)(-b)\n\nHence, by the transitivity of equality, we have\n\n      ab = (-a)(-b).
301328
Homework.\n\nLook at B=A. You should find something interesting about tr(A^tA).
939642
Counterexample.  Let u(x), v(x) be members of {sin x, cos x}.  Both of those are solutions, but you do not get zero by adding one to the other, or by doubling one of them.
811861
KCl(s)  ---->  K+(aq)  +  Cl-(aq)
798431
limit x>0 [(3x)sin3x/3x*(4x)sin4x/4x]\n/[(6x)tan6x/6x(x)sinx/x]\nlimit x>0 3x*4x/6x*x\nlimit x>0 12x^2/6x^2\n=2 (answer)
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Lim { [ sin(x-1)  ] / [ x^2 + x - 2] }\nx-->1\n\nAdd the following to the denominator\n-2x + 2x +1 -1\n\nLim { [ sin(x-1)  ] / [ x^2 + x - 2 -2x +2x +1 -1] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ x^2 -2x + 1 +3x -3] }\nx-->1\n\nNote: (x^2-2x+1) =(x-1)^2    and  (3x-3) = 3(x-1)\n\nThen\nLim { [ sin(x-1)  ] / [ (x-1)^2 +3(x-1)] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ (x-1)^2 +3(x-1)] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ (x-1)[ (x-1)+3 ] }\nx-->1\n\nLim { [ sin(x-1) / (x-1) ] * [ 1/ (x-1+3 )] }\nx-->1\n\nLet u= x-1 then as x-->1, u--> 0\nand\nLim { [ sin u / u ] * [ 1/ (u+3 )] }\nu-->0\n\nSince\n\nLim  [ sin(u) / (u) ]=1\nu-->0\n\nThen \nLim { [ sin u / u ] * [ 1/ (u+3 )] } = 1 * [1/(0+3)] = 1*1/3 = 1/3\nu-->0
909136
Let I = ∫e^s sin(t - s) ds\n= e^s sin(t - s) - ∫e^s *( -1) * cos(t - s)ds\n= e^s sin(t - s) + ∫e^s cos(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s * (-1) * -sin(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s sin(t - s) ds\n= e^s sin(t - s) + e^s cos(t - s) - I\nSo 2I = e^s (sin(t - s) + cos(t - s)\nThus I = ½(e^s (sin(t - s) + cos(t - s)) + K\nSo ∫[s = 0 to s = t] e^s sin(t - s) ds\n= ½[e^s (sin(t - s) + cos(t - s)] from s= 0 to s = t\n= ½[(e^t (sin(t - t) + cos(t - t)) - (e^0 (sin(t - 0) + cos(t - 0))]\n= ½[(e^t(1) - 1(sint + cost)]\n= ½(e^t - sint - cost)
794327
A pretty trivial one would be f(x) = absolute value of x and g(x) = -x\n\nf(g(x)) = abs (-x) = x\n\nbut g(f(x)) = - (abs x) = -x
768256
This is true additionally it is less(cannot be equal)\n\nreason is f(x) = g(x)\n\ningegartion is arera under curve = sum f(x)(h etc\n\narea under curve for each range is less for f than g\n\nso avarage of f < avarage of g
496092
let x equal (tant)^2=>dx=(sect)^2dt. then simply the integrand, u will have something like integral of (cost)(sint)^2 dt. let u be sint. then u know it :D
966444
Sorry I'm not english spoken\n\nIl suffit de choisir deux vecteurs u et v non nuls et non liés.\nu ∧ v ≠ 0\nOn constate alors que :\n(u ∧ u) ∧ v = 0 ∧ v = 0\nSupposons que :\nu ∧ (u ∧ v) = 0\nAlors\nu et u ∧ v sont liés\nCe qui est absurde car :\nu ∧ v ⊥ u\nDonc :\nu ∧ (u ∧ v) ≠ 0\nD'où :\n(u ∧ u) ∧ v ≠ u ∧ (u ∧ v)
880673
hint: mean value theorem
949116
set a linear combination of your 3 vectors =0 and prove that the numbers have to be =0:\nA sin(x) + B sin(2x) + C sin(3x) = 0\ni.e. A,B,C are numbers such that the linear combination is the constant zero.\nnow take x=pi/2,\nthen\n0= A sin(pi/2) + B sin(2pi/2) + C sin(3pi/2) \n= A - C, \nso A=C.\nNow x=pi/4\n0= A sin(pi/4) + B sin(2pi/4) + A sin(3pi/4) \n=  A/sqrt(2) + B + A/sqrt(2)\n= 2 A/sqrt(2) +B\n= sqrt(2) A + B, so B= -sqrt(2) A\nNow, x = 3pi/4\n0= A sin(3pi/4) + B sin(2(3)pi/4) + A sin(9pi/4) \n= A/sqrt(2) - B + A/sqrt(2)\n=sqrt(2) A - B, but then B = sqrt(2) A\nwhich means that -sqrt(2) A =  sqrt(2) A\nso A =0, therefore B=0 and C=0\n'
157474
I do not believe this is always true, only in the case of commuting matrices.\n\nLet A'=A\nLet B'=B\n\nThen (AB)' = B'A' = BA\n\nBut AB is  not always equal to BA unless A and B commute.
929836
v = b - u/k\nv - b = u/k\nk(v-b) = u