Rolle Theorem prove for cont and diff,How to do that like for 1-x^2/3 on [-1,1],I know that poly always proved
547668
--- An informal definition of a continuous function is a graph that can be drawn without picking up your pencil.\n--- This graph, however, is NOT differentiable at the point x = 0 because it is a cusp, i.e., the slope of the graph from the left is not equal to the slope of the graph from the right.\n--- If you take the derivative, you get -(2/3)x^(-1/3). If you substitute in x = 0, you get an undefined derivative. i.e., not differentiable at x = 0.
823238
I wont give you the entire proof, just give you the key arguments.\n\nknowing the joint distibution of (X, Y) allows to compute the marginal densities of X and Y.\nIndeed, let f_X be the marginal density of X, then:\n\nf_X(x) = &#92;int f(x, y) d_y\n\nf_Y(y) = &#92;int f(x, y) d_x\n\nNow, you might compute E(Y| X = x) = &#92;int y f_{Y|x} (y) d_y\n\nwhere f_{Y| x} is the conditionnal density of Y knowing the event X = x\n\nrecall that f_{Y| x}  = f(x, y) / f_X(x), and you just have to compute the resulting integrals\n\nregards,\n\njfp
267038
You might try reading Whitehead and Russell's Principia Mathematica.  I can't promise it will help, as I haven't read it, but I've been told that this issue is addressed.
993026
1. yes.\n\n2. one\n\n3. 2 equal segments (this one's a theorem)
702823
want the proof? Here it is ( i am assuming you know the dervatives of sinx and cosx)\n\ny = tanx\ny = sinx/cosx\ny' = [cosx(cosx) - sinx(-sinx)]/(cosx)^2 --> quotient rule\ny'= [(cosx)^2 + (sinx)^2]/(cosx)^2\ny' = 1/(cosx)^2\ny' = (secx)^2\n\nvoila
739421
x<1/2(x+y)<y can be proved by splitting it in two parts and proving those:\nx<1/2(x+y) is true because 1/2(x+y) > 1/2(x+x) = x\n1/2(x+y)<y is true because 1/2(x+y) < 1/2(y+y) = y
965748
|x - x²| \n= x - x² for x - x² ≥ 0\n= x² - x for x - x² < 0\n\nx - x² = x(1 - x) = 0 when x = 0 and x = 1\n\nSo x - x² ≥ 0 when 0 ≤ x ≤ 1\nand x - x² < 0 when x < 0 and x > 1\n\nSo  dx under the interval [-1,3]\n\n= ∫(x² - x) dx under the interval [-1, 0] +\n∫(x - x²) dx under the interval [0, 1] +\n∫(x² - x) dx under the interval [1, 3]\n\n=  [⅓x³ - ½x² ] [-1, 0] + [½x² - ⅓x³] [0, 1] + [⅓x³ - ½x² ] [1, 3]\n\n= [(⅓0³ - ½0²) - (⅓ (-1)³ - ½ (-1)²) ] + [(½(1)² - ⅓(1)³) - (½0² - ⅓0³)] + [(⅓ 3³ - ½ 3²) - (⅓1³ - ½1²) ]\n\n= [5/6] + [1/6] + [9/2 + 1/6]\n\n= 5 2/3
266936
Counterexample: 1\n\nJay's proof is false because x(x-1) = 0 just says that x could be either 1 or 0 to satisfy that equation. x was 1 in that case. It does not mean that 1=0.\n\nJZ-LP's proof is false because you can't say let 1=0.\n\nBianca is just plain wrong.
851180
acc to de moiver's theorem\n(cosx+jsinx)^k=coskx+jsinkx\ntherfore\ncos2x-jsin2x=cos(-2x)+jsin(-2x)\n=(cosx+jsinx)^(-2)\ntherfore\np=3*(cosx+jsinx)/(cosx+jsinx)^-2\n=3(cosx+jsinx)^2\n=3cos2x+jsin2x\nthanks & good luck
553253
http://en.wikipedia.org/wiki/Liskov_substitution_principle click the link to know in detail.
762587
use remainder theorem\nhttp://en.wikipedia.org/wiki/Remainder_theorem\n"When p(x) is divided by (x-a), the remainder is p(a)"\nUR QUESTION:\np(x) = x^3+3x+k\n(x-a) = (x-1)\np(a) = 6\nhence, p(1)=6\nso, (1)^3 + 3(1) + k = 6\nhence, 1+3+k = 6\nso, k = 2\n\nhence, k = 2
898977
log(2x+1)-(1/3)log(x+1)\n=log(2x+1) - log(  (x+1)^(1/3)  )\n=log( (2x+1) /   (x+1)^(1/3)  )
852487
all are possible\n\nanttderivative is x^2/2+3x +c\n1st one ok\n2nd ok ok\n3rd one x^2/2+3x + 9 ok\n4th one x^2/2+3x -8\nall differe only by a constant\n\nso all are possible
970176
Not without a fallacious proof like the one above.  (Notice how the "proof" includes a division by (a-b) which is a division by zero.  This is not allowed and leads to the seeming contradiction that 0 = 1.\n\nZero does *not* equal One.\n\nYou could use similar fallacious methods:\n\n1^0 = 1\n1^1 = 1\n\nTake the log base 1 of both sides:\n0 = log 1 (1)\n1 = log 1 (1)\n\nTherefore 0 = 1.
833002
Let u = x², so du = 2xdx\n\nSo, ∫ (x/1+x^4) dx =\n\n1/2 ∫ (1/1+u²) du =\n\n1/2 arctan(u) + C = \n\n1/2 arctan(x²) + C\n\n~ ♥ ~
904672
let u = x^2 + 1, then you have du = 2xdx, => dx = 1/2x du,\n\nsubstitute back you are on the way  to solve it
456169
There are many proofs but with fallacies. Please visit:\nhttp://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html\n\nhttp://www.math.toronto.edu/mathnet/falseProofs/second1eq2.html