Let f:[0,1]-[0,1] be an continuous function s.t f((x+y)/2)<=(f(x)+f(y))/2. show f is convex?
448359
A convex function is a continuous function whose value at the midpoint  of every interval in its domain does not exceed the average  of its values at the ends of the interval.\n\nIn other words, a function f(x) is convex on an interval [a,b] if for any two points x_1 and x_2 in [a,b] and any  β where 0< β<1,\nf[ βx_1+(1- β)x_2)]<= βf(x_1)+(1- β)f(x_2)\n\nTake ur function and prove the above.
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
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 |. ∎
857286
Im not sure don't right this at your Homework.........\nwhere f(x) > 0. a. (–3, 1) b . (–3, 0) ∪ (1, ∞) c . (1, 3) d . (0, 1) ∪ (3, ∞) If f(x) = x(x – 1)(x – 4)2, use interval ..............??????\n\n\nSORRY I DON'T HELP TOO MUCH I TRY MY BEST......
329632
{x: x inequal to 2}\nor\n{x: x=/= 2}
565556
This is counterexample is posted on Wikipedia; the link is below.  They don't explain all of the details, so I'll do that here.\n\nConsider the function f: R^2 -> R defined by\n\nf(x,y) = x y (x^2 - y^2) / (x^2 + y^2)\n\nI'm going to show that D_x (D_y f) is different from D_y (D_x f) at (x,y) = (0,0).\n\nThis function is continuous, because it is easy to check that lim_{(x,y) -> (0,0)} f(x,y) = 0.  Its derivatives are\n\nu(x,y) = D_x f(x,y) = y (x^4 + 4 x^2 y^2 - y^4) / (x^2 + y^2)^2\nv(x,y) = D_y f(x,y) = x (x^4 - 4 x^2 y^2 - y^4) / (x^2 + y^2)^2\n\nAgain, these functions are continuous, because lim_{(x,y) -> (0,0)} u(x,y) = lim_{(x,y) -> (0,0)} v(x,y) = 0.  Now to compute the mixed derivatives at (x,y) = (0,0), we need to compute the following limits.\n\nD_x (D_y f) @ (0,0) = lim_{h -> 0} ( v(h,0) - v(0,0) ) / h = lim_{h -> 0} h/h = 1\nD_y (D_x f) @ (0,0) = lim_{k -> 0} ( u(0,k) - u(0,0) ) / k = lim_{k -> 0} -k/k = -1\n\nHence these partial derivatives are *not* equal at (x,y) = (0,0) as claimed.\n\n-------------\n\nBy the way, I should point out that this gives a counter example to the famous statement\n\ndiv . curl (F) = 0\n\nIndeed, take F(x,y,z) to be that vector valued-function which is f(x,y) in the z-direction.  Then\n\ndiv . curl(F) = D_x (D_y f) - D_y (D_x f)\n\nIf you choose f(x,y) as above, then this expression is not zero at (x,y) = (0,0).
277316
let |f|=k (some constant).let u,v be the real & img parts of 'f'.\n            the we have |f|=u^2+v^2, u^2+v^2=k^2\n      differentiating partially,wrt to x&y ,and \n\n    USING CAUCHY-REI EQ'S FROM TWO RESULTING EQ'S \n\nu finally get     (u^2+v^2)(du/dx)=0.\n                 &  (u^2+v^2)(du/dy)=0. \nfrom above two eq's   we have \n                     k^2=u^2+v^2=0  implies u=0&v=0\n\n                  so 'f' is constant.
931612
f^{-1}(y) = (y-42) / 7\ns
442777
Suppose that the range of f does not intersect some neighborhood of w. Show that 1/(f-w) is a bounded entire function. Louiville will finish it off.
22349
For all real numbers, f(x) is the sum of two continuous components (as x^2 is always non-negative) and one component discontinuous at 0. Therefore the function is only discontinuous at x=0. For x>0, f(x)=x^2; for x<0, f(x)=x^2-2x.\n\nBoth x^2 and sin(x) (as well as sin^2(10x)) equal zero when x=0 and are continuous across all reals. Therefore no Squeeze Law is necessary. This expression evaluates to 0.
740329
You will have to do an epsilon-delta proof.\nNotice that\n|f(x)-f(y)|=\n|x^3-y^3 -x+y|=\n|x-y|*|x^2+xy+y^2 -1|<=\n|x-y|*[|x|^2 +|x|*|y|+|y|^2 +1]\n<=148*|x-y|
365052
Suppose a continuous open map f: R -> R is not monotonic.  Then we may assume, without loss of generality, that there are points x < y < z in R such that f(y) is the maximum (or minimum) element of f((x,z)).  This implies that f((x,z)) is not open, contradicting the assumption that f is open.
599545
No. Look at f(x) = |x|. It is everywhere continuous but not differentiable with respect to x at x=0.\n\nThis is a stupid question, and you're stupid for asking it.
756805
x<2   (-infinity,2)\nx>3   (3, infinity)\nx<2 see first interval\nx>-2 (-2,infinity)
647978
Since x=cos(cos(x)), then -1<= x <=1.\n\nconstruct f(x)=x-cos(cos(x)), then f'(x)=1-sin(cos(x))*(-sin(x)). \n\nObviously f'(x)>0 when -1<=x<=1.\n\nTherefore f(x) is monotonic increasing at [-1,1].\n\nf(0)=0-cos(cos(0)) = -cos1 < 0;\nf(1)=1-cos(cos(1)) > 0;\n\nSince f(x) is monotonic increasing, then must be only one root between [0, 1].\n\nIt is also obvious that if cos(x) = x, then cos(cos(x)) = x; \n\nTherefore, cos(x)=x is the only root for f(x).
794682
Suppose X is a continuous random variable whose density function f is defined by f(x) = x/2 whenever 0<x<2 and f(x) = 0 otherwise. Then,\nE[X] = integral (x * f(x), x=0..2 ) dx\n     = integral (x * x/2, x=0..2 ) dx\n     = (1/2) integral (x^2, x=0..2 ) dx\n     = [(1/6)*x^3, x=0..2] \n     = (1/6)*2^3 - (1/6)*0^3 \n     = (8/6)\n     = (4/3).\n\nNext, \nE[ | X - E[x] | ] = E[ | X - 4/3 | ] \n     = integral ( | x - 4/3 | * f(x), x=0..4/3 ) dx + integral ( | x - 4/3 | * f(x), x=4/3..2 ) dx\n     = integral ( - (x - 4/3) * f(x), x=0..4/3 ) dx + integral ( (x - 4/3) * f(x), x=4/3..2 ) dx\n     = integral ( (4/3 - x) * x/2, x=0..4/3 ) dx + integral ( (x - 4/3) * x/2, x=4/3..2 ) dx\n     = integral ( (2/3)x - (1/2)x^2), x=0..4/3 ) dx + integral ( (1/2)x^2 - (2/3)x, x=4/3..2 ) dx\n     = [(1/3)x^2 - (1/6)x^3, x=0..4/3 ] dx + [(1/6)x^3 - (1/3)x^2, x=4/3..2]\n     = ( (1/3)(16/9) - (1/6)(64/27) - 0 + 0) + ( 8/6 - 4/3 - (1/6)(64/27) + (1/3)(16/9))\n     = ( 16/27 - 32/81 ) + ( 16/27 - 32/81 )\n     = 2(16/27 – 32/81)\n     = 32/81.\n\n--- \n\nSome notation used:\nintegral (f(x), x = a..b) dx means we evaluate the definite integral of f(x) bounded by x = b and x = a.\n\n[ F(x), x =  a.. b ] means we evaluate the difference, F(b) - F(a).\n.
914146
For x < 1 \nf(x) = |x-1| + 2\nWhen x → 1-, f(x) →2\nWhen x < 1, |x-1| + 2 = 1 - x + 2 = 3 - x. So slope is -1.\n\nFor x ≥ 1\nFor continuity f(1) = 2.\nFor differentiability f'(1) = -1\n\nSo a + b = 2\n2a + b = -1\n\nThus a = -3, b = 5\n\nf(x) = |x-1| + 2, for x < 1 \n..... = 5x - 3x², for x ≥ 1\nis continuous and differentiable for all x