Find all numbers c between a and b for which:\n(f(b)-f(a))/(b-a) = f'(c)\n\nThe Problem:\nf(x) = (x)arcsin(x)    on [0,1]\n\nAny help would be greatly appreciated. Please explain steps also.
129851
Let's start off with a definition.\n\nThe Mean Value Theorem is one of the most important theoretical tools in Calculus. It states that if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that \n f '(c) = ( f(b) - f(a) ) / ( b -  a)  \n\nOK.\nSo in our problem here, a = 0 and b = 1. Thus,\n    f(a) = f(0) = 0\nand f(b) =  f(1) = arcsin(1) = (pi)/2\n\nf '(c) = ( f(b) - f(a) ) / ( b -  a)  \n      \n       = ( f(1) - f(0) ) / ( 1 -  0)\n\n       = ( (pi)/2 - 0 )\n   \n       = (pi)/ 2\n\nNow, \n  f'(x) = x [ 1 / (1 - x^2)^{1/2} ] + arcsin(x)\n\nusing the fact if g(x)  = arcsin(x) Then, \n                  g'(x) = 1 / (1 - x^2)^{1/2}   \nand the multiplication rule in differentiation.\n\nHence, f'(c) = c [ 1 / (1 - c^2)^{1/2} ] + arcsin(c)\nand f'(c) = (pi)/2\n\nNow, note another fact that \narcsin(x) = x / (1 - x^2)^{1/2} \n\nHence, \n\nc [ 1 / (1 - c^2)^{1/2} ] + c / (1 - c^2)^{1/2} = (pi)/2\n\nThat is,\n\n       ( 2c ) / ( 1 - c^2 )^{1/2} = (pi)/2\n\nManipulating using algebra, we get,\n\n       ( 1 - c^2 ) = [ 16 c^2 ] / [(pi)]\nThus,\n        c^2 = [ (pi) ] / [ (pi) + 16 ]\n\nFor our answer for the value c, we only take the positive square root because, the question asked for c in the interval [0,1]. \n\nHence, using a calculator, you should get...\n\nc = 0.405122068...\n\nHope this helps,\nRegards
461696
I am guessing your function to be one of the following:\n(1) f(x) = ½x - 3 or (2) f(x) = -½x -3\nIn function (1):\nf(0) = ½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = ½(2) - 3 = 1 - 3 = -2 ... or... (2, -2)\n\nIn function (2):\nf(0) = -½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = -½(2) - 3 = (-1) - 3 = -4 ... or ... (2, -4)
556109
nested function\nf(g(x))=f'(x)g'(x)\ng(x) = 7x\nf(x) = ln [7x]\n\nf'(x) = 7\ng'(x) = 1/x\n\nd(ln 7x)/dx = 7/x\n\nNow that I think about it, I'm not sure I like that answer either.  Maybe it's 7/7x or 1/x.
259902
I do not know about F.But I know (x).Does she have something like(3,4)(3,5)?\n\nx_|y_ <-- or does she have that?\n3, 4\n\ny=2f(x)\n4=2f(3)                          \n\n      ^\n      That's what she is supposed to do.\n      To solve the function.\n     But still i am sorry about not knowing F.\n    (Those # is example)
375415
1. f(x)= 3x^4 + 4x^3, -2 <_ x <_2\nf'(x)= 12x^3 + 12x^2 = 0\n12x^2(x+1)=0\nso the critical points are: x=0 or x=-1\nf''(x)= 36x^2 +24x\nf''(0)=0, so this test is inconclusive.\nf''(-1) = 36 -24 >0 so -1 is a local minimum.\n\nNow, since you want an absolute max and min over the closed interval, you need to compare the values of f at the end points with the values of f at the critical points:\nf(-2) = 3(-2)^4 +4(-2)^3 = 16\nf(2) = 3(2)^4 +4(2)^3 = 80\nf(-1) = 3(-1)^4 +4(-1)^3 = -1\nf(0) = 0\nthe largest value happens when x=2, which means this is the abs. maximum over the interval\nand the smalles value happens when x=-1, which means this is the abs. min over the interval.\n\n'
476094
You know that f(x) = 1 + 1/x\nf(1/x) = 1 + 1/(1/x) = 1 + x\nf(x) + f(1/x) = 1 + 1/x + 1 + x\n= 2 + 1/x + x\n\nf(x) * f(1/x) = (1+ 1/x)(1+x)\n= 1 + x + 1/x + 1\n= 2 + 1/x + x\n\nThus, it has been proven.
608780
You have that f(0) = 0, f(2) = 0, f(5) = 0, f(8) = 0, so you have two points of 0 that are 2 apart, so the period length divides 2, and also you have two that are three apart, so the period length divides 3.  This means that the period length must divide 1.  So we can just let the period length be 1.  So let f(x) = a*sin(pi*x).  This will work for any a.
174019
1) [1/x] <= 1/x, for all x>0\n2) [1/x] = 1/x, iff 1/x is an integer, so x=1/n for some integer n.\n3) [1/x] < 1/x, for all x != 1/n for any n (x>0).\n4) So, f(x) = [x*[1/x]] = [x*1/x] = [1] = 1 for all x=1/n for some integer n.\n5) Also, x*[1/x] < x*1/x = 1 for all x != 1/n for any n (x>0),\nthen f(x) = [x*[1/x]] = [<1] = 0 for all x != 1/n for any n (x>0).\n\nlim f(x)(x->a) = b, iff for ALL sequences {a_n->a} f(a_n)->b.\n\nLet's take 2 sequences:\n{a_n = 1/n} f(a_n)=1 for all n (see 4), so f(a_n)->1.\n{b_n = 2/(2*n+1)} f(b_n)=0 for all n (see 5), so f(b_n)->0.\nEasy to see that {a_n}->0, {b_n}->0.\n\nThus, there is NO limit of f(x) when x->0.\n\nPS: "iff" means "if and only if"
210187
Use any recursive algorithm.\n\nFor instance:\n\nf(x) = f(x) + 1
464650
a)  f(0) = f(0-0) = f(0) - f(0) = 0.\n\nb)  Fix k in the reals, and define f(x) = kx.\n\nc)  Suppose you have f(x)=f(y) for some reals x and y.  Then f(x)-f(y) = 0, and since f preserves subtraction, f(x-y) = 0.  Since 0 is the only element which gets sent to zero, this implies that x-y = 0, and thus x = y.\n\nI have a challenge for you: find a nontrivial function f which preserves subtraction but is not injective.  There are such functions, but they are not necessarily easy to find.  Cheers!
594476
f(x): 'f' is usually a defined function.\n\nf(x) = |4|, would mean the the function returns the absolute value of 'x', which in this case is four.\n\nei: f(-4) = 4. If function is defined as above. written out would be in some programing languages would be f(abs(x)).\n\nf(x) can be any formula you want to define it as.\n\nf(x*y) or f(sin(x))....\n\n\nreally unsure of the how you have stated the problem.
401471
(f(2) - f(0))/(2 - 0) = 2, but f'(x) < 0 for every x, except x=1 where it is undefined.\n\nThe function has to be continuous in [0,2] and differentiable in (0,2).  For this f(x), it is undefined at x=1, so the conditions of the MVT are not met.  Math is still safe!\n\nCheers!
571284
Please explain what is ~f.  I have curiosity, because neither inverse multiplicative, complex conjugate or inverse function could do that result...\n\nIt should be interesting!! Please post what is ~f !
451578
f(x) ≥ -9, or [-9,  ∞)\nThe function has a minimum at (-1, -9), and has no upper bound.\n\n{calculus derivation of the minimum:}\nf '(x) = 2x + 2\n2x + 2 = 0\nx = -1 defines the minimum point\nf(-1) = (-1)^2 + 2(-1) - 8 \n= 1-2-8\n= -9
648581
Okay this is one of those problem where you are kinda working your way backwards so here we go:\n\nFirst recall that a point of inflection is where a fuctions second derivative equals zero, in our case we are told this happens at x=3. Okay so first lets find the second derivative:\n\nIm gonna rewrite  f(x) =square root{x+1} + b/x\nas f(x)=(x+1)^(1/2) +bx^(-1)               recall that a square root is a power of 1/2 and having an x in the denominator is having it eing raised to a negative 1\n\nOkay so f'(x)=1/2(x+1)^(-1/2) - bx^(-2)\n\nand f''(x)=-1/4(x+1)^(-3/2) + 4bx^(-3)\n\nNow using the fact that f(x) has an inflection point at x=3, meaning that f''(x)=0 when x=3, we can substitute those two things into the above equation. Here we go:\n\nf''(x)=-1/4(x+1)^(-3/2) + 4bx^(-3)\n0=-1/4(3+1)^(-3/2) + 4b(3)^(-3)      Now simplify to solve for b\n0=-1/4(4)^(-3/2) + 4b(1/27)\n0=-1/4(1/8) +4/27b\n0=-1/32 + 4/27b\n\n1/32=4/27b\n1/32(27/4)=4/27(27/4)b\n27/128=b    \n\nThere it is, if my multiplication and stuff doesnt fail me b=27/128.\n\nHope that helped!
317639
1) f'(x) = 40x^4+6x^2+4\nTo find local max's and min's, set this equal to zero, but you will see that all solutions will be imaginary, so there are no local max's and min's on the real line.\n\nf''(x) = 160x^3+12x\nTo find the points of inflect, set this equal to zer, but you will see that the only place where there is an inflection is a x = 0.\n\nSo what have we learned?  f(x) is a strictly increasing function, meaning tha there could only be one real root.\n\n2. Let u = 1+t and use substituion.
74211
1) What is the difference between F(n) and f(n)? \n\n2) I cannot comprehend the second condition.\n\nEdit your question, please.