If m,n >=2 are integers, find and classify the critical points of f(x) = x^m(1-x)^n.?
824813
f(x) = x^m(1-x)^n\nm zeros at  x = 0 and n zeros at x = 1 \nSo x-axis is tangential to the curve at x = 0 and x = 1\n\nf'(x) = mx^(m - 1)(1 - x)^n + x^m * -n(1 - x)^(n - 1)\n= x^(m - 1) * (1-x)^(n - 1) * (m(1 - x) - nx)\n=  x^(m - 1) * (1-x)^(n - 1) * [m - (m + n)x]\n= 0 for stationary points.\n\nSo (m - 1) zeros at x = 0, (n - 1) zeros at x = 1 and a zero at x = m / (m + n)\n\nNow there has to be a turning point between x = 0 and x = 1 as f(x) is a polynomial and therefore continuous and differentiable for all x\n\nSo examine the possibilities\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... - .......... + ........... + ... Stationary inflexion at (0, 0)\n(1 - x)^n .............. + .......... + ........... - ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ....................... - .......... + ........... - ... Stationary inflexion at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... - .......... + ........... + ... Stationary inflexion at (0, 0)\n(1 - x)^n .............. + .......... + ........... + ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ...................... - .......... + ........... + ... Local minimum at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m ..................... + .......... + ........... + ... Local minimum at (0, 0)\n(1 - x)^n ............... + .......... + ........... - ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ....................... + .......... + ........... - ... Stationary inflexion at (1, 0)\n\n\n... Domain → ... x < 0 ... 0<x<1 ... x>1 \nm .... odd    \nn ..... odd    \nx^m .................... + .......... + ........... + ... Local minimum at (0, 0)\n(1 - x)^n .............. + .......... + ........... + ... Local maximum at (m/(m+n), f(m/(m+n))\nf(x) ...................... + .......... + ........... + ... Local minimum at (1, 0)
504243
I'll assume you want n to be a positive integer. Consider F[a], the field obtained by adjoining a. Then a^n is in F[a] and the index of F[a] over F is finite. But every element of a field of finite index over F will be algebraic over F.
778163
The problem is to show that f is identically 0 using the density of the rational numbers. I would strongly suggest to the asker that they do their own homework at this level. If you don't understand the questions or techniques, you will have a *lot* of trouble later on.
785253
So are you looking for a function with a singularity on (0,1)?  That would address the question I think.\nmaybe (x-(1/2))^(1/2)\n\nTHE FIRST ANSWER IS WRONG, I THINK, BECAUSE YOUR DOMAIN IS OPEN AND DOESN'T INCLUDE 0.
877408
1/[x-m]+1/[x-n]=2/x\nx[x-n]+x[x-m]=2[x-m][x-n]\nx^2-nx+x^2-mx=2[x^2-mx-nx+mn]\n2x^2-[m+n]x=2x^2-2[m+n]x+2mn\n[m+n]x=2mn\nx=2mn/[m+n]
892605
(constant)/(2^n) when n->infinity => the limit is 0(zero)\n\nPS\nNot to be a a stickler for rules but is this really laptop related ? :))
811402
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,
532934
INT[(1 - x)/x dx]\nINT[x^-1 (1-x) dx]\nINT[x^-1 - 1 dx]\nln|x| - x + C
499879
I'm trying to guess what the problems you were asked to do were, because the information you gave in posing the question was incomplete.\n\nIf f(x) = log(x)  (for log of any base)\n\nthen\n\nf(1/b) = log(1/b) = log(b^(-1))=-log(b)=-f(b)\n\n\nlog(MN^2) = log(M) + 2 log(N)\n\nor\n\nlog((MN)^2) = log(M^2) + log (N^2) = 2log(M) + 2log(N)\n\nThese are applications of the laws of logarithms.
983225
remember the rule nx^n-1\n2x^2-1\n \nSo in this case it is 2x
717892
This question isn't clear. When you say critical points, what does that mean? This equation is = x(x - 4(1 - y)) + y(y - 16) if that's what you're asking.
1086560
f(-2) = 5. What's so difficult about this question?
1048436
As a^2 is contained in F(a), clearly F(a^2) is contained in F(a). So we need to show the reverse inclusion.\n\nWe have the chain of fields F < F(a^2) <= F(a).\n\n(where I use <= and < for subset and proper subset containment) \n\nand hence [F(a) : F] = [F(a) : F(a^2)][F(a^2) : F]. Set n = [F(a) : F] and m = [F(a^2) : F]. The preceeding equation implies that m divides n. It will suffice to that m = n. Let f(X) = X^m + c_{m-1}X^{m-1} + ... + c_1X + c_0 denote the minimal polynomial of a^2 over F. Then f(X^2) = X^{2m} + c_{m-1}X^{2(m-1)} + ... + c_1X^2 + c_0 has a as a root. This implies that the minimal polynomial of a has degree less than or equal to 2m. So we get m divides n and n <= 2m. The only ways for this to occur are when m=n or when 2m = n. But n is odd, so n does not equal 2m. Hence m = n. This completes the proof.
1146893
x^2+mx+1/x^2+x+1<3\n\ncase 1 : for x>0\n=> (x+1/x)^2 +(m+1)x <4\n=> (m+1)x < 4- (x+1/x)^2\n\nNow, we find the maximum value of the RHS:\nthe minimum value of (x+1/x)^2 is 2 when x=1 and for all other values of x it is >2 \nThis makes max(RHS)=0 or m+1 <0 \nor m<-1\n\ncase 2: x<0\nthe same argument as above(but x=-1)  leads us to m+1 >0 \nor m>-1\n\ncase 3: x=0\nThe given expression is undefined \n\nI guess its a home work problem, which indeed makes you lazy!
468001
x+2 must not be 0\nso x must not be -2\ndomain is \nall real numbers - (-2)\nall real no except -2
813081
a=1, a=0\n\nTO ANSWER YOUR EMAIL QUESTION:\n\na^m * a^n = a^(m+n)\n\nm+n ≠ mn for all m's and n's\n\nTherefore we need to look at the base, a.\n\nSince 1 to any power is always 1, we can state for certain that 1^m * 1^n = 1^mn for any m and n.\n\nSimilarly, since 0 to any power is always 0, we can state for certain that 0^m * 0^n = 0^mn for any m and n.\n\nHope that helped explain it a bit!
55379
I would say, \n\nonly in the case when n=0 or 1, then this equation stands good! If n is not equal to zero or one, then everytime the answer would definitely be above 1 (even when u use negative integers)