What is the smallest positive value for X where Y = Sin 2X reaches it's maximum?
831435
The graph of sin x goes up to 1 when x = pi/2. When you graph y = sin (2x), it shrinks the x-values by half - so it will be 1 when x = pi/4.
909799
gopal's solution:\n\nf(x)=x^3-6x^2+9x+2\nf'(x)=3x^2-12x+9=0 for extremas\n3(x^2-4x+3)=0\n3(x-3)(x-1)=0\nx=3 or 1\nf''(x)=6x-12\nat x=1 6(1)-12=-ve so max\nat x=3 6(3)-12=+ve so min\nso the absolute maximum is at x=1 and\n=6\nabsolute min is at x=3 and is=2  <------ However still need to check endpoints of the interval [-1,4] as these turnng points are actually LOCAL maxima and minima!\n\nSo f(-1) = -1 - 6 - 9 + 2 = -14 < 2\nSo ABSOLUTE minimum = -14 at x = -1\n\nAlso f(4) = 64 - 96 + 36 + 2\n= 6\nSo ABSOLUTE maximum = 6 and occurs at both x = 1 AND at x = 4
749429
x = -7\n\nMultiply two negatives and you get a positive
833635
2)\nx and y can increase infinitely as long as the value of x is 1 more than the value of y
427245
You can only take the log of numbers that are positive, so\n\nx^2 + 6x + 5 > 0\n\nSo (x+5)(x+1) > 0\n\nSo x would have to be greater than -1 and less than -5.
869289
f(x) = x² + sinx\nf'(x) = 2x - cos x\n= 0 for stationary points\nie 2x - cosx = 0\nBy Newtons method\nX2 = X1 - F(X1)/F' (X1)\n= X1 - [(2X1 - cos(X!)]/[2 + sin(X1)]\n\nStarting at x = 1 you get x = 0.450184 using Excel:\n\nx ..................................... f(x) ......................... f'(x)\n1 .......................... 1.459697694 ......... 2.841470985\n0.486288017 ......... 0.088502303 ......... 2.467347449\n0.450418605 ......... 0.000572265 ......... 2.435342427\n0.450183622 ......... 2.4856E-08 ........... 2.435130868\n0.450183611 ......... 0 .......................... 2.435130859\n\nSo to 6 decimal places x = 0.4501836\n\nAs f"(x) = 2 + sinx > 0 for all x; then x = 0.4501836 gives the absolute minimum value\n\nSo absolute minimum value = (0.4501836)² + sin(0.4501836)\n= 0.637796 (to 6 decimal places)
793658
min of f(t) = 0 when t = 0 and t = 8\nmax occurs at t=2\n\nI assume you mean the maximum and minimum of the absolute value of f(t). If that's the case then the minimum of f(t) occurs when t= 0 or 8 where f(t) =0\n\nTo determine the max we find when the derivative f ' (t) = 0 \n\nlet g(t) = cuberoot(t) = t^(1/3) and h(t) = (8-t)\n\nThe derivative of f(t) (i.e. f ' (t)) is given by\n\nf ' (t) = g'h+h'g  (the product rule)\n\nf ' (t) = (8-t)(1/3)t^(-2/3) - t^(1/3)\n\nThe max, min and inflexion points occur when f ' (t) =0\n\nt^(1/3) = (8/3)t^(-2/3) - (1/3)t^(1/3)\n\n(4/3)t^(1/3) = (8/3) t^(-2/3)\n \nt =2\n\nf(2) = 7.56 apprx is the max
956637
Sorry tried a lot, searched a lot & surfed a lot but found no solution, I think Pascal is right.\n\nP.s- Saket it is ∫e^sinx dx & not ∫(e^x)*sin x dx, if it was that, it would have been trivial.
471125
Hi!\nI'm suprised nobody used the easiest way that i will demonstrate now.\n\nLets assume further that x is positive real.\n\nWe start with a simple fact that ANY squares are nonnegative. The idea is we "choose" a square intelligently to suit our purpose. Also not that x does NOT equal zero.\n\nBy experience, \nI choose the square ( x^(1/2) - 1 / [ x^(1/2)] )^2.\nIt must be nonnegative for whatever positive real values of x.\nThus,\n\n    ( x^(1/2) - 1 / [ x^(1/2)] )^2 >= 0\n\nExpanding and regrouping, we get,\n\n    x - 2 + 1/x >= 0 \n      \nor equivalently,\n\n      x + 1/x >= 2\n\nThus, minimum is 2. \nWith equality when x^(1/2) = 1 / [ x^(1/2)], or\n x = 1.\n\nDone.\n\nAlternative Method.\n\nOne can use two variable AM - GM,\nFor any two positive reals m and n, we have\n\n ( m + n )/ 2 >= (mn)^(1/2)\n\nWith equality if and only if m = n.\n\nJust set m = x and n = 1/x, to get,\n\n( x + 1/x )/ 2 >= 1\n\nor  x + 1/x >= 2\n\nAgain, the minimum values is 2 with equality if and only if x = 1 .\n\nRegards
919933
I assume this is equal to y?\n\nIf it is, it's definitely a maximum because it's a parabola that opens downward\n\nNow I'll work on the vertex
1183576
just pick two points from the graph\nif you wanna check plug them in the equation\n\n\n\nand when the line crosses the Y axis Y does NOT =0\nX=0 when line crosses Y axis\n\nchoose any set of #s that will not give you the square root of any negative. because that would result in an imaginary #, not a real
709175
If you need a positive result, then y will only be a real number when x - 1 is >= 0.\n\nx - 1 >= 0\nx >= 1\n\nSo x must be greater or equal to 1.
68097
do you mean things as far as max and mins, asymptotes and things like that?\n\nif you are a little more specific i can help you better, or e-mail me, it would be a little easier  matttlocke@yahoo.com
612056
the local maxima can be found at the points where dy/dx is equal to zero and (d^2.y/(dt)^2) is less than zero. The local minima can be found at the points where dy/dx is equal to zero and (d^2.y/(dt)^2)(read as dee sqared y by dee tee squared) is greater than zero.\n\nThe explanation is that if u were to draw the graph of watever function u want to find the maxima and minima of,.... at the point where its max, the tangent is zero and the differential of the tangent is sloping downward!! And it slopes upward for the point of minima....\n\nHope this helps.\n\nYou better gimme best answer for this.
725972
You mean the graph of the function?  The second term is a sine wave with frequency 2x, meaning it goes through a full cycle in pi radians instead of 2*pi radians.  The max amplitude of the second term is +/- 1/2, since the max of the sine is +/- 1.  That's the second term.  Now, offset that by 4 units down at every point.  Got it?  Okay, your turn.
1081487
A parabola opening down with a max at the point 0,-2.  it will also be twice as wide as a standard parabola
879179
There is no limit.\n\nYou can see this by approaching from two different directions, along the x-axis and along the y-axis.If x is zero and y->0, then the formula becomes:\n\n   (siny)^2/y^2\n\nSince sin y/y -> 1 as y->0, this also approaches 1 as y approaches zero.\n\nOn the other hand, if y=0, and x->0, then the formula becomes:\n\n    x^2/(2x^2)\n\nwhich has a limit of 1/2 as x->0.