I need help with a function question...?
355199
Im not entirely sure if this is correct but, I know that a relation is a partial ordering if it is reflexive, transitive and anti-symmetric.  \n\nSo to satisfy the reflexive part we need that x>=x, which is trivially satisfied by f(x)>=f(x).  For the transitive part we need that if x>=y and y>=z then x>=z, which will be true if for f(x)>=f(y) and f(y)>=f(z) then f(x)>=f(z).\nFor the anti-symmetric part we need that if x>=y and y>=x then x=y, so when f(x)>=f(y) and f(y)>=f(x) then x=y\n\nSo this would suggest that the necessary and sufficient conditions on f are that: (1) when f(x)>=f(y) and f(y)>=f(z) then f(x)>=f(z).  (2) if f(x)>=f(y) and f(y)>=f(x) then x=y.  So f would appear to be one to one.\n\nI hope this helps
612325
length = 9\nwidth = 9
635332
It sounds like a function that was expecting two or more parameters, was only passed one.\n\nfunctionx(x,y)\n\nwas only passed 1 parameter:\n\nfunctionx(x)\n\nFind the function and give it what it needs!\n\nGood luck.
921732
If you look at a picture (aka a graph) of the function, it will all become clear...\n\nhttp://upload.wikimedia.org/wikipedia/en/3/3e/1-over-x.png
769871
1.  The function decreases on the interval [-infinity,4].  (Your function is a parabola opening upwards, with one bottom vertex at the point (4,0).)\n\n2.  This function is also a parabola opening upwards, with its vertex at (2,-4).  On the interval [-infinity,4], the local min. is at (2,-4) and there is no max.\n\n3.  You have local max at x = -13/3, and local min at x = 1\n\n4.   f(x) is increasing on [-infintity,1], decreasing on [1,5], and increasing again at [5,infinity]\n\nHope that helped.  Email me if something's not clear!\n\n.
860906
it is called akermen function. search web about this
779791
(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!
1055471
You need to graph this function. \nIt is a line.\ny = (x^2)^ 1/2 - 16 \ny = x - 16\nx = 0, y = -16, \nx = 16, y = 0
1129152
The answer is 10\nYou find the average rate of change of the function f(x) over a given interval, by computing the total change in the function f(x) over the interval, and then dividing this value by the size of the interval.\nYour interval size is 2- (-1) = 3. So you want:\n\n(f(2) - f(-1)) / 3  = (22-(-8))/3 = 30/3 =10\n\nf(2) = 2(2^3) - 2^2 +5*2 = 16-4+10 = 22\nf(-1) = 2(-1^3) - (-1^2) -5 = -2-1-5 = -8
781968
I'm not sure if your question meant f(x)=2x(x-1)(x-4) or f(x)=x(x-1)(x-4)^2, so I will answer both.\n\n1) f(x)=2x(x-1)(x-4)\n(0, 1) U (4, ∞)\n\n2) f(x)=x(x-1)(x-4)^2\n(-∞, 0) U (1, 4) U (4, ∞)
659256
I don't understand the question. Can you please type it over so I can understand it? Thanks
1149923
(2,-4) if f(x) is defined for x=2
437573
Functions that maps x to x\nWhere x is an element of the function
1055273
Think about your order of operations. If you were evaluting the function for some number, what would you do first? That function is your *inside* function.
887664
let u = x^2\ndu = 2x dx\n\nThis turns the function in to:\n1 / (u^2 + 1) - which you can work from.
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
737185
How about something simpler, like\n\nf(x) = 0 when x<1/2, 1 when x>=1/2\n\nThen the limit from the left is 0, the limit from the right is 1, so the regular limit doesn't exist.