how do i sketch this function f(2-x)??
958895
If you mean y = 2 - x then theabove answers are right.\n\nI'm wondering, though, if you might mean you have a graph of some function f(x) and want to draw a graph of the combined function g(x) = f(2 - x). If so you simply flip the original graph around the y axis (this gives you -x) and then shift the whole thing 2 to the right (this is the 2). You can check individual points if you want to make sure: g(2) = f(0), g(3) = f(-1), ...
712712
[on] [y=] [(] [x,t,o,n] [+] [2] [)] [/] [x,t,o,n] [enter] [5] [-] [2] [x,t,o,n] [enter] [(-)] [2] [x,t,o,n] [-] [3] [enter] [window] [(-)] [5] [enter] [5] [enter] [1] [enter] [(-)] [5] [enter] [5] [enter] [graph]\n\nYou should have on screen a vertical asymptote at x=0, horizontal asymptote at y=1, a curve going to negative infinity towards 0 on the left, a curve going to positive infinity towards 0 on the right, on the left tangent line y=-2x-3 touches point(-1,-1), and on the right tangent line y=5-2x touches point(1,3).
578027
f(x)= 7x-6\nf(0)= 7(0)-6= 0-6= -6\nf(1)= 7(1)-6= 7-6= 1\nf(-2)= 7(-2)-6= -14-6= -14+(-6)= -20\n\nJust pretty much substitute for x with each value.  Hope this was of some help to you.
1005272
f(4) means 'rewrite f(x) = 2x - 5 with a "4" instead of an "x"', like this:\n\nf(4) = (2 * 4) - 5\n=8 - 5\n=3
779526
How are you even sure that such a function exists? the numbers look very random.
853022
It should be a function because one x input value gives only one y output value. Try the vertical-line test. But the function does not have an inverse though.
813038
f(x) = x^2 / (x - 2 ) ^2?\n\n    (a) Find the vertical and horizontal asymtotes\nx=2\n    (b) FInd the intervals of increase and decrease\nfind f'(x) = ( (x - 2 ) ^2 2x - x^2 2(x-2) )/(x - 2 ) ^4\n= ( x-2) ( (x-2)2x-2x^2) / (x - 2 ) ^4\n= ( 2x^2 -4x -2x^2 ) / (x - 2 ) ^3\n= -4x/ (x - 2 ) ^3  \n=0\nif -4x =0, i.e. x=0 one critical point.\n    (c) Find the local and maximum and minimum values\n    (d) Find the intervals of concavity and the inflection points
333191
Consider the function f as mapping the points x of a set X to the points y of a set Y. Draw a horizontal line (the x axis) so that each point on the line represents an element of the set X. Draw a perpendicular line (the y axis) to intersect the x axis at the null point or zero. Each point on the y axis represents an element of the set Y. \n\nFor each x in X calculate the value of y= f(x) in Y. This gives a pair of co-ordinates (x,y). (Depending on the function, you may have more than on value of y for each x eg the function f(x)=square root of x will give + and - values.)\n\nFor any x on the x axis draw a line parallel to the y axis to intersect the x axis at x. Draw another line to intersect the y axis at the point y=f(x) and parallel to the x axis. Where these lines intersect is a point on your graph. The complete graph consists of all such points.\n\nWhere the set X is infinite (eg the set of real numbers) you cannot calculate the value of f(x) individually for every x but by calculating the values for a series of points at suitable intervals you will see a pattern emerging to the graph and you will be able to fill in the gaps by joining the (x,y) points, hopefully with a smooth and continuous line.
918808
wouldnt it be like 3<x<-3 f(x)\nso like x would be 1 so it would be just f if you really want to simply that far done.
530638
only could work with known answeres.\nassuming f== pt i,2 , f*==-1.-2, x=0, mid point.\ndf*-1/2dx(2)= df, because the equation blances out the negotive and positve,\nf(x) =x*5 + 2x -1,x as 0. f(0) = o*8+2(0) - 1 answer points are -i,1theni culdn't figure out how to find the answers for line y,x. or the solpes\nthe last one i tried was\ndf*-1/dx (2)  i got =0 df = points 1,2 + -1/2 cancels out the equation.\n\ni thnk, maybe, might be so? could you e-mail me the correct answer?
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.
791940
In order to a od function the function have to obay:\nf(x) = - f (-x) <=>\n-3x^9 + 4x^5 = -(-3(-x)^9 + 4(-x)^5) <=>\n-3x^9 + 4x^5 = 3(-x)^9 - 4(-x)^5 <=>\n-3x^9 + 4x^5 = -3x^9 + 4x^5 ( True)\nthe function is od.
80570
f(x) = x^2 - 1\nIf you replace x with x + h, then\nf(x + h) = (x + h)^2 - 1\nYou will NOT do any operations when changing the function. Instead, just SUBSTITUTE. (A technique is putting parentheses in the variable to change, except if it already has its own parentheses. Then you may exchange the variables inside the parentheses with what you want to substitute)\nf(x) = x^2 - 1\nPutting the parentheses, we get\nf(x) = (x)^2 - 1\nThen replace x with x + h\nf(x + h) = (x + h)^2 - 1\nHope this helps ;-)
187139
Yes (-1,0) and (1,-2) share the tangent line y=-x-1\n\nwhich y=-x-1 is tangent to the function y= f(x)=x^4-2x^2-x\n\nNote that (-1,0) and (1,-2) satisfy y=x^4-2x^2-x
840688
Not sure what you were doing. I would rewrite f(x) in this form:\n\nf(x) = (2x-1)^(-1/5)\n\nThen you're dealing with a u^n situation, so\n\nf'(x) = (-1/5)(2x-1)^(-6/5)*(2)\n\nf'(x) = (-2/5)(2x-1)^(-6/5) \n\nThat's equivalent to what you had, but a little simpler.\n\nLet the underboss know that this is a "site" and I did answer your question.
916581
To write any equation in "function form", just solve for y (get y by itself on one side of the =), and replace y with f(x).\n\nI'll let you do that.  :)
361536
Let's call f(x) = x^2 - 4\n\nThe graph of f(x) = x^2 - 4 looks like a smile ( U )  with the graph passing the y - axis at -4.\n\nThe roots are at x = 2 and x = -2.  This can be found by solving for x^2 - 4 = 0\nx^2 - 4 = 0\n(x + 2)(x - 2) = 0\nso x + 2 = 0, or x - 2 = 0\nso x = -2 or x = 2\n\nsince you're int'ing from x = -1 to x = 3, you have to break up the int into\n[x = -1 to x = 2, where f(x) is negative and take the absolute value] \nand\n[x = 2 to x = 3, where f(x) is positive]\n\nso int of f(x)\n= x^3/3 - 4x + c\n\nso int of f(x) from x=-1 to x=3\n= {+ve value of int of f(x) from x = -1 to 2}\n                +\n  int of f(x) from x = 2 to 3\n\nSeparately\nint of f(x), x = -1 to 2\n= | (2^3)/3  - 4(2) - [((-1)^3)/3  - 4(-1)] |\n= | (8/3) - 8 - [(-1/3) + 4] |\n= | (8/3) - 8 + (1/3) - 4 |\n= | (9/3) - 12 |\n= | 3 - 12 |\n= | -9 |\n= 9\n\nint f(x), x = 2 to 3\n= (3^3)/3  - 4(3) - [(2^3)/3  - 4(2)]\n= 27/3 - 12 - [(8/3) - 8]\n= 9 - 12 - (8/3) + 8\n= 5 - (8/3)\n= 7/3 (or 2 1/3)\n\nAdd them together and you get:\n9 + 2 1/3\n= 11 1/3 which is appromixately 11.333_