Give the function f(x) = 3(x-1)^2, identify the value of f(-3/5) and the same for f(.25)?
138578
You should do your own homework, or simply ask for how to do it, not the answers.
618071
simplify to f(x)=(sqrt(x-2)(x-3))/(x-4)(x+5)\nthen sub in 1/x\nto get (sqrt(1/x -2)(1/x -3))/(1/x -4)(1/x +5)
1082751
1)   f(x) = x+1/x =(x^2+1)/x  \n      g(x) = x-2/x =(x^2-2)/x\n(f+g)(x) = (x^2+x^2+1-2)/x = (2*x^2-1)/x = 2*x -1/x\n\n2)  f(x) = -2/x\n     g(x) = (x+1) / (x-3)\n(f+g)(x) = -2/x + (x+1)/(x-3) = {-2*(x-3)+x*(x+1)}/x*(x-3)\n              = (-2x+6+x^2-x)/(x^2-3x) =(x^2-x+6)/x^2-3x
1149923
(2,-4) if f(x) is defined for x=2
754182
Put 2 in the function to get the first iteration\nf(2) = 1 / (1-2) = -1\nPut this result back in the  function to get the 2nd iteration.\nf(-1) = 1/(1 - (-1)) = 1/2\nDo it again for the 3rd iteration\nf(1/2) = 1/ ( 1 - (1/2)) = 2\n\nThis is what you started with.  So it will just repeat the same pattern every 3 iterations.\n\n999 is a multiple of 3 so this will be the same as iteration 3 which was 2.  iteration 1000 will be -1.
889441
Hello dear Neil \n\n5x^2+3x+1=0\n\n● Step 1 \nif F(x) = ax^2 + bx + c \n∆ = b^2 - 4(a)(c) & ∆ ≥ 0\nF(x) = 5x^2+3x+1\na = +5 , b = +3 , c = +1\n∆ = (3^2) - 4 (5)(1) \n∆ = 9 - 20 \n∆ = -11 < 0\nwell this function has so real solution\n\n● Step 2 \nx1 = 1/10 (-3 - i √11)\nx2 = 1/10 (-3 + i √11)\nso \nx 1 = -0.3 - 0.331662i\nx 2 = -0.3 + 0.331662i\n\nSo you are pretty right .\nGood Luck Neil ♣
649843
Does what you have written means f(x) = 4*x - 5/(16*x^2) - 25 or f(x) = 4*x - (5/16)*x^2 - 25? Make it clear.
965701
f(x) = x(x+2)(x-2)(x-4)\n\nMultiply it out if you need it in general form.\n\nThere could also be a constant in front of the x.
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 ;-)
881627
f(x)=x^3+5x^2+3x+1\nf'(x)=3x^2+10x+3\nsetting it to zero\n(3x+1)(x+3)=0\nx=-1/3 and x=-3\nf''(x)=6x+10\nat x=-1/3 f''(x)=+ve and so min\nat x=-3 f''(x)=-ve and so max
794664
1) If h(x) = f(g(x)) = (x^3 -1)/(x^3 +1), then \nf(x) = (x-1)/(x+1) and g(x) = x^3\n2) If h(x) = f(g(x)) = 3sqrt(x^2 -8), then\nf(x) = 3sqrt(x) and g(x) = x^2 -8
115638
You can't.  f(3) = 2 doesn't fully define the function. Right now, you only have one point, (3,2). You need at least one other point, and then you can calculate the slope, assuming f(x) represents a line, by taking the difference in the values of f(x) and dividing it by the difference in the values of x.  Say for example you're also told f(5)=8.  Then the other point is (5,8), and the slope is (8-2)/(5-3)= 6/2 = 3.\n\nNow, if you mean the line f(x) = 2, the slope of that line is 0, i.e. a horizontal line, since the function is always 2 regardless of what x is.\n\nIf f(x) does not represent a line, iow you're doing a calculus problem, then you need to take the derivative of f(x), evaluate it at the point x=3, and that is the value of the slope of the tangent line at that point.
166642
First subsitute -1 wherever you see x\n\nf(-1) = 4(-1) - 3\n      = -4 - 3\n      = -7
533058
Point slope form:\n\ny-(-3)=(1/2)(x-5)
554760
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)
818927
if you want f(3)\nyou will have f (2 +1) which leads us to know that 2 is x...\nsubstitute all x's with 2's and then your answer will be revealed
809444
ohh ya,,i know this,,\n\n1)f(x)=((x)^2)+2\n\n2)f(x)=( ( (x-3)^(-2) ) +2)\n\nbring on more of domain and range questions,,i love those problems,,