Given the function f(x)= 3(x-1)^2, identify the value of f(3).?
107384
f(3) = 3(3-1)^2\n= 3(2)^2\n= 3 x 4\n= 12
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
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
166642
First subsitute -1 wherever you see x\n\nf(-1) = 4(-1) - 3\n      = -4 - 3\n      = -7
974938
I believe this is how you do it:\n\n(f o g)(x) = f[g(x)] = 3(-x^2) - 4 = -3x^2 -4\n\nNow that this formula is found, sub in 2;\n\n(f o g)(2) = -3(2)^2 - 4 = -3(4) - 4 = -12 - 4 = -16
918705
Don't you see the pattern?\nIt's right in front of your eyes.  Let g(x) be what is in the parentheses.  So g(x) = x+2\nNow replace x+2 with just x, and that will be f(x).  \nSo f(x) = x^3 - 5x^2 + 3x - 1\nAgain, this is assuming that h(x) = f(g(x)) (based on your other problems)
915401
this means\n\nx^3 = -1\nor x^3+1 = 0\n\nf(x) = x^3+1\n\nf(-1) = 0\nso one root is -1\n\nfor the other root by division\n(x+1)(x^2-x+1)\nx^2-x+1 = 0 => x= (1+/-sqrt(3)i)/2
888735
Let f"(x) = 2x - 6. Then, \nf'(x) = integral ( f''(x) ) dx\n          = integral ( 2x – 6 ) dx\n          = 2 * (1/2)*x^2 – 6x + C\n          = x^2 – 6x + C.\n\nThus, \nf'(–3) = (–3)^2 – 6(–3) + C \n          = 9 + 18 + C \n          = 27 + C. \n\nBut f'(–3) = –4 as given above. So we can solve for C as follows\nf'(–3) = –4 = 27 + C.\n\nThis tells us that C = –4 – 27 = –31.\n\nTherefore, f'(x) = x^2 – 6x – 31.\n\nNext, we determine f(x). That is,\nf(x) = integral ( f'(x) ) dx\n          = integral ( f'(x) ) dx\n          = integral (x^2 – 6x – 31) dx\n          = (1/3)*x^3 – 6*(1/2)*x^2 – 31x + K.\n          = (1/3)x^3 – 3x^2 – 31x + K.\n\nSo we have, \nf(–3) = (1/3)*( –3)^3 – 3(–3)^2 – 31(–3) + K\n            = –9 – 27 + 93 + K\n            = 57 + K.\n\nSince f(–3)= –1, we equate this to the previous equation to obtain K, that is:\nf(3) = –1 = 57 + K.\n\nThis implies that K = – 58. Thus,\nf(x) = (1/3)*x^3 – 3x^2 – 31x – 58.\n\nMoreover, we have\nf(3) = (1/3)*(3)^3 – 3(3)^2 – 31(3) – 58\n          = 9 – 27 – 93 – 58 \n          = – 169.
741552
You have got it right.\nthe answer is\n\nf(x) = 3x
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)
936251
I think f^(-1) (x) means the inverse of f(x), not the reciprocal.\n\nLey y =f(x)\ny = (x^3 +1)^(1/3)\n\nsubstitute the x and y\nx = (y^3 +1)^(1/3)\n\nnow solve for y\nx^3 = (y^3 +1)\ny^3 =x^3 -1\ny =(x^3 -1)^(1/3)\n\nNow if you put 1.5 into this equation\ny = ((1.5)^3 -1) ^(1/3)\ny =(2.375)^(1/3) or approx. 1.3342
1161096
Note that f(x) = (26/e^3)x^(-2).  f'(x) = (-46/e^3)x^(-2-1) = (-46/e^3)x^(-3).  Thus, f'(2) = (-46/e^3)(2)^(-3) = -23/(4e^3).
831999
y = (x+3)(x-1)(x-2) = x^3 - 7x + 6
2302
(1) = x\n(2) = / \n(3) = x
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)
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 ;-)
1164377
There are zeros at 2, -3/2 and -6, so\n\nf(x) = (x-2)*(x+3/2)*(x+6)\n\nf(x) = x^3 + 11/2*x^2 - 6*x -18