Given f(X) = 3X^2-X+10 and g(X) = 1-20X find (f * g)(5)?
449247
f(x)=(3x^2)-x+10\ng(x)=1-20x\n\n(f°g)=f[g(x)]\n\nSo we have:\n\nf[g(x)]={[3(1-20x)^2]-(1-20x)+10}\n={3[1-40x+(400x^2)]-1+20x+10}\n=3-120x+(1200x^2)-1+20x+10\n=(1200x^2)-100x+12\n\n(f°g)(5) means that we substitue 5 for x.\n\n[(1200x^2)-100x+12](5)\n={[1200(5)^2]-[100(5)]+12}\n=[1200(25)]-500+12\n=30000-488\n=29512
935264
Actually, the answer is *not* 4+x.\n(f/g)(x) = (16-x^2)/(4-x)\nNow, factorise the top:\n(4+x)(4-x)/(4-x).\nSo *if x is not 4* then (f/g)(x) = 4+x. However, the function is not defined at x = 4.
797737
Ok, take your equation \nf(x) = 2x^2 - 10\n\nreplace f(x) with y\n\ny = 2x^2 - 10\n\nswap y and x\n\nx = 2y^2 - 10\n\nsolve for y\nto do this first add 10 to both sides\n\nx + 10 = 2y^2\n\ndivide both sides by 2\n\n(1/2)x + 5 = y^2\n\ntake the sqaure root of both sides\n\nsqrt((1/2)x + 5) = y\n\nchange y to f(x)^-1 [the inverse function notation]\n\nf(x)^-1 = sqrt((1/2)x + 5)\n\nhope this helps
36762
Factor by grouping:\nx^2 (3x - 5) - 3 (3x - 5)\nAnswer: (3x - 5)(x^2 - 3)
30589
2(5)-5(5)+2(5)=\n10-25+10\n20-25\n-5
560872
g(x) = (Sq root: x^2 -5x +6) / x^2 +x -20\n\nResolve the equation  when it is equal zero\n(Sq root: x^2 -5x +6) / x^2 +x -20 = 0\n \nPass the divider to multiply the other side of the equation\n(Sq root: x^2 -5x +6) = (x^2 +x -20)* 0\n\nNow this is a equation, you could resolve \n(Sq root: x^2 -5x +6) = 0\n\nTo resolve this equation remember ax^2 +bx + c \nis:    (-b + - Sq root(b^2 - 4ac))/2a\n\nthen:\n(-(-5)+ - sq root ( (-5)^2 - (4)(1)(6))/((2)(1)) = 0\n ( 5 + - sq root (25 - 24))/ 2 = 0\n  (5 + - sq root (1))/ 2 = 0\n  (5 + - 1)/2 = 0\n\nx = (5+1)/2 = 6/2 =3\nx = (5-1)/2 = 4/2 =2\n\nThere are 2 values for x, x=3 and x=2\n\nNow, you can evaluate g(1/x)
491386
Just substitute x+1 into f(x)\n\nf(x+1) = [ 6(x+1)^3 ] - [ 3(x+1)^2 ] + 4(x+1) - 9
904826
f(g(x)) = g(x) + 1 = (x^2 - 1) + 1 = x^2  - 1 + 1 = x^2\n\ng(f(x)) = (f(x))^2 - 1 = ( x + 1 )^2 - 1 = (x^2 + 2x + 1) - 1 \n                                                         = x^2 + 2x + 1 - 1 \n                                                          = x^2 + 2x
963450
20 + x = -4x + 15\n5x = -5\nx = -1\n\n5 = 5(x - 1) + 3 - 4x\n5 = 5x - 5 + 3 - 4x\n5 = x - 2\nx = 3\n\n-5 + 6 - 8x = -150\n5 - 6 + 8x = 150\n-1 + 8x = 150\n8x = 149\nx = 149/8
741312
If f(t)=(t-a)*(t-b) = t^2-(a+b)*t+ab; then \nI=(1/3)*t^3 - (a/2+b/2)*t^2 + a*b*t+C
893232
f(g(x)) = f(√5 * x)\n= cot(π/10 * (√5 * x)\n= cot(πx/(2√5))\nWhen x = 1\nf(g(x)) = cot(π/(2√5))\n≈ 1.1813
210475
f(g(x)) = 3/( (3/x)+4) -4)\nso that becomes\n3/ ((3/x)) because the + 4 and the - 4 cancel out\nand when you divide you multiply by recipricol so 3 divided by 3/x = 3 multiplied by x/3\n3 multiplied (x/3) = 3x/3 = x
785524
y = 3x-5\n\nx^2 +(3x-5)^2 = 25\nx^2 + 9x^2 -30x + 25 = 25\n10x^2 -30x = 0\n10x(x-3) = 0\n\nso x = 0 or x = 3\n\nWhen x = 0 y = -5\n\nWhen x = 3  y = 4
846363
3x² + 6 + 10 - (3x² - 8x - 5)\n\n3x² + 6x + 10 - 3x² + 8x + 5\n\n3x² - 3x² + 6x + 8x + 10 + 5\n\n0 + 14x + 15\n\n14x + 15\n\nThe answer is 14x + 15\n\n- - - - - - -s-
1019687
F(n)=F(n-1)+F(n-2), where F(0) = 0 and F(1) = 1\nn= 2 \nF(2) = F(2-1) + F(2-2)\nF(2) = F(1) + F(0) \nF(2) = 1 + 0 = 1\nn = 3\nF(3) = F(3-1) + F(3-2)\nF(3) = F(2) + F(1)\nF(3) = 1 + 1 \nF(3) = 2\nn = 4\nF(4) = F(4-1) + F(4-2)\nF(4) = F(3) + F(2)\nF(4) = 2 + 1\nF(4) = 3\nn = 5\nF(5) = F(4) + F(3)\nF(5) = 3 + 2\nF(5) = 5
1157065
(x+5)(2x^2+4x+2)\nx(2x^2+4x+2)+5(2x^2+4x+2)\n2x^3+4x^2+2x+10x^2+20x+10\n2x^3+14x^2+22x+10
465556
Given that f(x) is a polynomial of degree EIGHT such that\n\nf(m) = 1/m for m = 1,2,3,4,5,...,9\n\nFind f(10)  \n\n\n*Answer: 2/10\n\n*Solution:\n\nConsider another polynomial, g, defined as g(x) = x f(x) - 1\n\ng(1)=0, g(2)=0, …, g(9)=0\n\nSo g(x) = A(x-1)(x-2)…(x-9)\n\ng(0) = 0 f(0) - 1 = -1\n\nalso, g(0) = A(-1)(-2)…(-9), so A = 1/9!\n\ng(10) = A(9)(8)…(1) = (1/9!)(9!) = 1\n\ng(10) = 1 = 10 f(10) - 1, so f(10) = 2/10\n\n\n**Or try doing some research on "Rabin's method"