f (x) = abs(x-1) + 2 ,for x<1 and ax^2 + bx ,for x>=1?
914146
For x < 1 \nf(x) = |x-1| + 2\nWhen x → 1-, f(x) →2\nWhen x < 1, |x-1| + 2 = 1 - x + 2 = 3 - x. So slope is -1.\n\nFor x ≥ 1\nFor continuity f(1) = 2.\nFor differentiability f'(1) = -1\n\nSo a + b = 2\n2a + b = -1\n\nThus a = -3, b = 5\n\nf(x) = |x-1| + 2, for x < 1 \n..... = 5x - 3x², for x ≥ 1\nis continuous and differentiable for all x
708644
Use the quadratic formula.\nx1 = 1\nx2 = -1/2
852832
x = [-b +/- sqrt(b^2 - 4ac)] / (2a)\nf(x): a = 1, b = 0, c = -4\ng(x): a = 1, b = -1, c = -12
879937
ax^2 + bx+c = 0\nax^2 + bx = -c\nx^2+(b/a)x+ ... = -c/a + ...\nx^2+(b/a)x+ (b/2a)^2 = -c/a + (b/2a)^2\n{x - (b/2a)}^2 = -4ac/4(a^2) + b^2/4(a^2)\nx - (b/2a)}^2 = + or - root{(-4ac + b^2)/4(a^2)}\nx= {-b + or - sqrt (b^2-4ac)}/2a\n\nTh
114499
Break it down into three sections determined by the points where the absolute values change sign.\n\nfor 2 < x, g(x) = (x-2) + (x+2) = 2x\n\nfor x < -2, g(x) = -(x-2) + (-(x + 2)) = -2x \n\nfor -2 < x < 2, g(x) = -(x-2) + (x+2) = 4
46351
Sy   = |x| - x^2 + 2   \n     = (x^2/2) - (X^3/3) + 2X
604301
We just interchange the x's and the y's. then find what is y\nExample the inverse of f(x)=x+1\nSince f(x) is just equivalent to y... \nthen y=x+1. we then switch...\nx=y+1... then find for y... y=x-1. \nThat is the inverse of f(x)=x+1\nNow let's solve your problem\ny=f(x)=7x-2\ny=7x-2\nswitch x and y \nx=7y-2\nsolve for y\n(x+2)/7=y\ninverse is (x+2)/7
626693
ax - b = -c\nax = b - c\nx = (b - c) / a.
511872
ab+ac=0\na(b+c)=0\na=0 or b+c=0, b=-c
700872
d sin ax=a*cos ax dx\nd bx = b dx
879979
f(x)=(x^2-2x)(x^2+2.44x+1.4884)\n =(x^4+2.44x^3+1.4884x^2-2x^3-4.88x^2- 2.9768x)\n =(x^4 + 0.44x^3 - 3.3916x^2 - 2.9768x)\n\nF(x)=(x^5)/5+(0.44x^4)/4-(3.3916x^3)/3-             (2.9768x^2)/2+c\n\nam i really asked a Q?
464622
Use any recursive algorithm.\n\nFor instance:\n\nf(x) = f(x) + 1
931057
f(x) = sin½x + a \nSo x = sin(½f^(-1)( x)) + a\nx - a = sin(½f^(-1)( x))\nso ½f^(-1)( x) = arcsin(x - a)\nSo f^(-1)( x) = 2 arcsin(x - a) \nDomain: a -1 ≤ x ≤ a + 1\nRange: - π ≤ f^(-1)( x) ≤ π
276641
f(x)=3x^2-2x+8 for x=6\n\nf(6)=3*(6^2)-2*6+8\nf(6)=3*(36)-12+8\nf(6)=108-12+8\nf(6)=104
860831
I am guessing your function to be one of the following:\n(1) f(x) = ½x - 3 or (2) f(x) = -½x -3\nIn function (1):\nf(0) = ½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = ½(2) - 3 = 1 - 3 = -2 ... or... (2, -2)\n\nIn function (2):\nf(0) = -½(0) - 3 = 0 - 3 = -3 ... or... (0, -3)\nf(2) = -½(2) - 3 = (-1) - 3 = -4 ... or ... (2, -4)
547775
f(x)= x^3+x^2-22x-40\nf(x)= (x+4)(ax^2 + bx + c)\n\nComparing coefficients,\nx^3: 1=a\nconstant: -40=4c => c=-10\nx^2: 1=b+4a => 1=b+4 => b=-3\n\nf(x) = (x+4)(x^2-3x-10) = (x+4)(x+2)(x-5)\n\nThe other factors are (x+2) and (x-5).
809962
1. f(x)= 3x^4 + 4x^3, -2 <_ x <_2\nf'(x)= 12x^3 + 12x^2 = 0\n12x^2(x+1)=0\nso the critical points are: x=0 or x=-1 \nf''(x)= 36x^2 +24x\nf''(0)=0, so this test is inconclusive.\nf''(-1) = 36 -24 >0 so -1 is a local minimum.\n\nNow, since you want an absolute max and min over the closed interval, you need to compare the values of f at the end points with the values of f at the critical points:\nf(-2) = 3(-2)^4 +4(-2)^3 = 16\nf(2) = 3(2)^4 +4(2)^3 = 80\nf(-1) = 3(-1)^4 +4(-1)^3 = -1\nf(0) = 0\nthe largest value happens when x=2, which means this is the abs. maximum over the interval\nand the smalles value happens when x=-1, which means this is the abs. min over the interval.\n\n\n2. f(x)=(x^2 + x)^2/3, [-2,3]\n\n3. f(x)= [6-4x], [-3,3]\n\n/