y =(x^3)/((x^2)-1... Find the Maclaurin series of the function y as far as the term in x^9..?
485277
The Maclaurin series is a Taylor expansion around 0.\n\nSo you get:\n\nf(x) = sum_i=0:inf(f^(i)(0)/i!)   where f^(i) is the i'th derivative.\n\nSo you need to work out the first, second, and third derivatives of y(x) until you get the x^9 term.  Then you plug those terms into the Maclaurin series, and simplify.
766536
(x^4-y^4)\n=(x^2+y^2)(x^2-y^2)\n=(x^2+y^2)(x+y)(x-y)\n\n2.(x^4-1)\n=(x^2+1)(x^2-1)\n=(x^2+1)(x+1)(x-1)
890367
You divide the bottom into the top and then see what happens as the limit of x--> infinity.\n\nso x^3/(x^2-4) = x+4x/(x^2-4). Now notice that when x goes to infinity the second part of our answer is going to 0 because the x^2 part is going to dominate the 4x in the numerator. So when this function gets very large it is going to be closer and closer to y= x +0.\n\nThis is all that is meant by asymptotes. They are the functions that aproximate another function as x reacher infinity or a certain constant where the equation is undefined (0 in the denominator)\n\nhope this helps.
965753
sqrt(x) is the same as x^(1/2)\nso Sqrt(x^3y^9) is the same as \n(x^3y^9)^(1/2)\nand this is equal to (x^3)^(1/2) * (y^9)^(1/2).\nNow,\n(x^3)^(1/2)=\nx^(3/2)=\nx*x^(1/2)=\nx sqrt(x)\nand\n(y^9)^(1/2)=\ny^(9/2)=\ny*y*y*y*y^(1/2)=\ny^4 sqrt(y)\nso (x^3)^(1/2)(y^9)^(1/2)=\nxy^4sqrt(xy)
821547
f(x)=4x^2 + 9x - 2\n\nhere, this means the function (f) of x=4x^2 + 9x - 2\n\nso if there was f(2)=4x^2 + 9x - 2  that would mean solve the function if x=2
87719
x^2 + 9x + 10x + 90\n\n(x + 9)(x + 10)
948239
xy = y^½\n y = y^½ /x\n y = y^½*x^-½\n\ndy/dx = -½ y^½ x^ -3/2\ndy/dx = - [½ y^½ ]/x^+3/2\ndy/dx = - (y^½) / (2x^3/2)
916997
y+9=2x+12\n\nSolve for y:\n\n y+9=2x+12\n    -9=    - 9\n----------------\ny=2x+3\n\nsolve for x:\n\ny+9=2x+12\n -12=    -12\n--------------\ny-3=2x\n       ````\n        2\n\nx=y-3\n    ````\n      2\n\nx=y-3/2    y=2x+3
942027
The first step in these type of questions is to get the expression into index form.\n\nThis becomes x - 3 x^1/2.  Also, remember x is the same as x^1\n\nSo dy/dx =  1*x^(1-1) - 3*1/2 * x^(1/2 -1)\n\n               =  1 * x^0 - 3/2 * x^(-1/2)\n\n               =  1 - 3/2* x1/sq.root x     remember x^0 =1\n\n                = 1 - 3/(2sq.root x)
864521
y = 3/(1 - x)²\n\n1, (1 - x)² ≥ 0 for all x and therefore 3/(1 - x)² ≥ 0 for all x\n\n2. 3/(1 - x)² is undefined when x = 0. Therefore 3/(1 - x)² (ie y) → +∞ when x → 1 from the left of 1 and when x → 1 from the right of 1\n\n3. As x → ±∞, 3/(1 - x)² (ie y) → 0\n\n4. Plot a few other points and Bob's your uncle!!
3588
You've answered your own question.\nNo, f(x) is not defined at x=3 because the denominator would equal 0.  It doesn't matter what the numerator is.\n\nAnd btw, the expression simplifies to f(x)=x+3, where x<>3, \nbecause (x^2)-9=(x+3)*(x-3).
1068232
Use quotient rule to get the derivative, which, simplified, is:\n\n-30*(2x+3)/((3x^2+9x+7)^2)\n\nSet it equal to zero and solve.  Since the denominator cannot be zero, set the numerator equal to zero\n\n2x+3=0\nx=(-3/2)\n\nI checked it, and it indeed is the solution. \n\nBy the way, you don't have to worry about an undefined points of this function making problems in the graph or in the max/min values in the derivative.  The denominator has imaginary roots.
139329
calculus, fun stuff. To get the derivative, you need to drop the exponent then subtract one from the explonent. SO the X^2 would be 2x and 2x would just be 2 .    Final answer is 2x + 2.   Hm.. explained that poorly. In general, to find the derivative, Take the exponent on the X, drop it in front so its multipluing, than subract one from the exponent and have that number in the old exponents place (3x^3 would turn into 9x^2, 4x^2 would be just 8x) any number with X^1 is just the number, and just a number with no X is 0.
931778
Since you have conflicting answers, I'm just chiming in to say go with ManiacalAce -- he got the correct answer.\n\nCorrect second derivative is as follows:\n\n(-9/4)x^(-3/2) - (3/4)x^(-1/2)\n\n~ ♥ ~
860585
Very ingenious to have a "That's Me" account. You can give yourself ten points over and over.\n\nfor x = -3, x^2 = 9 and 3x = -9. Result = 0\nfor x = -2, x^2 = 4 and 3x = -6. Result = -2\nfor x = -1, x^2 = 1 and 3x = -3. Result = -2\nfor x = 0, x^2 = 0 and 3x = 0. Result = 0\nfor x = 1, x^2 = 1 and 3x = 3. Result = 3\n\nOnly your zero value is correct.\n\nThis is an upward-opening parabola with its minimum point at \n\nx = -1.5\n\nIt crosses the x-axis at (-3, 0) and (0, 0). Obviously, it crosses the y-axis at (0, 0) also.
744242
let y=4x(x+x^1/2)^1/2\nlny=ln4+lnx+1/2ln(x+x^1/2)\ny'/y=1/x+{1/(x+x^1/2)}*(1+1/2x^1/2)\ny'=y[1/x+{1/(x+x^1/2)}*(1+1/2x^1/2)]
745065
(3m/n)(n^2/9) \n=(3mn^2)/(9n)\n=(3mn)/(9)\n=(mn)/3