anonymous
  • anonymous
what is the domain of 4x/6x^2+3x-5
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Luigi0210
  • Luigi0210
Welcome to Openstudy :) First what you wanna do is set the denominator equal to zero and solve.
anonymous
  • anonymous
find the values of x for which denominator is 0 then domain=all real numbers except these points.
anonymous
  • anonymous
Solve for x over the real numbers: (2 x^3)/3+3 x-5 = 0 Write the left hand side as a single fraction. Bring (2 x^3)/3+3 x-5 together using the common denominator 3: 1/3 (2 x^3+9 x-15) = 0 Multiply both sides by a constant to simplify the equation. Multiply both sides by 3: 2 x^3+9 x-15 = 0 Write the cubic equation in standard form. Divide both sides by 2: x^3+(9 x)/2-15/2 = 0 Change coordinates by substituting x = lambda/y+y, where lambda is a constant value that will be determined later. If x = lambda/y+y then y = 1/2 (x+sqrt(x^2-4 lambda)) which will be used during back substitution: -15/2+9/2 (y+lambda/y)+(y+lambda/y)^3 = 0 Transform the rational equation into a polynomial equation. Multiply both sides by y^3 and collect in terms of y: y^6+y^4 (3 lambda+9/2)-(15 y^3)/2+y^2 (3 lambda^2+(9 lambda)/2)+lambda^3 = 0 Find an appropriate value for lambda in order to make the coefficients of y^2 and y^4 both zero. Substitute lambda = -3/2 and then z = y^3, yielding a quadratic equation in the variable z: z^2-(15 z)/2-27/8 = 0 Solve for z. Find the positive solution to the quadratic equation: z = 3/4 (5+sqrt(31)) Perform back substitution on z = 3/4 (5+sqrt(31)). Substitute back for z = y^3: y^3 = 3/4 (5+sqrt(31)) Move everything to the left hand side. Subtract 3/4 (5+sqrt(31)) from both sides: y^3-3/4 (5+sqrt(31)) = 0 Divide by an appropriate factor to make the constant term 1. Divide both sides by 3/4 (5+sqrt(31)): (4 y^3)/(3 (5+sqrt(31)))-1 = 0 Write (4 y^3)/(3 (5+sqrt(31))) with a common power. (4 y^3)/(3 (5+sqrt(31)))-1 = y^3 ((2^(2/3))/((3 (5+sqrt(31)))^(1/3)))^3-1 = ((2^(2/3) y)/((3 (5+sqrt(31)))^(1/3)))^3-1: ((2^(2/3) y)/((3 (5+sqrt(31)))^(1/3)))^3-1 = 0 Simplify ((2^(2/3) y)/(3 (5+sqrt(31)))^(1/3))^3-1 = 0 by making a substitution. Substitute u = (2^(2/3) y)/(3^(1/3) (5+sqrt(31))^(1/3)): u^3-1 = 0 Solve for u. By definition, the 3^rd roots of unity are the roots of u^3-1 = 0: u = 1 or u = -(-1)^(1/3) or u = (-1)^(2/3) Look at the first equation: Perform back substitution on u = 1. Substitute back for u = (2^(2/3) y)/(3^(1/3) (5+sqrt(31))^(1/3)): (2^(2/3) y)/(3^(1/3) (5+sqrt(31))^(1/3)) = 1 or u = -(-1)^(1/3) or u = (-1)^(2/3) Solve for y. Divide both sides by 2^(2/3)/(3^(1/3) (5+sqrt(31))^(1/3)): y = (3^(1/3) (5+sqrt(31))^(1/3))/2^(2/3) or u = -(-1)^(1/3) or u = (-1)^(2/3) Perform back substitution on y = (3^(1/3) (5+sqrt(31))^(1/3))/2^(2/3). Substitute back for y = x/2+sqrt(x^2+6)/2: x/2+sqrt(x^2+6)/2 = (3^(1/3) (5+sqrt(31))^(1/3))/2^(2/3) or u = -(-1)^(1/3) or u = (-1)^(2/3) Solve for x. Solve the radical equation in x: x = (3 (5+sqrt(31)))^(1/3)/2^(2/3)-3^(2/3)/(2 (5+sqrt(31)))^(1/3) or u = -(-1)^(1/3) or u = (-1)^(2/3) Look at the second equation: Perform back substitution on u = -(-1)^(1/3). Substitute back for u = (2^(2/3) y)/(3^(1/3) (5+sqrt(31))^(1/3)): x = (3 (5+sqrt(31)))^(1/3)/2^(2/3)-3^(2/3)/(2 (5+sqrt(31)))^(1/3) or (2^(2/3) y)/(3^(1/3) (5+sqrt(31))^(1/3)) = -(-1)^(1/3) or u = (-1)^(2/3) Solve for y. Divide both sides by 2^(2/3)/(3^(1/3) (5+sqrt(31))^(1/3)): x = (3 (5+sqrt(31)))^(1/3)/2^(2/3)-3^(2/3)/(2 (5+sqrt(31)))^(1/3) or y = -(3^(1/3) (-(5+sqrt(31)))^(1/3))/2^(2/3) or u = (-1)^(2/3) Perform back substitution on y = -(3^(1/3) (-(5+sqrt(31)))^(1/3))/2^(2/3). Substitute back for y = x/2+sqrt(x^2+6)/2: x = (3 (5+sqrt(31)))^(1/3)/2^(2/3)-3^(2/3)/(2 (5+sqrt(31)))^(1/3) or x/2+sqrt(x^2+6)/2 = -(3^(1/3) (-(5+sqrt(31)))^(1/3))/2^(2/3) or u = (-1)^(2/3) Solve for x. Solve the radical equation in x: x = (3 (5+sqrt(31)))^(1/3)/2^(2/3)-3^(2/3)/(2 (5+sqrt(31)))^(1/3) or x = -(-3 (5+sqrt(31)))^(1/3)/2^(2/3)-(-3)^(2/3)/(2 (5+sqrt(31)))^(1/3) or u = (-1)^(2/3) Look at the third equation: Perform back substitution on u = (-1)^(2/3). Substitute back for u = (2^(2/3) y)/(3^(1/3) (5+sqrt(31))^(1/3)): x = (3 (5+sqrt(31)))^(1/3)/2^(2/3)-3^(2/3)/(2 (5+sqrt(31)))^(1/3) or x = -(-3 (5+sqrt(31)))^(1/3)/2^(2/3)-(-3)^(2/3)/(2 (5+sqrt(31)))^(1/3) or (2^(2/3) y)/(3^(1/3) (5+sqrt(31))^(1/3)) = (-1)^(2/3) Solve for y. Divide both sides by 2^(2/3)/(3^(1/3) (5+sqrt(31))^(1/3)): x = (3 (5+sqrt(31)))^(1/3)/2^(2/3)-3^(2/3)/(2 (5+sqrt(31)))^(1/3) or x = -(-3 (5+sqrt(31)))^(1/3)/2^(2/3)-(-3)^(2/3)/(2 (5+sqrt(31)))^(1/3) or y = ((-1)^(2/3) 3^(1/3) (5+sqrt(31))^(1/3))/2^(2/3) Perform back substitution on y = ((-1)^(2/3) 3^(1/3) (5+sqrt(31))^(1/3))/2^(2/3). Substitute back for y = x/2+sqrt(x^2+6)/2: x = (3 (5+sqrt(31)))^(1/3)/2^(2/3)-3^(2/3)/(2 (5+sqrt(31)))^(1/3) or x = -(-3 (5+sqrt(31)))^(1/3)/2^(2/3)-(-3)^(2/3)/(2 (5+sqrt(31)))^(1/3) or x/2+sqrt(x^2+6)/2 = ((-1)^(2/3) 3^(1/3) (5+sqrt(31))^(1/3))/2^(2/3) Solve for x. Solve the radical equation in x: Answer: | | x = (3 (5+sqrt(31)))^(1/3)/2^(2/3)-3^(2/3)/(2 (5+sqrt(31)))^(1/3) or x = -(-3 (5+sqrt(31)))^(1/3)/2^(2/3)-(-3)^(2/3)/(2 (5+sqrt(31)))^(1/3) or x = 3^(2/3) (-1/(2 (5+sqrt(31))))^(1/3)+((-1)^(2/3) (3 (5+sqrt(31)))^(1/3))/2^(2/3)

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anonymous
  • anonymous
So the domain would be R (all real numbers)
goformit100
  • goformit100
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