Question

what is the domain of 4x/6x^2+3x-5

Answer 1

Welcome to Openstudy :)
First what you wanna do is set the denominator equal to zero and solve.

Answer 2

find the values of x for which denominator is 0 then domain=all real numbers except these points.

Answer 3

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)

Answer 4

So the domain would be R (all real numbers)

Answer 5

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