Question

What is the domain of the rational function f of x is equal to 2 x over the quantity 2 x cubed minus 5 x squared minus 12 x end quantity

x is an element of all real numbers such that x is not equal to 0
x is an element of all real numbers such that x is not equal to negative three halves comma 4
x is an element of all real numbers such that x is not equal to 0 comma negative three halves comma 4
x is an element of all real numbers such that x is not equal to 0 comma three halves comma negative 4

Answer 1

Using the domain concept, it is found that the domain of the function is: x is an element of all real numbers such that $x \neq (-\frac{3}{2}, 4)$, second option.

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• The domain of a function is all possible values that the input value x can assume.
• The domain of a fraction is all real values of x except the zeros of the denominator.

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The function is:

$f(x) = \frac{2x}{2x^3 - 5x^2 - 12x} = \frac{2x}{x(2x^2 - 5x - 12)} = \frac{2}{2x^2 - 5x - 12}$

The points outside the domain are the zeros of $2x^2 - 5x - 12$, which we find solving a quadratic equation.

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Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

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$2x^2 - 5x - 12$ is a quadratic equation with $a = 2, b = -5, c = -12$.

$\Delta = b^{2} - 4ac = (-5)^2 - 4(2)(-12) = 121$

$x_{1} = \frac{-(-5) + \sqrt{121}}{2(2)} = 4$

$x_{2} = \frac{-(-5) - \sqrt{121}}{2(2)} = -\frac{6}{4} = -\frac{3}{2}$

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The domain is: x is an element of all real numbers such that $x \neq (-\frac{3}{2}, 4)$, second option.

A similar problem is given at https://brainly.com/question/13136492

Answer 2

Answer:

2

Step-by-step explanation: