Respuesta :
The domain of the function is given by (–∞, –3) ∪ (–3, 2) ∪ (2, ∞)
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The function is given by:
[tex]g(x) = \frac{x^2 + 3x}{x^2 + x - 6}, x < 3[/tex]
[tex]g(x) = \log_{2}{x + 5}, x \geq 3[/tex]
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- For the second definition, the restriction is that x + 5 has to be positive. Since the definition is only valid for [tex]x \geq 3[/tex], this will always be true.
- In the second definition, the denominator cannot be zero, thus the zeros of the denominator will be outside the domain.
These zeros are found using Bhaskara, we have the quadratic function [tex]x^2 + x - 6 = 0[/tex], thus the coefficients are [tex]a = 1, b = 1, c = -6[/tex].
[tex]\Delta = b^2 - 4ac = (1)^2 - 4(1)(-6) = 25[/tex]
[tex]x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-1 + 5}{2} = 2[/tex]
[tex]x_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-1 - 5}{2} = -3[/tex]
Thus, x = 2 and x = -3 are outside the domain, which is given by:
(–∞, –3) ∪ (–3, 2) ∪ (2, ∞)
A similar problem is given at https://brainly.com/question/13136492
