Question

Which of the following describes the domain of the piecewise function g of x is equal to the piecewise function of the quantity x squared plus 3 times x end quantity over the quantity x squared plus x minus 6 end quantity for x is less than 3 and the function log in base 2 of the quantity x plus 5 end quantity for x is greater than or equal to 3 question mark
(–∞, ∞)
(–∞, 2) ∪ (2, ∞)
(–∞, 2) ∪ (2, 3) ∪ (3, ∞)
(–∞, –3) ∪ (–3, 2) ∪ (2, ∞)

Answer 1

The domain of the function is given by (–∞, –3) ∪ (–3, 2) ∪ (2, ∞)

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The function is given by:

$g(x) = \frac{x^2 + 3x}{x^2 + x - 6}, x < 3$

$g(x) = \log_{2}{x + 5}, x \geq 3$

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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 $x \geq 3$, 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 $x^2 + x - 6 = 0$, thus the coefficients are $a = 1, b = 1, c = -6$.

$\Delta = b^2 - 4ac = (1)^2 - 4(1)(-6) = 25$

$x_1 = \frac{-b + \sqrt{\Delta}}{2a} = \frac{-1 + 5}{2} = 2$

$x_2 = \frac{-b - \sqrt{\Delta}}{2a} = \frac{-1 - 5}{2} = -3$

Thus, x = 2 and x = -3 are outside the domain, which is given by:

(–∞, –3) ∪ (–3, 2) ∪ (2, ∞)

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