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timama [110]
2 years ago
13

I need this solved asap please do so

Mathematics
1 answer:
mafiozo [28]2 years ago
7 0

The <em>simplified</em> expression for h(x) = f(x)/g(x) is h(x) = \frac{1}{3}\cdot \frac{x-4}{x-\frac{2}{3} } and the domain of the function is all <em>real</em> numbers except x = 2/3.

<h3>How to determine the domain of a rational function</h3>

In this problem we have a <em>rational</em> function, the domain of <em>rational</em> functions is the set of all <em>real</em> numbers except the values such that the denominator becomes zero. First, we need to simplify the division of two <em>rational</em> functions:

h(x) = \frac{\frac{3\cdot x^{2} + 17\cdot x + 20}{3\cdot x^{2} + 10\cdot x - 8} }{\frac{3 \cdot x^{2} + 8\cdot x + 5}{x^{2}-3\cdot x - 4} }

h(x) = \frac{(3\cdot x^{2}+17\cdot x + 20)\cdot (x^{2}-3\cdot x - 4)}{(3\cdot x^{2}+10\cdot x - 8)\cdot (3\cdot x^{2}+8\cdot x + 5)}

h(x) = \frac{1}{3}\cdot \frac{\left(x^{2}+\frac{17}{3}\cdot x + \frac{20}{3}  \right)\cdot (x^{2}-3\cdot x - 4)}{\left(x^{2}+\frac{10}{3}\cdot x -\frac{8}{3}\right)\cdot \left(x^{2}+\frac{8}{3}\cdot x +\frac{5}{3}  \right)}

h(x) = \frac{1}{3}\cdot \frac{\left(x + \frac{5}{3} \right)\cdot (x+4)\cdot (x - 4)\cdot (x+ 1)}{\left(x-\frac{2}{3} \right)\cdot (x + 4)\cdot (x + 1)\cdot \left(x + \frac{5}{3} \right)}

h(x) = \frac{1}{3}\cdot \frac{x-4}{x-\frac{2}{3} }

The <em>simplified</em> expression for h(x) = f(x)/g(x) is h(x) = \frac{1}{3}\cdot \frac{x-4}{x-\frac{2}{3} } and the domain of the function is all <em>real</em> numbers except x = 2/3.

To learn more on rational functions: brainly.com/question/27914791

#SPJ1

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