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Sagot :
Answer:
Following are the solution to the given question:
Step-by-step explanation:
Given:
[tex]\to g(x)= -3+2x^2 \\\\\to f(x)=9-6x[/tex]
Calculating the value of [tex](\frac{g}{f})(2)[/tex]:
[tex]\to (\frac{g}{f}) (2)= (\frac{g(x)}{f(x)}) \times (2)[/tex]
[tex]=\frac{2x^2-3}{9-6x} \times 2\\\\=\frac{2\cdot 2^2-3}{3(3-2\cdot 2)} \\\\=\frac{2\cdot 4-3}{3(-1)}\\\\ =\frac{8-3}{-3}\\\\ =\frac{5}{-3}\\\\ = - \frac{5}{3}\\\\[/tex]
x=[tex]- \frac{\sqrt{3}}{2}[/tex] and x=[tex]\frac{\sqrt{3}}{2}[/tex] are not in the domain.
Domain: [tex]x \belong (- \infty, -\frac{\sqrt{3}}{2})\cup (- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2} )\cup (\frac{\sqrt{3}}{2}, \infty)[/tex]
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