Question

Suppose that the functions g and f are defined as follows.

Answer 1

Answer:

Following are the solution to the given question:

Step-by-step explanation:

Given:

$\to g(x)= -3+2x^2 \\\\\to f(x)=9-6x$

Calculating the value of $(\frac{g}{f})(2)$:

$\to (\frac{g}{f}) (2)= (\frac{g(x)}{f(x)}) \times (2)$

               $=\frac{2x^2-3}{9-6x} \times 2\\\\=\frac{2\cdot 2^2-3}{3(3-2\cdot 2)} \\\\=\frac{2\cdot 4-3}{3(-1)}\\\\ =\frac{8-3}{-3}\\\\ =\frac{5}{-3}\\\\ = - \frac{5}{3}$

x=$- \frac{\sqrt{3}}{2}$ and x=$\frac{\sqrt{3}}{2}$ are not in the domain.

Domain: $x \belong (- \infty, -\frac{\sqrt{3}}{2})\cup (- \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2} )\cup (\frac{\sqrt{3}}{2}, \infty)$