Question

Given $g\left(x\right)=\frac{3x-1}{2x+9},g\left(g\right(x\left)\right)=$

• A. $\frac{9x-4}{6x+7}$
• B. $\frac{7x-12}{24x+79}$
• C. $\frac{x-10}{21x+80}$
• D. $\frac{7x+6}{24x-79}$
• E. $\frac{9x^2-6x+1}{4x^2+36x+81}$

Answer 1

The correct option is B $\frac{7x-12}{24x+79}$

It is given that $g\left(x\right)=\frac{3x-1}{2x+9}$, therefore $g\left(g\right(x\left)\right)$ is given by:

$\therefore g\left(g\right(x\left)\right)=g\left(\frac{3x-1}{2x+9}\right)$

$=\frac{3\left(\frac{3x-1}{2x+9}\right)-1}{2\left(\frac{3x-1}{2x+9}\right)+1}$

$=\frac{\frac{9x-3}{2x+9}-1}{\frac{6x-2}{2x+9}+1}$

$=\frac{\frac{9x-3-2x-9}{2x+9}}{\frac{6x-2+18x+81}{2x+9}}$

$=\frac{9x-3-2x-9}{6x-2+18x+81}$

$=\frac{7x-12}{24x+79}$

Hence, $g\left(g\right(x\left)\right)=\frac{7x-12}{24x+79}$.