Solving inverse function problems The following is my (likely completely incorrect) attempt.
$$f(x)= 3x + 5, \qquad  \qquad g(x)= 1 - 2x$$


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*Show that $f{^{-1}}(2) < g(-3)$


$$\frac{1}{3\cdot 2+5} < (1-2\cdot(-3))\implies\frac{1}{11} < 7$$


*Find the value of $a$ such that $f(2a) = g{^{-1}}(a)-2$


$$6a+5 = \frac{1}{-1-2a}$$
$$6a = \frac{1}{4-2a}$$
$$8a = \frac{1}{4}\implies a = \frac{1}{32}$$


*Solve $\frac{f(x)}{2} + \frac{g(x)}{3} = -23$


$$\frac{3x+5}{2} + \frac{1-2x}{3} = -23$$
$$\frac{3(3x+5)}{6} + \frac{2(1-2x)}{6} = -23$$
$$2x+17 = \frac{-23}{6}\implies x = -\frac{125}{12}$$

I think $$f^{-1}(x)=\frac{1}{3}(x-5)$$ so $$f^{-1}(2)=\frac{2}{3}-\frac{5}{3}=-1$$
