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


*

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


$$\frac{3x+5}{2} + \frac{1-2x}{3} = -23$$
$$\frac{3(3x+5)}{6} + \frac{2(1-2x)}{6} = -23$$
$$2x+17 = \frac{-23}{6}\implies x = -\frac{125}{12}$$
 [/INST] The first one is already stated in other answers.
For the second, the incorrect step is:
$$g(x)=1-2x\Rightarrow x=\dfrac{1-g(x)}{2}\Rightarrow g^{-1}(x)=\dfrac{1-x}{2}$$
So we need to solve $6a+5=\dfrac{1-a}{2}-2\Leftrightarrow 12a+14=1-a\Leftrightarrow a=-1.$
For the third, the incorrect step is:
$$\frac{3(3x+5)}{6} + \frac{2(1-2x)}{6} = -23 (2)$$
$$2x+17 = \frac{-23}{6}$$
This is because from $(2)$
$\Leftrightarrow 9x+15+2-4x=-23\times6$
$\Leftrightarrow 5x+17=-138$
$\Leftrightarrow x=-31$
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