Solving inverse function problems

Question

The following is my (likely completely incorrect) attempt.

$$f(x)= 3x + 5, \qquad \qquad g(x)= 1 - 2x$$

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

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

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

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

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

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

Answer 1

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 $(2)$

$\Leftrightarrow 9x+15+2-4x=-23\times6$

$\Leftrightarrow 5x+17=-138$

$\Leftrightarrow x=-31$

Answer 2

You are correct that you have incorrectly answered $(1)$. Note that $f^{-1}$ most likely denotes the inverse of $f$, not its reciprocal.

Since $$f(x)-5=3x\implies x=\frac{f(x)-5}3$$ the inverse function of $f$ is $$f^{-1}(x)=\frac{x-5}3$$ so $$f^{-1}(2)=\frac{2-5}3=-1<g(-3)$$ as required.

Answer 3

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

Answer 4

• $f$: "Take $x$, multiply it by $3$ and add $5$".

• Inverse of $f$, $f^{-1}$ "Take a $y$, subtract 5, then divide by $3$"