Finding constants a and b, given a function and its inverse

Question

Given a function and its inverse, where a and b are constants, find the constants a and b.
$$h(x) = x + a $$ $$h^{-1}(x) = b(2x + 3)$$

---

I tried simultaneous equation(not quite sure is it done like this):

I created an inverse from the first function, i.e h(x) = x + a 

$h(x) = x + a$
$h^{-1}(x) = x - a$

For the second inverse function, I rearrange it back to a function 

$h^{-1}(x) = b(2x + 3)$
$h(x) = \frac{x - 3b}{2b}$

Then I equate the inverse with the inverse function I created and the original function with the function

$x - a = b(2x + 3)$
$x + a = \frac{x - 3b}{2b} $

---

From h(x), I rearrange to get b and substitued b into the inverse equation, which got me a = 0 or a = -1.5

$x + a = \frac{x - 3b}{2b} $
$b = \frac{x}{2x + 2a + 3}$

I'm stuck till this step. Can anyone please point out where my mistake is? Thank you.

Answer 1

both of these equations are same. because the inverse is unique checking $h^{-1}(x)=b(2x+3)$ is same as checking $h(x) = \frac{x-3b}{2b}$. you chould solve linear system of first degree polynomial (line equation): $$ x-a = b(2x+3) \rightarrow (2b-1)x+(3b+a)=0 \quad \text{for all x} \\ $$ coefficients of $x$ and constant term should be zero in order to this equation be always zero. $$ 2b-1=0 \rightarrow b = \frac{1}{2} \\ (3b+a)=0 \rightarrow a = -3b = -\frac{3}{2} $$

Answer 2

From $$x = id(x) = (h\ \circ\ h^{-1})(x) = h\left(h^{-1}(x)\right) = \left(h^{-1}(x) + a\right) = \left(b(2x+3) + a\right) = 2bx + (3b+a)$$ you directly derive by comparison of coefficients $$ b=1/2\\ a=-3b=-3/2$$

--- rk