# Inverse function on the interval $[1, 9.5]$

I am struggling to find the inverse of the following function $$f(x) = \frac{10}{3}\exp\big(-0.06x\big)\log\bigg(\frac{1}{5}\big(2x + 3\big)\bigg).$$ I noticed that this function is not one-to-one, so I restricted its domain to the interval $$[1, 9.5]$$. I am aware that there may not be an elementary answer, so while I may prefer an elementary answer, I would be very glad to get an answer in the form of something like a series.

• I fail to see why $f(x)$ is not one-to-one on its entire domain. Commented Jul 2, 2020 at 13:31
• @Carlo graph it up Commented Jul 2, 2020 at 13:32
• That is the first thing I did. Commented Jul 2, 2020 at 13:32
• @Carlo it does not satisfy the horizontal line test Commented Jul 2, 2020 at 13:32
• Where did this function come from? Commented Jul 3, 2020 at 2:49

Suppose you want to find $$f^{-1}(a)$$ for some $$a$$. Define $$g(x) = f(x) - a$$. The problem is now one of finding the root of $$g$$. Newton's method sets up the recurrence
$$x_{n+1} = x_n - \dfrac{g(x_n)}{g'(x_n)} = x_n - \dfrac{f(x) - a}{f'(x)}$$
Which hopefully will converge to the root of $$g$$, which is $$f^{-1}(a)$$. For your function, if you start with $$x_0 = 1$$, it will converge very quickly for $$a < 2.5$$. As $$a$$ gets close to its maximum (a little under $$3$$), convergence will slow, but the sequence should still converge instead of blowing up, as occasionally happens with Newton's method.
$$f'(x) = \frac{10}3e^{-0.06x}\left[(-0.06)\log\left(\dfrac{2x+3}5\right) + \dfrac2{2x+3}\right]$$
So the method becomes (with some simplification) \begin{align}x_0 &= 1\\x_{n+1} &= x_n - \dfrac{\dfrac{50}3\log\left(\dfrac{2x_n + 3}5\right) - 5a\,e^{0.06x_n}}{\dfrac {100}{6x_n+9} - \log\left(\dfrac{2x_n+3}5\right)}\end{align} which will converge to $$f^{-1}(a)$$ for $$a \in (-\infty, 2.7949)$$, giving values in $$\left(-1.5, 9.6483\right)$$.