# What is the inverse of $f(x) = a⋅e^{bx} + cx + d$

Does an inverse function for

$f(x) = a⋅e^{bx} + cx + d$

exist where a, b, c, d are constants? If so, what is it?

I've tried lots of methods, but they've all failed.

What I ended up doing to approximate an inverse:

For very small (very negative) values of y, I ended up using the following function:

$x = \frac{ln(\frac{y-cx-d}{a})}{b}$

to assist in a recursive approximation by which, after seeding x with a guess, I feed the function back on itself a few times. For greater values of y, I used a similar method/function inspired by binary search algorithms. My implementations converged quickly for my purposes.

Thanks everyone who commented.

• There isn't a solution in general. Some might reference the W Lambert function but I am not a fan of that argument. It's almost like saying "the solution is the solution!" The W Lambert function is constructed to be a solution to a related problem so it's circular to me. – Cameron Williams Feb 19 '15 at 1:29
• @CameronWilliams But distinguish between the two cases (1) There is no inverse because the function is not $1-1$ and (2) There is an inverse, but not one with a closed form. – Thomas Andrews Feb 19 '15 at 1:38
• @ThomasAndrews Oh doh. I meant to say "no solution in elementary functions." Slipped my mind. You're completely right. – Cameron Williams Feb 19 '15 at 1:40
• @CameronWilliams The one nice thing about special functions like the Lambert W function is that they reduce a large family of problems to a smaller family of problems. This makes robust numerical implementation a lot easier. Solving an equation like that in the OP with Newton's method, the convergence properties may depend very strongly on the parameters. – Ian Feb 19 '15 at 1:51
• @CameronWilliams: the benefit of "standardizing" the problem to the Lambert function is that the latter is a well-studied function. As soon as you reduce a problem to it, you immediately "inherit" all the knowledge about it. For instance, we know that it has no closed form. – Yves Daoust Feb 19 '15 at 13:55

Consider $$y = a⋅e^{bx} + cx + d$$ and make a change of variable $e^{bx}=z$. This then leads to $$y=a.z+\frac cb .\log(z)+d$$Any equation which can write $A+B x+C\log(Dx+E)=0$ has solution(s) in terms of Lambert function. For the particular case in the post, we should then obtain $$z=\frac{c }{a b}W(t)$$ where $$t=\frac{a b }{c}e^{\frac{b (y-d)}{c}}$$ and, back to $x$ (taking into account the fact that $t=W(t).e^{W(t)}$), $$x= \frac{y-d}{c}-\frac{1}{b}W(t)$$ The advantage of this is that there are very good approximations of Lambert function.
For illustration purposes, let us use $a=1,b=2,c=3,d=10^6$ and let us try to solve the equation for $y=2.10^6$. So, replacing, we have $$x=\frac{1000000}{3}-\frac{1}{2} W\left(\frac{2 e^{2000000/3}}{3}\right)$$ Now, let us use for the approximation $$W(t)\approx L_1-L_2+\frac{L_2}{L_1}+\cdots$$ where $L_1=\log(t)$,$L_2=\log(L_1)$. This then leads to $x=6.907744917$ which is the exact solution for ten significant figures.
In this case, $f(x)$ is $1-1$ if and only if $f'(x)\neq 0$ for all $x$ in the (interior of the) domain, which means if $abe^{bx}\neq -b$ has no solution, so if $\frac{-d}{ab}$ is negative.
However, combinations of $x$ and $e^x$ generally make it impossible to find a formula for the inverse function.
We can easily find a way to compute the inverse when $f$ is increasing or decreasing - just apply a binary-search algorithm to get a sequence of approximations to the inverse.