Why can't this mixed function be inverted?

Question

Given the function

$$y=Ax + B\sqrt x$$

where $A$ and $B$ are real constants, $x$ is real and $x > 0$

I want to find the inverse where $x$ is a function of $y$. ButI don't believe that's possible without approximating the square root term, right?

Either $$y=B\sqrt x$$ or $$y=Ax $$

by themselves are invertible but the mixed function is not.

My question though is why can't it be inverted?

Is there something fundamental in the theory of functions that says such mixed functions (transcendental and linear in this example) cannot be inverted?

Answer 1

It may not work in a very pretty way, but you could write $$0=Ax+B\sqrt{x}-y$$ and solve this for $\sqrt{x}$ using the quadratic formula to find $$\sqrt{x}=\frac{-B\pm\sqrt{B^2+4Ay}}{2A}$$

---

EDIT: from there, simply square and take the domain of definition based on $A$, $B$, and $y$. You get $$x=\frac{2B^2\pm2B\sqrt{B^2+4Ay}+4Ay}{4A^2}$$

---

ALSO EDIT: One will note that the expressions I have given both include a $\pm$, but this is not helpful when we are looking for an inverse function. The way to resolve this depends upon the values of $A$ and $B$. In some cases, the expression will be dual-valued with two non-negative values (such as when $A>0$ and $B<0$). In others, it will be non-negative ONLY for the $+$. As a rule, you should take the operator that agrees on the interval of relevance to your particular problem.

Example: $A=1$ and $B=-1$. This results in an expression that has dual non-negative values for the interval $[0,1]$, so you have to consider $$x=\frac{1-1\sqrt{1+4y}+2y}2$$ for $x\in[0,\frac14]$ and $$x=\frac{1+1\sqrt{1+4y}+2y}2$$ for $x>\frac14$

Answer 2

One approach is to "complete the square" by treating $\sqrt x$ as the variable quantity. Then you have

$$y=Ax+B\sqrt x\\ =Au^2+Bu$$

By the quadratic formula, we then have

$$u={-B\pm\sqrt{B^2+4Ay}\over 2y}=\sqrt x$$

Continuing the inversion should be easy from this point.

Answer 3

It depends on the values of $A$ and $B$.

To get the trivial objections out of the way first, if $A = 0$ or $B = 0$ you can easily invert the function, but if $A = B = 0$ then clearly you cannot invert the function, as there is no inverse function for $y = 0$.

So assume that neither $A$ nor $B$ is zero. Try plotting $y= Ax + B\sqrt x$ where $A$ and $B$ have the same sign (for example, $A = 1$, $B = 1$). Now try it where $A$ and $B$ have opposite signs (for example, $A = 1$, $B = -1$).

If $A$ and $B$ are both positive then then $f(x) = Ax + B\sqrt x$ describes a one-to-one and onto function whose domain and range are both $\mathbb R^+$ (the positive real numbers). To find the inverse of this function, use the fact (shown in other answers) that

$$\sqrt x=\frac{-B\pm\sqrt{B^2+4Ay}}{2y}.$$

For the inverse function, we assume $y > 0$ (that is, $y$ must be in the range of the original function). Therefore $B^2+4Ay > B^2$, so $\sqrt{B^2+4Ay} > B$, so $-B + \sqrt{B^2+4Ay} > 0$. On the other hand, $-B - \sqrt{B^2+4Ay} < 0$, which is an impossible value for $\sqrt x$. So only the $+$ case of the $\pm$ can occur, and

$$\sqrt x = \frac{-B + \sqrt{B^2+4Ay}}{2y}.$$

The inverse function is given by the formula

$$x = \left(\frac{-B + \sqrt{B^2+4Ay}}{2y}\right)^2.$$

If $A$ and $B$ are both negative, then $f(x) = Ax + B\sqrt x$ is always negative when $x > 0$. We can say that $f(x)$ is a one-to-one and onto function with domain $\mathbb R^+$ and range $\mathbb R^-$ (negative real numbers). I will skip a few steps here that are similar to what we did before for positive $A$ and $B$, but we can find an inverse function with domain $\mathbb R^-$ and range $\mathbb R^+$:

$$x = \left(\frac{-B - \sqrt{B^2+4Ay}}{2y}\right)^2.$$

But when $A$ and $B$ have opposite signs, the function $f(x) = Ax + B\sqrt x$ with domain $\mathbb R^+$ is not one-to-one. There will be some pairs of numbers $x_1$ and $x_2$, $0 < x_1 < x_2$, such that $Ax_1 + B\sqrt{x_1} = Ax_2 + B\sqrt{x_2}$. In order to make an invertible function, you will have to limit the function's domain further so that only one possible value of the domain remains for each function value that is produced in the range.