simple inverse function question

Functions in the form of $y = f(x)$ describe various sorts of line.

In a quadratic line, for every extra unit in $x$, then $y$ increases by roughly $2x$.

A line where for every extra unit in $x$, then $y$ doubles is exponential, $y = 2^x$.

Thease can be inversed, for example:

For every doubling of $x$, then $y$ increases by $1$ is exponential, y = $log2(x)$.

What is this called with quadratic equations?

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So, square roots? – J. M. Apr 29 '11 at 13:50
Yes, thanks. Square roots for quadratics, and more generally roots for polynomials I suppose. – alan2here Apr 29 '11 at 14:16

Your question seems to be directed at inverse functions. Your statement "For every $1$ in $x$ then $y$ increases by $x$ is quadratic, $y = x^2$" is off by a factor of $2$, and even then it's only approximately true that $y$ increases by $2x$ when $x$ increases by $1$. The corresponding inverse statement is about the inverse function of $y=x^2$, which is $y=\sqrt{x}$: For every increase of $2y$ in $x$, $y$ increases by $1$, roughly.

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These are functional equations in words:

a) $f(x+1)=f(x)+x$

b) $f(x+1)=2f(x)$

c) $f(2x)=f(x)+1$

These all have many solutions. For (a) it can be $f(x)=x(x+1)/2$ though not $x^2$; for (b) as you say $2^x$; and for (c) $\log_2(x)$. But there are many other solutions. For example for (b) it could be $17 \cos(2\pi x) 2^x$. EqWorld has a large number of examples.

If you want $f(x)=x^2$ than a possible functional equation is $f(x+1)=f(x) + 2x + 1$; another is $f(x+1)=f(x) + 2\sqrt{f(x)} + 1$.

If you want $f(x)=\sqrt{x}$ than a possible functional equation is $f(x+1)=\sqrt{f(x)^2+1}$.

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