# Is $x^{-1}$ a linear function?

I want to know if $\frac1x$ is a linear function or not.

I read that a linear function can also be defined as a function of the first degree. Since $\frac1x=x^{-1}$, the function is of the first degree, isn't it?

But the function's graph does not look linear at all. What am I missing?

• What do you mean by "does not look like a function at all"? Sep 7 '16 at 9:19
• A linear function satisfies $f(a+b)=f(a)+f(b)$ and $f(\lambda a)=\lambda\,f(a)$. Does your function have those properties?
– lulu
Sep 7 '16 at 9:19
• @lulu No,But I was told that any function of the first degree is a linear function,Is that True?Does all first degree functions satisfy those conditions? Sep 7 '16 at 9:27
• @lulu: this works for linear functions as defined in Linear algebra. Linear functions in PreCalculus are usually $f(x)=mx+p$, and they don't satisfy these properties when $p\neq 0$. Sep 7 '16 at 9:27
• Your function has degree $-1$. More importantly, "degree" is really only defined for polynomials and your function isn't one.
– lulu
Sep 7 '16 at 9:29

"Linear" refers to a function having only terms of a positive first degree: $x^{+1}$. However, $\frac1x=x^{-1}$ is a term of negative first degree, so is not linear. It is correctly called an inverse linear or reciprocal function.
For the same reason, $\frac1{x^2}=x^{-2}$ is not a second-degree/quadratic equation, but an inverse square function, and so on for higher powers.
• $x^{+1} \cdot \frac{1}{x} = 1$ (for nonzero $x$), not $x^{-1}$. Sep 7 '16 at 13:42
As a property, linearity can be defined as: a function is linear if $f(ax+by) = af(x)+bf(y)$, with $a,b,x,y$ defined in the suitable domain. This collides with the notion of a linear function $y=f(x)=ax+b$, which is not linear in the above sense if $b\neq0$.
Your function is not linear with respect to the first meaning, for instance as in general $\frac{1}{ax}\neq a\times \frac{1}{x}$. For the second one, assuming $a\neq0$ and $b\neq0$ for simplicity, you should have $f(-b/a)=0$, but $\frac{1}{\frac{-b}{a}} = \frac{-a}{b}\neq0$. Cases $a=0$ or $b=0$ can be treated as special cases, and yield the conclusion.