In mathematics, the reciprocal or multiplicative inverse of a number, $n$, is ${n}^{1}equals;\frac{1}{n}$, because this satisfies the multiplicative identity: $n\cdot {n}^{1}equals;\frac{n}{n}equals;1$. For a rational number $\frac{a}{b}$, the reciprocal is given by $\frac{b}{a}$.
Following this definition, for a function $f\left(x\right)$, the reciprocal function is $yequals;\frac{1}{f\left(x\right)}$. If $f\left(x\right)$ is a rational function of the form $f\left(x\right)equals;\frac{h\left(x\right)}{g\left(x\right)}$, its reciprocal function will be $yequals;\frac{g\left(x\right)}{h\left(x\right)}$.

Interesting Properties of Reciprocal Functions


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The reciprocal function has vertical asymptotes wherever the original function has xintercepts, and xintercepts wherever the original function has vertical asymptotes. If ${x}_{0}$ is an isolated root of the original function, that is, if $f\left({x}_{0}\right)equals;0$ and $f\left(x\right)\ne 0$ for other values of x near ${x}_{0}$, then the reciprocal function will approach ± infinity at these points, creating vertical asymptotes. Conversely, if a vertical asymptote occurs in the original function at ${x}_{0}$, that is, its value approaches ± infinity as x approaches a given value ${x}_{0}$, then the reciprocal function will have a root at ${x}_{0}$.

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Intervals, in which the original function is increasing, correspond with intervals in which the reciprocal function is decreasing. Meanwhile, intervals in which the original function is decreasing, correspond with intervals in which the reciprocal function is increasing. This occurs because as $f\left(x\right)$ increases, $yequals;\frac{1}{f\left(x\right)}$ must decrease. Similarly, as $f\left(x\right)$ decreases, $yequals;\frac{1}{f\left(x\right)}$ must increase.

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The original and reciprocal functions will intersect at the points where $f\left(x\right)equals;1$ and $f\left(x\right)equals;1$. This occurs because the reciprocal function will have the same value as the original, since $yequals;\frac{1}{1}equals;1$ and $yequals;\frac{1}{1}equals;1$.


