Reciprocal Function is Strictly Decreasing/Proof 1
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Theorem
The reciprocal function:
- $\operatorname{recip}: \R \setminus \set 0 \to \R$, $x \mapsto \dfrac 1 x$
- on the open interval $\openint 0 \to$
- on the open interval $\openint \gets 0$
Proof
Let $x \in \openint 0 {+\infty}$.
By the definition of negative powers:
- $\dfrac 1 x = x^{-1}$
From Power Rule for Derivatives:
\(\ds \frac \d {\d x} x^{-1}\) | \(=\) | \(\ds -x^{-2}\) |
From Square of Real Number is Non-Negative:
- $-x^{-2} < 0$
for all $x$ within the domain.
Thus from Derivative of Monotone Function, $\operatorname{recip}$ is strictly decreasing.
The proof for $x \in \openint {-\infty} 0$ is similar.
$\blacksquare$