Prove: Let $a,b\in R$ such that $a\lt b$, and $f:[a,b]\rightarrow R$ be monotonic, then $\frac{1}{f}$ is also monotonic

Question

Prove or disprove: Let $f:[a,b]\rightarrow \mathbb{R}$ be monotone. If $f(x)\ne 0$ for all [a,b], then $1/f$ is also monotone on [a,b].

I've been sitting on this for quite a while trying to find a function to disprove this.

If $x\le y$, then $f(x) \le f(y)$ (increasing) or $f(x)\ge f(y)$ (decreasing).

Soft-Question I need to either prove for all functions, or give one example where this is false. When given questions like this, what's the approach/ process when trying to prove/disprove? I tried approaching it by testing various elementary functions (square root, linear,polynomials, trig functions) but nothing has come up, meaning it has to be some sort of combination function, but I'm still having difficulties.

Answer 1

I believe a counter-example (check) is $$ f(x)=\begin{cases} x & \text{if }x<0;\\ x+1 & \text{if }x\geq0. \end{cases} $$ You can take your interval to be $[-1,1]$, but anything including zero suffices.

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As for your soft question, my thought process was as follows: I proved it for continuous functions, then I noticed that it seemed "too good to be true" for any function. I also noticed that the discontinuity should "avoid" the value zero on the $y$-axis, so I tried to think of a function with a single simple discontinuity.

Answer 2

If $f(x) \leq f(y)$ dividing both sides by $f(x) f(y)$ we get $$\frac{1}{f(x)} \geq \frac{1}{f(y)} \mbox{ whenever } f(x) f(y) > 0 \\ \frac{1}{f(x)} \leq \frac{1}{f(y)} \mbox{ whenever } f(x) f(y) < 0 $$

This tells you that $\frac{1}{f(x)}$ is monotonic exactly when $f(x) f(y)$ doesn't change sign (and it is easy to see that this happens exactly when this is always positive, meaning that $f(x)$ doesn't change sign).

Conclusion:

• If $f(x)$ doesn't change sign the claim is true.
• If $f(x)$ changes sign the claim is false.