Monotonicity of Product of Functions

Calculus Level 3

A function $f(n)$ is monotonically increasing on some interval $[a,b]$ if for every $m, n \in [a,b]$ , $m \leq n$ implies $f(m) \leq f(n)$ .

Is the following statement TRUE ?

If $f(n)$ and $g(n)$ are monotonically increasing functions on $[a,b]$ , then $fg(n)$ is also monotonically increasing on $[a,b]$ .

Notations:

$[a,b] = \{ x \in \mathbb{R} : a \leq x \leq b\}$ , where $a, b \in \mathbb{R}$ with $a\leq b$ .
$fg(n) = f(n) \times g(n)$ , for every $n$ belonging to the intersection of the domains of $f(n)$ and $g(n)$ .

No Yes

1 solution

Muhammad Rasel Parvej
Dec 1, 2016

Consider $f(n)=n-4$ and $g(n)=n-2$ who are monotonically increasing on $[0,100]$ .

But " $1 \leq 3 \implies fg(1) \leq fg(3)$ " is false, hence, the statement is $FALSE$ .

$fg(1)= f(1) \times g(1)=3 > fg(3)=f(3) \times g(3)= -1$ .

By the way, you can verify this fact:

If $f(n)$ and $g(n)$ are monotonically increasing functions on $[a,b]$ AND each of $f(n)$ and $g(n)$ possibly touches the $x-axis$ but doesn't cut it over the interval $[a,b]$ , then $fg(n)$ is also monotonically increasing on $[a,b]$ .

Yes, for a function $f(n)$ , " $f(n)$ possibly touches the $x-axis$ but doesn't cut it over the interval $[a,b]$ " is equivalent to to saying exactly one of the following:

$f(n) \geq 0$ for every $n \in [a,b]$ . (All the way Non-negative )
$f(n) \leq 0$ for every $n \in [a,b]$ . (All the way Non-positive )

Sign change should not be there

Kushal Bose - 4 years, 6 months ago

Yeah. It's so.

Muhammad Rasel Parvej - 4 years, 6 months ago

The trivial example is $f(n) = g(n) = n$ . However, your answer seems correct.

Michael Huang - 4 years, 6 months ago

Monotonicity of Product of Functions

1 solution

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