piecewise ceiling function and what $g(x) = f(1/x)$ means

36 Views Asked by At

I have some limits I am supposed to find and I'm having trouble understanding the question.

Given:

$$f(x) = \begin{cases} 2 & \text{ if } \left \lceil x \right \rceil \text{is even}\\-1 & \text{ if } \left \lceil x \right \rceil \text{is odd} \end{cases}$$

$g(x) = f\left(\frac{1}{x}\right)$

what does $g(x) = f\left(\frac{1}{x}\right)$ mean? (maybe I'm making it more complicated than it should be in my head, it's supposed to be a 'challenge' problem.)

example: $g\left(\frac{1}{3}\right)$, does that mean $g\left(\frac{1}{\frac{1}{3}}\right)$ so then $g(3) = -1$ ?

or like a composite function where $g\left(\frac{1}{3}\right)$ is odd so $f\left(\frac{1}{3}\right) = -1$ then $g(x) = \frac{1}{-1} = -1$?

I realize both come out to $-1$, but I think it makes a difference for when I have to find the limits. Or maybe not?

2

There are 2 best solutions below

0
On

$g(x)=f(\frac 1x)$ is just as you suspect-if I ask you for $g(x)$ you compute $y=\frac 1x$ and ask somebody for $f(y)$. For any $x \gt 1, 0 \lt \frac 1x \lt 1, \lceil \frac 1x \rceil=1$, so $f(\frac 1x)=-1$.

0
On

$g(x) = f(\frac 1x)$ means just what it looks like:

$g:x\to f(\frac 1x)$.

So for example: If $x = \frac {\sqrt 3}4$ then

$g(x)=g(\frac {\sqrt 3}4) = f(\frac 4{\sqrt 3}) = \begin{cases}2&\text{if }\lceil \frac 4{\sqrt 3} \rceil \text{ is even}\\-1&\text{if }\lceil \frac 4{\sqrt 3} \rceil \text{ is odd}\end{cases}$

$=2$

Or $g(52) = f(\frac 1{52}) = \begin{cases}2&\text{if }\lceil \frac 1{52} \rceil \text{ is even}\\-1&\text{if }\lceil \frac 1{52} \rceil \text{ is odd}\end{cases}$

$=-1$

......

Or we could simple say $g(x) =\begin{cases}2&\text{if }\lceil \frac 1{x} \rceil \text{ is even}\\-1&\text{if }\lceil \frac 1{x} \rceil \text{ is odd}\end{cases}$

.....

Might, or might not, be worth noting:

If $0< x \le 1$ then $f(x) = -1$ and if $-1 < x \le 0$ then $f(x) = 2$.

So if $x \ge 1$ then $0< \frac 1x \le 1$ so $g(x) = -1$. And if $x \le -1$ then $-1 \le \frac 1x < 0$ so $g(x)=2$.

And $g(0)$ is not defined.