If the range of $y = f(x)$ is $-1\leq y\leq 2$, what is the range of $y = 1/f(x)$ If the range of $y = f(x)$ is $-1\leq y\leq 2$, what is the range of $y = 1/f(x)$
Could someone explain why is it not $-1\leq y\leq 1/2$?
 A: When you take the reciprocal of the sides of this inequality, you need to flip the inequality signs. Because $y = \frac{1}{f(x)}$ is undefined when $f(x) = 0$ and $y$ is not monotonic over it's entire domain, let's seperate the inequality into two inequalities as following:
\begin{align}
\tag{1}
0 > f(x)\geq -1
\newline 
\tag{2}
0 < f(x) \leq 2
\end{align}
Then, from $(1)$ and $(2)$, you get:
\begin{align}
\tag{3}
\frac{1}{f(x)}\leq-1
\newline
\tag{4}
\frac{1}{f(x)}\geq \frac{1}{2}
\end{align}
We can say that the range of $y = \frac{1}{f(x)}$ is $(-\infty,-1] \cup [\frac{1}{2},\infty)$
Note: A way you can intuitively see this is for $-1\leq f(x)<0$, as $f(x)$ get closer to zero, the value of $\frac{1}{f(x)}$ gets bigger, starting from $-1$. You can use a similar method to see for $0<f(x)\leq 2$, this time the value of $\frac{1}{f(x)}$ will get closer to $1/2$ when the value of $f(x)$ gets closer to $2$, and the value of $\frac{1}{f(x)}$ will approach $\infty$ while $f(x)$ approaches $0$.
A: *

*
Using $f(x)=x$ (the red line) as an example, notice that  its
reciprocal (the blue curve) has a greater absolute value than itself
on $(-1,1).$
Also, notice that its reciprocal gets infinitely large on the right
side of $0$ and infinitely small on the left side of $0.$
This shows why the reciprocal of a function with range $[-1,2]$
cannot possibly lie within $[-1,\frac12].$


*In general, when applying a decreasing function to an inequality and flipping its inequality signs, the new inequality is equivalent to the original one.
Notice that $\displaystyle\frac1x$ is a decreasing function if and only if its domain lies on either side of the $y$-axis. So, the inequality $$a\leq f(x)\leq b$$ is equivalent to $$\frac1a\geq \frac1{f(x)}\geq \frac1b$$ if and only only $a$ and $b$ have the same sign.
A: Your question is equivalent to finding the range of $g(x)=\frac{1}{x}$ over the domain [-1,2]. Notice that the function is not defined at $x=0$ so we instead find its domain over $[-1,0)\cup(0,2]$.
Now, the $g(x)$ is strictly decreasing over the intervals $(-\infty,0)$ and $(0,\infty)$ which means that

*

*for any $a,b\in (-\infty,0)$  where $a<b$, $g(a)>g(b)$

*for any $c,d\in (0,\infty)$  where $c<d$, $g(c)>g(d)$
You can see this by either plotting $g(x)=1/x$ , or by differentiating to find $g'(x)=\frac{-1}{x^2}<0$ for all $x\in \mathbb{R}≠0$
So for $$0<x≤2$$
$$g(0)>g(x)≥g(2)$$
Αs stated before, $g(0)$ is not defined, but what you can do is see that $$\lim_{x\rightarrow0^+}\left(\frac{1}{x}\right)=+\infty$$
so , roughly speaking, $$+\infty>g(x)≥\frac{1}{2} $$
$$\Rightarrow \text{range of g over }(0,2]=\left[\frac{1}{2},\infty \right)$$
Similarly,for $$-1≤x<0$$
$$g(-1)≥g(x)>g(0)$$
and again,$$\lim_{x\rightarrow0^-}\left(\frac{1}{x}\right)=-\infty$$
so , roughly speaking, $$-1≥g(x)>-\infty $$
$$\Rightarrow \text{range of g over }[-1,0)=\left(-\infty,-1\right]$$
Note: The reason we had to divide the domain into 2 parts was to ensure that the upper and lower bounds lay in a domain in which the function was strictly decreasing.
You can use this strategy to solve many inequalities in general. For example, since $\ln(x)$ and $e^x$ are monotically increasing over their entire domains, you can take natural logs/exponentials of both sides while preserving the inequality. Similarly, if sides of the inequality are both positive or  both negative, you can square with both while preserving the inequality, as $x^2$ is strictly increasing over $(-\infty,0]$ and $[0,\infty)$.
