Basic doubt regarding reverse of wavy curve I am able to solve inequality ,
$$(x-3)(x-1) \leq 0$$
Which gives ,
$$x \in [1,3]$$
But
How to do reverse ,
If it is given
$$1\leq x \leq 3$$ .
How can I write it in form ?
$$(x-3)(x-1) \leq 0$$
Please help , I hope sir / mam you got, what I want to ask ? Thank you for your time and sorry if this doubt seems silly to you ...
 A: If you have a union of disjoint intervals $E=[a_1,b_1]\cup [a_2,b_2]\cup\ldots\cup[a_n,b_n]$ with $a_1<b_1<a_2<\ldots<a_n<b_n$ you may consider the function $f(x)=(x-a_1)(x-b_1)\cdot\ldots\cdot(x-a_n)(x-b_n)$. This is a polynomial with degree $2n$ and simple roots at the endpoints of the intervals. The sign of $f(x)$ changes each time $x$ crosses an endpoint, so the solution of $f(x)\leq 0$ can only be
$$ (-\infty,a_1]\cup[b_1,a_2]\cup\ldots\cup[b_n,\infty) $$
or $E$. Since for any sufficiently large $x$ we have that $f(x)>0$, $E$ is the solution set of $f(x)\leq 0$.

Extended answer: $f(x)$ is a continuous function which vanishes only at $a_1,b_1,\ldots,a_n,b_n$. Since its root are simple, the sign of $f(x)$ is constant over any interval of the form $(a_n,b_n)$ or $(b_n,a_{n+1})$, and the sign is alternating over consecutive intervals, even including $b_0=-\infty$ and $a_{n+1}=+\infty$. It follows that the sign over $(a_n,+\infty)$ establishes the sign over the whole real line.
This is a principle I try to teach almost every day. The common technique for solving inequalities like
$$ \frac{(x-1)(x-3)}{(x-2)(x-4)}\geq 0 $$
is to consider the sign of each factor of the numerator/denominator and put toghether these informations through a diagram (my students call it the cemetery, due to the usual abundance of $+$ signs interpreted as crosses), but such technique is usually highly inefficient. Except for $x\in\{1,2,3,4\}$ the sign of $\frac{(x-1)(x-3)}{(x-2)(x-4)}$ is the same as the sign of $(x-1)(x-2)(x-3)(x-4)$. This function is positive for $x>4$, so the ratio $\frac{(x-1)(x-3)}{(x-2)(x-4)}$ is positive for $x>4,2< x < 3$ and $x<1$ due to alternating intervals and signs. By considering what happens for $x\in\{1,2,3,4\}$ it follows that $\frac{(x-1)(x-3)}{(x-2)(x-4)}\geq 0$ holds for $x\leq 1, 2< x \leq 3$ and $x > 4$. No diagram needed (it is also possible to completely avoid the use of paper and solve these inequalities "at first sight", with sufficient training).
A: As an addition to the fine and more general answer by Jack D'Aurizio, note that the following holds
$$(x-3)(x-1) \leq 0 \iff 1\leq x \leq 3$$
and more in general for $b\le a$
$$(x-a)(x-b) \leq 0 \iff b\leq x \leq a$$
therefore we can easily convert one statement in the other one.
