How to know that we have found the correct solution set of an inequality? AND why is "my method" not working? Background: I was solving this inequality: $(1-4x)^{-1}\geq 7$.
I did it as follows:
$$
\begin{array}{l}
\quad(1-4 x)^{-1} \geq 7 \\
\Rightarrow \frac{1}{1-4 x} \geq 7 \\
\Rightarrow \frac{1}{7} \geq 1-4 x \\ 
\quad\left[\begin{array}{l}
1-4 x \neq 0 \text { and is  positive as if }\\
 \frac{a}{b}>0 
\text { and } a>0 \Rightarrow b>0.\\
\text {Hence I can multiply} \text{the sides with 1-4x,}\\
\text{without reversing sign. }
\end{array}\right] \\
\Rightarrow \frac{-6}{7} \geq-4 x \\
\Rightarrow x \geq\frac{6}{28} \\
\Rightarrow x \geq\frac{3}{14} \\
\therefore \text { Soln set }=\left[\frac{3}{14}, \infty\right)
\end{array}
$$
But this is not the answer; the answer is:
$
\left\{x \mid \frac{3}{14} \leq x<\frac{1}{4}\right\} \text { or }\left[\frac{3}{14}, \frac{1}{4}\right)
$
After thinking some time I realised that the other end point can be found if we solve the question as follows:
$$
\begin{aligned}
& \frac{1}{1-4 x} \geq 7 \\
\Rightarrow & 1-4x>0 \quad[\text { The reason is same }] \\
\Rightarrow & x<\frac{1}{4}
\end{aligned}
$$
Question: This post has two main questions:
First question: I know that the maximum number of roots an equation, in one variable of any degree, can have, will be same as its degree.
So for the inequalties, how could I be sure that I have found all the end point(s)? In some question the solution set is union of other two disjoint sets, So in this cases how could I be sure that I have found all the disjoint sets?
For this specific question, how could I be sure that this is finally the solution set? Can't it be that the author(of the book the problem is from) missed considering other inequalities, like me, which could have given us another set(s), so finally the answer would be union of those sets?
So I can sum up the above question in this question: How to know that we have found the correct solution set of an inequality?
Second question: Adding, subtracting and multiply(by non-zero number/polynomial) both side by same number/polynomial always led us to equivalent equation and inequality. In this case I multiplied by $1-4x$, which I know for sure is not zero. So according to the logic I must get an equivalent inequality i.e. the solution set of the new inequality must be same as that of the original inequality, but apparently this is not the case with this question i.e. solution set of $(1-4x)^{-1}\geq 7$ is not same as that of $\frac{1}{1-4x}\geq 7$. Why? I can get the correct answer by solving as others have mentioned in the answer, but, the second question is, why is "my method" not working?
 A: The reason for the correct answer is right here in your calculations:
$$ \frac{1}{7} \geq 1-4 x \quad\left[\begin{array}{l}
1-4 x \neq 0 \text { and is  positive as if }\\
 \frac{a}{b}>0 
\text { and } a>0 \Rightarrow b>0.\\
\text {Hence I can multiply} \text{the sides with 1-4x,}\\
\text{without reversing sign. }
\end{array}\right]$$
Specifically, it's this excerpt from the first line in the brackets:

$1-4 x \neq 0$  and is  positive ...

The $\neq 0$ part is redundant actually; all you really need is the part
that says it "is positive".
In short, you have shown that $$1-4 x  > 0. \tag1$$
The part on the left side of the brackets also is correct, of course:
$$ \frac17 \geq 1-4 x. \tag2 $$
From Inequation $(1)$ you get $$x < \frac14,$$
and from Inequation $(2)$ you get $$x \geq \frac3{14}.$$
Your error was that after proving $1-4 x  > 0$ you ignored that fact in the final solution set.
A: For inequalities when you multiply with a number you need first to verify it is not negative because if it then the direction will be flipped!
The approach to solve such a problem is move all terms to one side then you will have either $f(x)>0$ or $f(x)<0$ so you need to study the sign of the function and to do so you need to find it is zeros and point of discontinuities
A: When you multiply by $1-4x$, we are still not sure if we could have introduced additional solutions, implicitly, you have assume that $1-4x > 0$ and it should be incorporated to the solution directly and you should note it down and take intersection at the end.
What I do usually is rather than multiplying $1-4x$, I multiply by $(1-4x)^2$  and note that $x \ne \frac14$.
$$\frac1{1-4x} \ge 7, x\ne \frac14$$
$$1-4x\ge 7(1-4x)^2, x\ne \frac14$$
$$(1-4x)(7-28x-1) \le 0, x\ne \frac14$$
$$(1-4x)(6-28x) \le 0, x\ne \frac14$$
$$(1-4x)(3-14x) \le 0, x\ne \frac14$$
$$\frac3{14} \le x < \frac14$$
Once you convert a fractional inequality to polynomial form, you can sketch the region.
Edit:
Note that $(1-4x)^{-1}=\frac1{1-4x}$. You have to take intersection of the condition that you imposed.
A: To make sure of not missing any solution sets, you can do the calculations as follow:
$$\frac1{1-4x}\ge7\quad\Rightarrow\quad\frac1{1-4x}-7\ge0\quad\Rightarrow\quad\frac{1-7+28x}{1-4x}\ge0$$
We have $\dfrac{28x-6}{1-4x}\ge0$, hence $\frac3{14}\le x<\frac14$.
