Solving $\frac{1}{x} < 4$ gives me $x > \frac{1}{4}$. The book however states the answer is: $x < 0$ or $x > \frac{1}{4}$.
My questions are:
Why does this inequality has two answers (preferably the intuition behind it)?
When using Wolfram Alpha it gives me two answers, but when using $1 < 4x$ it only gives me one answer. Aren't the two forms equivalent?
You have to be careful when multiplying by $x$ since $x$ might be negative and hence flip the inequality. Suppose $x>0$. Then $$\frac{1}{x}<4\iff4x>1\iff x>1/4.$$ If $x>0$ and $x>1/4$, then $x>1/4$.
Now suppose $x<0$. Then $$\frac{1}{x}<4\iff4x<1\iff x<1/4.$$ If $x<0$ and $x<1/4$, then $x<0$. So the solution set is $(-\infty,0)\cup(1/4, \infty).$
Here is the solution $$\frac { 1 }{ x } <4$$$$ \frac { 1-4x }{ x } <0$$$$ \frac { x\left( 1-4x \right) }{ { x }^{ 2 } } <0$$$$ x\left( 1-4x \right) <0$$$$ x\left( 4x-1 \right) >0 $$
so $$x\in \left( -\infty ,0 \right) \cup \left( \frac { 1 }{ 4 } ,+\infty \right) $$
I am following the suggestion given by @quid in the comments because I like pictures:
The orange/red line is the $x$-axis. The yellow line is the line $y=4$. The two blue curves are the graph of $y=1/x$. The solution to the inequality is the set of $x$ values for which the blue curve is below the yellow line. As @quid predicted, this picture gives some intuition for why there are two "zones" in the solution.
Note: originally I had $y=\frac{1}{4}$ which was incorrect, so I changed the answer to reflect the fact that it should be $y=4$ and re-plotted the graph.
The accepted answer is good, but I feel like you're really asking: why is there only one piece to the question, but two pieces to the answer?
This is actually a great question. Sometimes it happens that one piece turns into two (or more), like when you try to solve $x^2 = 9$ (which has "one piece") and get the two-piece solution $x = 3, -3$. Here, to figure out why one piece becomes two, you have to think about how the equation $y = x^2$ works.
So in our case we should think about how the equation $y = 1/x$ works. And when you think about it, you realize that you didn't really start with one piece. No matter what you plug in for $x$, the value of $1/x$ can never be zero. And that means when you write $1/x < 4$, really this gives you the TWO pieces
$$ 0 < 1/x < 4 $$
and
$$ 1/x < 0 $$
Basically, everything smaller than 4 but with zero removed. And that's why you end up with two pieces at the end -- because that's actually how many you started with!
Just draw a graph of $1/x$ and you'll see 'why'.
Here is an image by WolframAlpha, with appropriate parts enhanced:
English doesn't have good words for this, so exactly what's going on can be a bit tricky to describe if you don't already understand the meaning.
It is true that every $x$ satisfying $x > \frac14$ does in fact satisfy $\frac 1x < 4$.
However, the converse fails: there are some $x$ that satisfy $\frac 1x < 4$ that do not satisfy $x > \frac 14$.
So, you have found a simple description of some of the $x$ that satisfy $\frac 1x < 4$ — but (presumably) you were being asked to describe all of the $x$ that satisfy $\frac 1x < 4$.
And there are indeed more of them: every $x$ satisfying $x < 0$ satisfies $\frac 1x < 4$.
Now, what is true is the following: if $\frac 1x < 4$, then it follows that at least one of the two statements "$x > \frac 14$" and "$x < 0$" is true. Thus, this gives a complete description of the solutions to $\frac 1x < 4$.
Put differently, for every $x$, the following bullet points are either both true or both false:
Regarding your solution method, you forgot that multiplying by negative numbers reverses the sign of an inequality, and multiplying by zero turns any inequality into an equality. Here, we know that $x$ can't be zero, but it still could be either positive or negative, so you don't know the effect that multiplying by $x$ will have on the inequality.
The typical way to fix this problem is to break the problem into two parts: one part where you solve the case with the assumption $x<0$, and one part where you solve the case with the assumption $x>0$, and then you put the results together.
we have $$\frac{1}{x}<4$$ is true if $$x<0$$ and if $x>0$ we get $$x>\frac{1}{4}$$
No, $$\frac 1x<4\text{ and }1<4x$$ are not equivalent.
You could think so just multiplying by $x$. But a rule says that an inequality is preserved when you multiply by a positive number and inverted with a negative one. So the right thing is
$$\begin{cases}x>0\to1<4x,\\x<0\to1>4x.\end{cases}$$
$$\frac{1}{x} < 4,$$ $$\frac{1}{x} - 4 < 0,$$ $$\frac{1 - 4x}{x} < 0,$$ $1 - 4x < 0$ or $x > 0$. Because if $1 - 4x$ is negative then $x$ must be positive. So we must write the solution in that way.
Here is an important aspect which should be always considered. If someone asks me:
Problem: Find the solution of \begin{align*} \frac{1}{x}<4 \end{align*} I would not answer the problem, but instead ask: What is the domain of $x$?
Please note the problem is not fully specified if the domain of $x$, the range of validity, is not given. This is crucial to determine the set of solutions.
Some examples:
Find the solution of
\begin{array}{lcl} \text{domain of }x\qquad&\qquad\text{inequality}\qquad&\qquad\text{solution}\\ \hline\\ \{x|x\in\mathbb{R}\setminus\{0\}\}\qquad&\qquad\frac{1}{x}<4\qquad&\qquad (-\infty,0)\cap(1/4,\infty)\\ \{x|x\in\mathbb{R}^{+}\}\qquad&\qquad\frac{1}{x}<4\qquad&\qquad (1/4,\infty)\\ \{\pi\}\qquad&\qquad\frac{1}{x}<4\qquad&\qquad \{\pi\}\\ \{x|x\in(0,1/4)\}\qquad&\qquad\frac{1}{x}<4\qquad&\qquad \emptyset \end{array}
Note: If a domain is not explicitly stated in the problem section of a book we should expect a corresponding statement somewhere else at the beginning of the chapter.
we have $\frac{1}{x}>4$ then $ \frac{1}{x}-4>0$
that is , $$\frac{1-4x}{x}>0$$ but the domain of definition is $x\neq 0$
first of all you need to find the zeros and then study the its signs $$1-4x=0$$ then $$x=1/4 $$
$\begin{align} & \underline{\left. x\,\,\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,-\infty \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1/4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\infty } \\ & \underline{\left. 1-4x\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \,\left. \, \right| \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ & \underline{\left. x\,\,\,\,\,\,\,\,\,\,\,\, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \left. \, \right| \right|\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ & \left. \frac{1-4x}{x}\, \right|\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \,\left. \, \right| \right|\,\,\,\,\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \, \right|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \end{align}$
as we observe from the above table that inequality is positive only when $0<x<\frac{1}{4}$
I would like to harp on the meaning of your "two answers". When a question asking for a number satisfying some conditions leads to a unique number having that property we are fine. When there are two numbers having the required property then we can say two answers.
Here we have uncountably many real numbers $x$ such that $\frac1x < 4$. So even the region $x>4$ is infinitely many answers. One possible interpretation that could justify "two" could be number of connected components of the solution space of the question. This happens many times. The set $GL(n,\mathbf{R})$ of non-singular real $n\times n$ matrices has two connected components.
