Solve the inequality: $\displaystyle \frac{1}{x-6}\le 3$
solution:
\begin{align*}\frac{1}{x-6}& \le 3 \\ x-6& \le \frac{1}{3} \\x& \le 6+\frac{1}{3}\\ x&\le19/3\end{align*}
but, for values of $x\le 6$ also, inequality holds true as left hand side provides some negative value. but, i don't find this from the solution. what am i missing?
Hint: $$\frac{1}{x-6} \leq 3 \iff \frac{1}{x-6} - 3 \leq 0 \iff \frac{19-3x}{x-6} \leq 0 \iff \frac{3x-19}{x-6} \geq 0$$
Can you take it from here?
Your first step is wrong. If $A\le 3$, that doesn't mean $\dfrac 1 A\le \dfrac 1 3$. It does mean $\dfrac 1 A \ge \dfrac 1 3$ if $A$ is positive, but we can't assume it's positive in this case.
If you write this as $\dfrac 1 {x-6} - 3\le 0$, then use a common denominator to get just one fraction, then do the routine simplifications, you'll be well on your way toward solving this.
If $x < 6$, then $x - 6$ is negative, so multiplying both sides by $x-6$ reverses the inequality. Thus $1 \ge 3(x - 6)$, or $19 \ge 3x$, which is true for all $x < 6$.
Hint:
$a<b\iff\dfrac1a>\dfrac1b$ is true only if $a$ and $b$ have the same sign.
Hence you have to consider two cases.
With one variable, simplest is cases. If $x=6$ it is not defined. If $x<6$, as you noted LHS is negative, and the inequality trivially holds.
If $x>6$, we may multiply throughout by the positive quantity $x-6$, and then it should be easy.
In the end take the Union of the allowable regions you get.
To complete the given answer by user11235813:
$$\frac{3x-19}{x-6} \geq 0\to 3x\geq 19\ \&\ x\geq 6 \text{ or } 3x\leq 19 \ \&\ x\leq 6$$
So, the answer is $$x\geq \frac{19}3\ \cup\ x\leq 6$$
that is $$\mathbb{R}-(6,\frac{19}3)$$
Find the critical values: The values where the domain is undefined and when the function $f(x)= \frac{1}{x-6} -3$ is equal to $0$.
Here, the critical values are $6, \frac{19}{3} $
Use these values to divide the $x-axis$ into intervals: $(-\infty, 6), (6, \frac{19}{3}), (\frac{19}{3}, \infty) $
Test each intervals by plugging in the number within the interval and see if the value is lesser or equal than 3. If it is, then it's answer. If not, then you can eliminate that interval.
For example: $(-\infty, 6]$ pick $x=0$ then $ \frac{1}{0-6} = \frac{1}{-6} < 3$ so this is an answer.
$[6, \frac{19}{3}]$, pick $x= \frac{19}{3} $, then $\frac{1}{\frac{19}{3}-6} = 3$ So this is not an answer
Similarly, $[\frac{19}{3}, \infty) $, pick $x=7$, you can see that $\frac{1}{7-6} =1 < 3 $. Thus, this is an answer!
Therefore, the answer are the intervals: $(-\infty, 6)$ and $(\frac{19}{3}, \infty) $
Another way to do this is by graphing! Graph the function $f(x) = \frac{1}{x-6}$. This is the graph of the funcion $f(x) = \frac{1}{x}$ shifted right by 6 units. From the graph, you can see where the values are below the line $y=3$.
