Name: Algebraic Operations: Functions
Uploaded: 2019-07-19
Duration: 7 min 10 s
Description: Now that we know what functions are, what can we do with them? Can we do operations with them? Yes! What won't these functions do? Let us learn more here.

Question

Find domain of

f (x) = \sqrt{\frac{\log_{2} (x - 2)}{\log_{1 / 2} (3 x - 1)}}

.

Accepted Answer

Given: $f (x) = \sqrt{\frac{\log_{2} (x - 2)}{\log_{1 / 2} (3 x - 1)}}$

$f (x)$ , will be defined when $\frac{\log_{2} (x - 2)}{\log_{1 / 2} (3 x - 1)} \geq 0 . . . (1)$ .

So the solution set of the inequality $(1)$ , will be the domain of the function.

From $(1)$ , we have, $(-) \frac{\log_{2} (x - 2)}{\log_{2} (3 x - 1)} \geq 0$

(change the base of $\log_{1 / 2} (3 x - 1)$ to $2$ , using base change property)

$\frac{\log_{2} (x - 2)}{\log_{2} (3 x - 1)} \leq 0 . . . (2)$ (if inequality is multiplied by $(- 1)$ , the inequality will reverse)

Now to solve $(2)$ , take

$\log_{2} (x - 2) = 0$ $\Rightarrow x = 3$ and $x > 2 . . . (3)$

Similarly, take $\log_{2} (3 x - 1) = 0$ $\Rightarrow x = \frac{2}{3}$ and $x > \frac{1}{3} . . . (4)$

Now plot these numbers $\frac{2}{3}$ , $3$ , in number line we get,

Take any value in the $(3, \infty)$ , let say $x = 4$ , and substitute this value in $(2)$ , we get

$\frac{1}{\log_{2} (11)} > 0$ , so the sign of this interval is positive and the next interval will be negative as shown in the figure.

So the solution set of $(2)$ , is $\frac{2}{3} < x \leq 3 . . . (5)$ .

Now take the intersection of all the inequalities, we get the final domain as,

So, from figure final intersection is $2 < x \leq 3$ ,

So, the domain of the given function is $x \in (2, 3] .$

Find domain of $f (x) = \sqrt{\frac{\log_{2} (x - 2)}{\log_{1 / 2} (3 x - 1)}}$ .

Important Questions on Functions

Learn and Prepare for any exam you want !

Find domain of fx=log2x-2log1/23x-1.

Important Questions on Functions

Learn and Prepare for any exam you want !

Find domain of $f (x) = \sqrt{\frac{\log_{2} (x - 2)}{\log_{1 / 2} (3 x - 1)}}$ .