6,406 views
29 votes
29 votes

If $g(x) = 1 - x$ and $h(x) = \frac{x}{x-1}$, then $\frac{g(h(x))}{h(g(x))}$ is:

  1. $\frac{h(x)}{g(x)}$
  2. $\frac{-1}{x}$
  3. $\frac{g(x)}{h(x)}$
  4. $\frac{x}{(1-x)^{2}}$

4 Answers

Best answer
36 votes
36 votes
option a) is correct.

$g\left(h\left(x\right)\right) =g\left(\frac{x}{x-1}\right)$
    $=1-\frac{x}{x-1}$
    $= \frac{-1}{x-1}$

$h\left(g\left(x\right)\right) =h(1-x)$
  $=\frac{1-x}{-x}$

$\frac{g(h(x))}{h(g(x))} = \frac{x}{(1-x)(x-1)} = \frac{h(x)}{g(x)}$
       

option A)
edited by
1 votes
1 votes
Correct answer is A because of

(x/x-1)*(1/1-x)

g(x) = 1-x and h(x)=x/x-1

so h(x)/g(x)
Answer:

Related questions

25 votes
25 votes
2 answers
3
makhdoom ghaya asked Feb 13, 2015
6,306 views
The binary operator $\neq$ is defined by the following truth table.$$\begin{array}{|l|l|l|} \hline \textbf{p} & \textbf{q}& \textbf{p} \neq \textbf{q}\\\hline \text{0} & ...
38 votes
38 votes
5 answers
4