Question 1094396
<pre><font size = 4>
Let f(x) = 3x-2 and g(x) = {{{5/(3x)}}}, for x does not equal 0.
Find f<sup>-1</sup>(x).

1. In place of f(x), replace it by y

y = 3x-2

2. Interchange x and y

x = 3y-2

3. Solve for y
   x = 3y-2
3y-2 = x          <--swap sides
  3y = x+2
   y = {{{(x+2)/3}}}

4. In place of y, replace it by f<sup>-1</sup>(x)

f<sup>-1</sup>(x) = {{{(x+2)/3}}}  


Show that (g&#8728;f<sup>-1</sup>)(x) = {{{5/(x+2)}}}.

g(x) = {{{5/(3x)}}}

To avoid so many fractions, rewrite using "÷"

g(x) = {{{matrix(1,3,5,"÷",3x)}}}

In place of x, replace it by the right side of f<sup>-1</sup>(x),
which is {{(x+2)/2}}}

(g&#8728;f<sup>-1</sup>)(x) = {{{matrix(1,3,5,"÷",3((x+2)/3)))}}}

(g&#8728;f<sup>-1</sup>)(x) = {{{matrix(1,3,5,"÷",cross(3)((x+2)/cross(3))))}}}
(g&#8728;f<sup>-1</sup>)(x) = {{{matrix(1,3,5,"÷",x+2)}}}

(g&#8728;f<sup>-1</sup>)(x) = {{{5/(x+2)}}}
</pre>
Let h(x) = {{{5/(x+2)}}}, for x&#8807;0. The graph of h has a horizontal 
asymptote at y=0. 
Find the y-intercept of the graph of h. 
<pre>
We let x equal to 0

h(0) = {{{5/(0+2)}}}
h(0) = {{{5/2}}}

So the y-intercept is the point {{{(matrix(1,3,0,",",5/2))}}}

Hence, sketch the graph of h. 

Since x&#8807;0, the graph starts at the y-intercept and goes to the
right.  The graph does not extend left of the y-intercept:

{{{drawing(400,400,-2,8,-5,5,
locate(.2,3,(matrix(1,3,0,",",5/2))),circle(0,2.5,.1),
graph(400,400,-2,8,-5,5,(5/(x+2))*sqrt(x)/sqrt(x)))}}} 

You do the rest by yourself.  If you have trouble, tell me in
the thank you note form below, and I will get back to you by
email. 

Edwin</pre></font>