$f(y)=\frac4{y-2},\;g(x)=\frac5{3x-1}$
Find the graph of $f\circ g$?
I am stuck at this step. Since $f(x)$ is not given.
$f\circ g\;=\;f(g(x))\\f(\frac5{3x-1})=\frac4{y-2}$
I am thinking whether $f^{-1}(y)=f(x)$?
$f(y)=\frac4{y-2},\;g(x)=\frac5{3x-1}$
Find the graph of $f\circ g$?
I am stuck at this step. Since $f(x)$ is not given.
$f\circ g\;=\;f(g(x))\\f(\frac5{3x-1})=\frac4{y-2}$
I am thinking whether $f^{-1}(y)=f(x)$?
On
Variables are just place holders.
If $f(y)=\frac4{y-2}$ then $f(x) = \frac 4{x-2}$ and $f(c) = \frac 4{c-2}$ and $f(\text{fred the tap-dancing goat}) = \frac 4{\text{fred the tap-dancing goat}-2}$
And $f(g(x)) = \frac 4{g(x)-2}$.
So $f\circ g$ is the function so that $f\circ g(x) = f(g(x)) = \frac 4{g(x)-2} = \frac 4{\frac 5{3x-1}- 2}$
So just graph the graph $y = \frac 4{\frac 5{3x-1}- 2}$
Which can be rewritten as
$y=\frac {4(3x-1)}{5 - 2(3x-1)};3x -1 \ne 0$ or
$y = \frac {12x-4}{-6x+7}; x\ne \frac 13$ or as
$y = \frac {12x-14 + 10}{-6x+7}= -2- \frac {10}{6x-7}; x\ne \frac 13$
If you have that $y = g(x) := \frac{5}{3x-1}$ and $f(y) := \frac{4}{y-2}$ then you may construct the function $f \circ g$ by the following:
Since the output of $g$ is $y$ and the input of $f$ is $y$ we can write $$f(g(x)) := \frac{4}{\frac{5}{3x-1}-2} = - \frac{4(3x - 1)}{6x - 7}$$ since we know know what $y$ is, we can just perform the substitution above and then simplify the expression with some algebra to obtain the result. The graph of the function $f \circ g$ is shown below where there is a vertical asymptote at $x = \frac{7}{6}$, a horizontal asymptote at $y = -2$, a gap at $x = \frac{1}{3}$, and no oblique asymptotes