Question

How do you show that $f(x) = 7x + 1\;{\text{and}}\;g(x) = \dfrac{{x - 1}}{7}$ are inverse functions algebraically and graphically?

Answer 1

Hint: To show algebraically use composite function and show output of $fog(x) = gof(x)$ equals input that is $x$. And to proof it graphically, plot graphs of both the lines and show that they are symmetric about the line $x = y$

Complete step by step solution:
Proving that $f(x) = 7x + 1\;{\text{and}}\;g(x) = \dfrac{{x - 1}}{7}$ are inverse functions algebraically:
To show $f(x) = 7x + 1\;{\text{and}}\;g(x) = \dfrac{{x - 1}}{7}$ are inverse function algebraically, we will first find the composite function $fog(x)\;{\text{and}}\;gof(x)$
Composite function $fog(x)$ will be written in the way that we will take function $g(x)$ as the argument of the function $f(x)$ as follows
\[
  f(g(x)) = 7\left( {g(x)} \right) + 1 \\
   = 7 \times \dfrac{{\left( {x - 1} \right)}}{7} + 1 \\
   = x - 1 + 1 \\
   = x \\
 \]
Now similarly finding $gof(x)$, we will get
$
  gof(x) = \dfrac{{f(x) - 1}}{7} \\
   = \dfrac{{\left( {7x - 1} \right) + 1}}{7} \\
   = \dfrac{{7x - 1 + 1}}{7} \\
   = \dfrac{{7x}}{7} \\
   = x \\
 $
So we can see that the composite functions are giving the output equals input that is $fog(x) = gof(x) = x$
Therefore we can say that $f(x)\;{\text{and}}\;g(x)$ are inverse functions.
Again proving their inverse nature towards each other graphically,
We will plot the graph of $f(x) = 7x + 1\;{\text{and}}\;g(x) = \dfrac{{x - 1}}{7}$ and then see if they will be symmetric about the line $x = y$ or not. And if they will be symmetric then they are inverse function to each other.
Now to plot the graph we will rewrite the equations as
$y = 7x + 1\;{\text{and}}\;y = \dfrac{{x - 1}}{7}$
We will first find some points of both the equations, to plot their graph, so collecting points as follows

$x$	$y = 7x + 1$	$y = \dfrac{{x - 1}}{7}$	Coordinates for graph of first equation	Coordinates for graph of second equation
$0$	$1$	$ - \dfrac{1}{7}$	$\left( {0,\;1} \right)$	$\left( {0,\; - \dfrac{1}{7}} \right)$
$1$	$8$	$0$	$\left( {1,\;8} \right)$	$\left( {1,\;0} \right)$

So we get $\left( {0,\;1} \right)\;{\text{and}}\;\left( {1,\;8} \right)$ as points of graph for first equation and $\left( {0,\; - \dfrac{1}{7}} \right)\;{\text{and}}\;\left( {1,\;0} \right)$ as points of graph for second equation. So plotting their graph:

Blue line represents $f(x) = 7x + 1$
Green line represents $g(x) = \dfrac{{x - 1}}{7}$
And black line represents $x = y$
So we can clearly see that the green and blue lines are symmetric about black line. Hence proved

Note: When finding the complex function then simply put the other function in the parent’s function argument (or replace it with $x$ ) then solve further to get the composite function.
Also consider such points to plot graphs which are whole numbers and at reasonable distance from each other.