If #g(x)# has an inverse function, which we call #g^-1(x)#, then #g[g^-1(x)]=x# (for all #x# in the domain of #g#).
A function #f# is like a "little black box". It takes in an input #x#, does something to it, and returns an output #f(x)#.
When we seek an inverse function, what we're looking for is another little black box that, in essence, undoes the action of #f#.
#stackrel"input"stackrel""x ->stackrel"function"stackrel""f->stackrel"output"stackrel""f(x)" "->->" "stackrel"input"stackrel""f(x)->stackrel"inverse"stackrel"function"stackrel""(f^-1)->stackrel"output"stackrel""x#
The same will hold true if we start with the inverse function. Meaning, if we start with an input #x# and plug it into #g^-1#, we'll get an output #g^-1(x)#. Then, if we take that output and give it to #g# as input, #g# will undo the effect of #g^-1#, and we get plain old #x# as our output.
For this question, we've been given #g(x)=x/8#. We also know that for #g# and #g^-1# to be inverses, #g[g^-1(x)]=x#. Combining these two statements gives us
#" "g(color(red)x)=color(red)x/8#
#=>g[color(red)(g^-1(x))]=color(red)(g^-1(x))/8#
#=>" "x=g^-1(x)/8#
#=>" "8x=g^-1(x)#
Look at that—we have an equation that says, "The function #g^-1# takes in an input #x#, and returns #8x#." This is the inverse function of #g.#
We can even test it:
#g(x)=x/8" "=>" "g^-1[g(x)]=g^-1(x/8)#
#color(white)(g(x)=x/8" "=>" "g^-1[g(x)])=8 * x/8#
#color(white)(g(x)=x/8" "=>" "g^-1[g(x)])=x#
This shows that it always works; no matter what input we give #g#, when we take that output and feed it into #g^-1#, we get the original input back.
