What point does it lie on?

It generally helps to identify the equation for #f(x)# (although it is not necessary). First, we'll try this without an equation, and then we'll try this by finding an equation.

The two graphs superimposed onto each other look like this:

graph{((x-1)^2 - 3 - y)(sqrt(x+3)+1 - y)(-sqrt(x+3)+1 - y) = 0 [-17.44, 23.11, -10.89, 9.39]}

METHOD 1

An inverse is defined so that some coordinate #(x,y)# in #f(x)# is found as #(y,x)# in the inverse, #f^(-1)(x)#. That is, the inversion of #f(x)# moves a point #(x,y)# to #(y,x)#.

So, to work backwards, select each answer and invert its coordinates from #(y,x)# in #f^(-1)(x)# to #(x,y)# in #f(x)# to see if it lies on #f(x)#.

• #(3,1) -> (1,3)#, which is not on #f(x)#.
• #(2,-2) -> (-2,2)#, which is not on #f(x)#.
• #(1,-3) -> (-3,1)#, which is not on #f(x)#.
• #color(blue)((-3,1) -> (1,-3))#, which is on #f(x)#.

To be clear, this means that #(-3,1)# is on #f^(-1)(x)# and #(1,-3)# is on #f(x)#.

METHOD 2

Or, we could construct an equation for #f(x)#. By shifting the equation back to the origin, we shift it left 1 and up 3 to get an equation where #y = ax^2#.

This means #f(x)# is of the form that shifts it right 1 (subtract 1 in parentheses) and down 3 (subtract 3 outside parentheses):

#f(x) = a(x-1)^2 - 3#

remembering that #a(x + h) + k# shifts left by #h# units and up by #k# units, sign included.

So now, given one point #(3,1)# on #f(x)# we can solve for #a#:

#1 = a(3 - 1)^2 - 3#

#4 = 4a#

#=> a = 1#

and the equation should be #f(x) = (x-1)^2 - 3#:

graph{(x-1)^2 - 3 [-10, 10, -5, 10]}

The more mathematical approach then is to take

#y = (x-1)^2 - 3#

and swap #x# and #y#, solving for #y# again.

#x = (y-1)^2 - 3#

#x + 3 = (y - 1)^2#

#=> color(blue)(y = f^(-1)(x) = pm sqrt(x + 3) + 1)#

which looks like this:

graph{(sqrt(x+3) + 1 - y)(-sqrt(x+3) + 1 - y) = 0 [-4.96, 15.04, -3.88, 6.12]}

From here you can see that since #(1,-3)# is on #f(x)#, #(-3,1)# is on #f^(-1)(x)#:

#(1) stackrel(?" ")(=) cancel(pmsqrt((-3) + 3))^(0) + 1#

#=> 1 = 1#

which shows that #(-3,1)# is on #f^(-1)(x)#.