What is the inverse of h?

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To find the inverse function of any function you switch the variables and solve for the initial variable:

#h(x)=6x+1#

#x=6h+1#

#6h=x-1#

#h^-1(x)=1/6(x-1)#

To find the inverse of #h(x)#, substitute #h^-1(x)# for every x within #h(x)#; this will cause the left side to become x . Then solve for #h^-1(x)# in terms of x. To verify that you have obtained the correct inverse, check that #h(h^-1(x)) =x# and #h^-1(h(x))=x#

Given: #h(x) = 6x+1#

Substitute #h^-1(x)# for every x within #h(x)#

#h(h^-1(x)) = 6(h^-1(x))+1#

The left side becomes x, because of the property #h(h^-1(x)) =x#:

#x = 6(h^-1(x))+1#

Solve for #h^-1(x)# in terms of x:

#x -1 = 6(h^-1(x))#

#h^-1(x) = 1/6(x-1)#

To verify that this is the correct inverse, check that #h(h^-1(x)) =x# and #h^-1(h(x))=x#.

#h(x) = 6x+1#

#h^-1(x) = 1/6(x-1)#

#h(h^-1(x)) = 6(1/6(x-1))+1#

#h^-1(h(x)) = 1/6((6x+1)-1)#

#h(h^-1(x)) = x-1+1#

#h^-1(h(x)) = 1/6(6x)#

#h(h^-1(x)) = x#

#h^-1(h(x)) = x#

Selection D) is the inverse

The way shown below is similar, but has some insight on visual verification.

The simplest way as shown by the others is to rewrite in terms of #x# and #y#

#y = 6x + 1#

and switch #x# and #y#, re-solving for #y#.

#=> x = 6y + 1#

#=> x - 1 = 6y#

#=> color(blue)(y = 1/6(x - 1))#

The graph of #h(x)# and #h^(-1)(x)# are superimposed here:

graph{(6x+1-y)(1/6(x-1) - y) = 0 [-2.798, 3.362, -1.404, 1.676]}

Notice how it is basically reflected over #y = x#. If you want to visually verify it, you can treat #y = x# as a reflection axis and generate #h^(-1)(x)# that way.