How do you find the inverse of #f(x)=root3(x-1)# and graph both f and #f^-1#?

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Answer 1 · 2017-05-25T13:00:32

#f^-1(x)=x^3+1#

To find #f^-1(x)#, take the formula for #f(x)# and replace #f(x)# with #x#, and #x# with #f^-1(x)#.

#f(x) = root(3)(x-1)#

#x = root(3)(f^-1(x)-1)#

Now, solve for #f^-1(x)#.

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

#x^3 + 1 = f^-1(x)#

So our equation for the inverse is #f^-1(x)=x^3+1#.

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The graph of #f^-1(x)# will be the graph of #x^3# shifted up 1. The best way to graph this would be to plot a few points and then connect them.

Plugging in -2, -1, 0, 1, and 2, we get

#x = -2, y = -7#
#x = -1, y = 0#
#x=0,y=1#
#x=1,y=2#
#x=2,y=9#

Connecting these points gives you the graph of #f^-1(x)#:
enter image source here

Finally, to graph #f(x)#, switch the #x# and #y# coordinates of the 5 points above, and connect those new points.