How do you find the inverse of #f(x)=x^3+1# and graph both f and #f^-1#?

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Answer 1 · 2017-01-23T15:27:28

THe answer is #root(3)(x-1)#

Let #y=x^3+1#

#x^3=y-1#

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

Therefore,

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

Verification :

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

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

#=x-1+1=x#

The graphs of #f(x)# and #f^-1(x)# are symmetric wrt #y=x#

graph{(y-x^3-1)(y-root(3)(x-1))(y-x)=0 [-6.243, 6.243, -3.12, 3.123]}