How do you find the inverse of #f(x)=x^5-2# and graph both f and #f^-1#?

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Answer 1 · 2017-01-26T11:03:09

The answer is #f^-1(x)=root(5)(x+2)#

Let #y=x^5-2#

#x^5=y+2#

#x=root(5)(y+2)#

Therefore,

#f^-1(x)=root(5)(x+2)#

Verification

#f(f^-1(x))=f(root(5)(x+2))=(root(5)(x+2))^5-2#

#=x+2-2=x#

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

graph{(y-(x^5-2))(y-x)(y-root(5)(x+2))=0 [-10, 10, -5, 4.995]}