If #f(x) = 5-2x^3#, what is #f^-1(x)#?

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Answer 1 · 2018-02-24T01:08:56

#f^-1(x)=root(3)(-1/2(x-5))#

#f(x)=y#

#y=5-2x^3#

Switch the places of #x# and #y#:

#x=5-2y^3#

Now, solve for #y#:

#x-5=-2y^3#

#y^3=-1/2(x-5)#

Take the cube root of both sides:

#root(3)y=root(3)(-1/2(x-5))#

#y=root(3)(-1/2(x-5))#

Now that we have solved for #y# after switching the places of #x# and #y#, #y=f^-1(x)#

#f^-1(x)=root(3)(-1/2(x-5))#