If f(x)=2x^3/5+7 ,find f^-1(x)?

1 Answer
KillerBunny
Jun 21, 2018

#f^{-1}(x) = \root[3]{5/2(x-7)}#

Explanation:

Given a function #y=f(x)#, you can find its inverse #x = f^{-1}(y)# by solving the equation for #x#. By doing so, you change the roles of dependant and independant variables.

So, if we start from

#y = 2/5x^3+7#

We subtract #7# from both sides:

#y-7 = 2/5 x^3#

We multiply both sides by #5/2#:

#5/2(y-7)=x^3#

And finally extract the third root:

#\root[3]{5/2(y-7)} = x#

So, given #f(x)=2/5x^3+7#, we have #f^{-1}(x) = \root[3]{5/2(x-7)}#

As you can see here, the graphs of a function and its inverse are symmetrical with respect to #y=x#.

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