Given #f(x)=1/8x-3# and #g(x)=x^3#, how do you find #(f^-1og^-1)(1)#?

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Answer 1 · 2017-02-16T04:44:37

#(f^-1 @ g^-1)(1) = 2root(3)(4)#

To find the inverse functions, switch the #x# and #y# values.

Finding #g^-1(x)#

#x = y^3#

#y = root(3)(x)#

Finding #f^-1(x)#

#x = 1/8y - 3#

#x + 3 = 1/8y#

#y = 8x + 24#

Now, the notation #(f @ g)(x)# is equivalent to #f(g(x))#.

#(f^-1 @ g^-1)(1) = f^-1(g^-1(1))#

Evaluate now using function notation.

#g^-1(1) = 8(1) + 24 = 8 + 24 = 32#

#f^-1(32/3) = root(3)(32) = root(3)(8 * 4) = 2root(3)(4)#

Hopefully this helps!