Given #f(x)=1/8x-3# and #g(x)=x^3#, how do you find #(f^-1of^-1)(6)#?

1 Answer
Steve M
Jun 16, 2018

# (f^(-1) @ f^(-1))(6) = 600 #

Explanation:

We have:

# f(x) = 1/8x-3 #

And so we can construct the inverse, #f^(-1)(x)# by writing:

# y = 1/8x-3 #

So that:

# 1/8x = y+3 => x = 8y+24#

Thus we have:

# f^(-1)(x) = 8x+24 #

So that:

# (f^(-1) @ f^(-1))(x) = f^(-1)( 8x+24 )#

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 8( 8x+24 ) + 24 #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 64x + 194 + 24 #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 64x + 216 #

Hence:

# (f^(-1) @ f^(-1))(6) = 64*6+216 = 600#

Noting that #g(x)# is not required.

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