Given #f(x)=1/8x-3# and #g(x)=x^3#, how do you find #(f^-1of^-1)(6)#?
1 Answer
Jun 16, 2018
# (f^(-1) @ f^(-1))(6) = 600 #
Explanation:
We have:
# f(x) = 1/8x-3 #
And so we can construct the inverse,
# 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