Given f(x)=1/8x-3f(x)=18x−3 and g(x)=x^3g(x)=x3, how do you find (f^-1of^-1)(6)(f−1of−1)(6)?
1 Answer
Jun 16, 2018
(f^(-1) @ f^(-1))(6) = 600 (f−1∘f−1)(6)=600
Explanation:
We have:
f(x) = 1/8x-3 f(x)=18x−3
And so we can construct the inverse,
y = 1/8x-3 y=18x−3
So that:
1/8x = y+3 => x = 8y+2418x=y+3⇒x=8y+24
Thus we have:
f^(-1)(x) = 8x+24 f−1(x)=8x+24
So that:
(f^(-1) @ f^(-1))(x) = f^(-1)( 8x+24 )(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
