Determine wether f(x) = 3x + 8 and g(x) = x - 8 / 3 are inverse functions.
For f(x) = 3x - 19, find the inverse f-¹(x). Minus sign is supposed to be uppercase..
Thanks, skeeter! Sorry, I'm kind of new to this.. but I'm kind of stuck with proving g[f(x)]=x. The first one should be the following, right?
1. Determine wether f(x) = 3x + 8 and g(x) = x - 8 / 3 are inverse functions.
f[g(x)] = f[(x - 8) / 3]
= 3x [(x - 8) / 3] + 8
= x - 8 + 8
= 8
g[f(x)] = g(3x + 8)
..and then how would you continue? Or would you've done it differently?
2. I understood this one.. should be the following?
f^-1(x) = (x + 19) / 3
Br,
Evan
$\displaystyle f[g(x)] = 3g(x)+8 = \cancel{3}\left(\dfrac{x-8}{\cancel{3}}\right)+8 = (x-8)+8 = x$
$\displaystyle g[f(x)] = \dfrac{f(x)-8}{3} = \dfrac{(3x+8)-8}{3} = \dfrac{3x}{3} = x$
So, $\displaystyle f^{-1}(x) = g(x)$.
For the second one, yes, $\displaystyle f(x) = 3x-19 \Longrightarrow f^{-1}(x) = \dfrac{x+19}{3}$