Is this supposed to be the solution to finding the inverse of
y = 1/3x - 5 ???
I suggest a 2-step process
1. interchange the x and y variables
----> x = 1/3y - 5
(you had that)
2. now solve this new equation for y
x = 1/3y - 5
x+5 = 1/3y
times 3
3x + 15 = y
so f^-1 (x) = 3x+15
Why did you involve a square root ??
Find the inverse of each function, if it exists.
f(x)=1/3x-5
y=1/3x-5
x=1/3y-5
x-5=1/3y
y=the square root of x-5/3
1 answer