Let f(x) = (3x - 7)/(x + 1. Find the inverse f^-1(x).
Once again, thank you so much for helping me!
let y = (3x - 7)/(x + 1)
for the inverse , exchange the x and y's
x = (3y-7)/(y+1)
now solve this for y ...
xy + x = 3y - 7
xy - 3y = -x - 7
y(x-3) = -x - 7
y = (-x - 7)/(x - 3) = (x+7)/(3-x)
f^-1 (x) = (x+7)/(3-x)
check:
f(1) = (3-7)/(2
= -2
f^-1 (-2) = (-2+7)/(3-(-2))
= 5/5 = 1
my answer is reasonable
Thank you Reiny for helping! I understand the problem now! :)
To find the inverse function of f(x), denoted as f^-1(x), we need to follow a step-by-step process:
Step 1: Replace f(x) with y.
So, y = (3x - 7)/(x + 1).
Step 2: Swap x and y.
So, x = (3y - 7)/(y + 1).
Step 3: Solve for y in terms of x.
To do this, we will cross-multiply and simplify the equation:
x(y + 1) = 3y - 7.
xy + x = 3y - 7.
xy - 3y = -x - 7.
Factor out y on the left side:
y(x - 3) = -x - 7.
Divide both sides by (x - 3):
y = (-x - 7)/(x - 3).
Step 4: Replace y with f^-1(x).
Therefore, the equation f^-1(x) is given by:
f^-1(x) = (-x - 7)/(x - 3).
So, the inverse function of f(x) is f^-1(x) = (-x - 7)/(x - 3).