Let f(x)= 3x-7/x+1. Find the inverse f^-1(x). Thank you.
y = (3x-7) / (x+1)
inverse
x = (3y-7) / (y+1)
x y + x = 3 y - 7
3 y - x y = x + 7
y (3 - x) = (x+7)
y = -(x+7)/(x-3)
f^-1(x) = -(x+7)/(x-3)
x = (3y-7)/(y+1)
x(y+1) = 3y-7
xy + x = 3y - 7
y(x-3) = -(x+7)
y = (7+x)/(3-x)
To find the inverse of a function, we need to express the function in terms of its dependent variable (in this case, y) and then solve for x.
Given: f(x) = (3x - 7)/(x + 1)
Step 1: Replace f(x) with y:
y = (3x - 7)/(x + 1)
Step 2: Swap x and y:
x = (3y - 7)/(y + 1)
Step 3: Solve for y:
Multiply both sides by (y + 1) to eliminate the denominator:
x(y + 1) = 3y - 7
Expand:
xy + x = 3y - 7
Rearrange terms:
xy - 3y = -x - 7
Factor out y:
y(x - 3) = -x - 7
Divide both sides by (x - 3) to solve for y:
y = (-x - 7)/(x - 3)
Now we have expressed the inverse function f^(-1)(x) in terms of x:
f^(-1)(x) = (-x - 7)/(x - 3)