Inverse function explanation please?

let y = 3/2 x - 5

for inverse, reverse x and y ==> x = 3/2 y - 5

now solve for y

x + 5 = 3/2 y

2(x + 5) = 3y

2/3(x + 5) = y

inverse:

y = f^-1(x) = 2/3 x + 10/3

or y = f^-1(x) = (2x + 10) / 3

or y = f^-1(x) = 2/3 (x + 5) , take your pick!

(remember, f^-1(x) is the inverse of f(x) , NOT 1 / f(x))

as a check: f(4) = 1, or the point (4 , 1) is on the original function

that means (1 , 4) must be on the inverse

check: f^-1 (1) = 2/3 (1 + 5) = 2/3 (6) = 4

yep

inverse function is to switch your variables and solve

f(x) = 3/2(x) - 5 => (x) = 3/2f(x) - 5

x + 5 = 3/2f(x)

2/3(x+5) = f(x)

f(x) = 2/3x + 10/3

first you remedy for x... ax - b = y (cx +d) ax - b = ycx + dy ax - ycx = dy + b x(a - yc) = dy + b x = (dy+b)/(a-yc) then you definately swap all the x's to y's and y's to x's.... so the inverse functionality is y = (dx + b)/(a - xc)