# algebra 2

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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

• algebra 2 -

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

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 ??

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