Question

Another inverse function: f(x) = 3 sqrt x- 2

Answer 1

this will sound silly, but I know exactly how to do this if that 3 was a cube, not a real number. The real number of 3 is throwing me off.

Answer 2

this will sound silly, but I know exactly how to do this if that 3 was a cube, not a real number. The real number of 3 is throwing me off.

Answer 3

f(x) = 3 sqrt ( x- 2)
->( y/3 )^2 = x -2
=> x = y^2/9 + 2
Thus f ^-1(x) = x^2/9 + 2

Answer 4

let \[x = 3\sqrt{y-2}\]

make y the subject

\[(x/3)^2 +2 = y\]

or \[x^2/9 + 2 = y\]

Answer 5

awesome, thank you so much. This makes sense now