Question

Inverse of (2x-3)^2............pls help

Answer 1

Let\[y = (2x-3)^2\]\[\sqrt{y} = \pm (2x-3)\]\[\pm \sqrt{y} + 3 = 2x\]\[x = \frac{\pm \sqrt{y} + 3}{2}\]

Now since an invertible function must be one-one and onto that is every y image must have a Unique x pre-image, we take only either + or -.

So finally,
\[x = \frac{\sqrt{y} + 3}{2}\]

Getting this?

Answer 2

To obtain the inverse function, you just have to replace y by x and x by y\[y=(2x-3)^2\]replacing\[x=(2y-3)^2\]now we have to isolate the y again\[\sqrt{x}=2y-3\]then we got\[y=\frac{\sqrt{x}+3}{2}\]

Answer 3

Another thing, what I wrote there is \(g(y)\) and not \(f^{-1} (x)\).
@D3xt3R has explained it there! ^

Answer 4

ok, thanks a lot guys @AkashdeepDeb and @D3xt3R .....

Answer 5

:)

Answer 6

;D