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

Question

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

akashdeepdeb · Answer

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?

anonymous · Answer

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$$y=\frac{\sqrt{x}+3}{2}$$

akashdeepdeb · Answer

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

driftracer305 · Answer

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

akashdeepdeb · Answer

:)

anonymous · Answer

;D