Inv.Functions

**ADY** · October 28th 2009, 04:57 AM

Ok bit of help please guys - i think i'm just twisting myself in unnecessary knots!

The function: $g(x) = \frac{1}{9}(x - 3)^2 - 4 (6 \leq x \leq 9)$

Is the Domain [6,9] ?

How can i then find the image set of $g^-1$ and its rule?

Thank you

**e^(i*pi)** · October 28th 2009, 05:25 AM

Originally Posted by ADY

Ok bit of help please guys - i think i'm just twisting myself in unnecessary knots!

The function: $g(x) = \frac{1}{9}(x - 3)^2 - 4 (6 \leq x \leq 9)$

Is the Domain [6,9] ?

How can i then find the image set of $g^-1$ and its rule?

Thank you

Yes the domain is [6,9].

The range can be found by finding f(6) and f(9).

$f(6) = -3$ and $f(9) = 0$ so the range is [-3,0]

When taking the inverse the the domain becomes the range and the range becomes the domain. Therefore the domain of $g^{-1}(x)$ is [-3,0] and the range is [6,9]

Let $y = g(x)$

$9(y+4) = (x-3)^2$

$\pm 3\sqrt{y+4} = x-3$

$x = 3\pm 3\sqrt{y+4}$ .

$g^{-1}(x) = 3\pm 3\sqrt{x+4}$

To determine the sign find g^{-1}(0) = 9 = 3+3\sqrt{4} = 9. So we want the positive root:

$g^{-1}(x) = 3+3\sqrt{x+4}$

Domain: [-3,0]
Range : [6,9]

**ADY** · October 28th 2009, 05:47 AM

So is range the same as image set?

**ADY** · October 29th 2009, 12:04 PM

Just to check in my calculator...graphing

$f(x) = \frac{1}{9}(x - 3)^2 - 4$ $(6 \leq x \leq 9)$

this is a parabola, so the inverse function, how should this look as when i put in

$g^{-1}(x) = 3+3\sqrt{x+4}$

why doesnt it look like the reverse?

**corleone2463** · October 29th 2009, 12:15 PM

if u have the graph of the function then just consider ur y axxis as xaxis and viceversa (i.e rotate ur copy) and now see the graph thats how it wud luk like