# Inv.Functions

• Oct 28th 2009, 04:57 AM
Inv.Functions
Ok bit of help please guys - i think i'm just twisting myself in unnecessary knots!

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

Is the Domain [6,9] ?

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

Thank you :)
• Oct 28th 2009, 05:25 AM
e^(i*pi)
Quote:

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

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

Is the Domain [6,9] ?

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

Thank you :)

Yes the domain is [6,9].

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

$\displaystyle f(6) = -3$ and $\displaystyle f(9) = 0$ 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 $\displaystyle g^{-1}(x)$ is [-3,0] and the range is [6,9]

Let $\displaystyle y = g(x)$

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

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

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

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

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

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

Domain: [-3,0]
Range : [6,9]
• Oct 28th 2009, 05:47 AM
So is range the same as image set?
• Oct 29th 2009, 12:04 PM
Just to check in my calculator...graphing

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

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

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

why doesnt it look like the reverse?
• Oct 29th 2009, 12:15 PM
corleone2463
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
• Oct 29th 2009, 12:18 PM
Thats what i thought - so is e^(i*pi) answer wrong?
• Oct 29th 2009, 12:20 PM
corleone2463
answer to wat ....e^(i*pi) is equal to -1
• Oct 29th 2009, 12:28 PM
yes?
• Oct 29th 2009, 12:31 PM
Plato
Quote:

Thats what i thought - so is e^(i*pi) answer wrong?

No, his/her second answer is correct.
$\displaystyle g^{-1}(x)=3\sqrt{x+4}+3$

The range of $\displaystyle g$ is $\displaystyle [-3,0]$ so that is the domain of $\displaystyle g^{-1}$
• Oct 30th 2009, 04:45 AM
HallsofIvy
Quote:

Originally Posted by corleone2463
answer to wat ....e^(i*pi) is equal to -1

I think he meant the person using the forumname $\displaystyle e^{i\pi}$!(Rofl)