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

The function:

Is the Domain[6,9]?

How can i then find the image set ofand its rule?

Thank you:)

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

**The function:**

**Is the Domain**[6,9]**?**

**How can i then find the image set of****and its rule?**

**Thank you**:) - Oct 28th 2009, 05:25 AMe^(i*pi)
Yes the domain is [6,9].

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

and 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 is [-3,0] and the range is [6,9]

Let

.

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

Domain: [-3,0]

Range : [6,9] - Oct 28th 2009, 05:47 AMADY
So is range the same as image set?

- Oct 29th 2009, 12:04 PMADY
Just to check in my calculator...graphing

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

why doesnt it look like the reverse?

- Oct 29th 2009, 12:15 PMcorleone2463
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 PMADY
Thats what i thought - so is e^(i*pi) answer wrong?

- Oct 29th 2009, 12:20 PMcorleone2463
answer to wat ....e^(i*pi) is equal to -1

- Oct 29th 2009, 12:28 PMADY
yes?

- Oct 29th 2009, 12:31 PMPlato
- Oct 30th 2009, 04:45 AMHallsofIvy