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Inverse Function watch

1. Find inverse function of: f(x) = e^(x-2) - 1

So I did:

f(x) + 1 = e^(x-2)
f(x) + 1 = e^x multiplied by e^-2
lnf(x) + ln1 = -2x

gives x=-lnf(x)/2

Then swap f(x) for x and make x become f-1

f-1(x) = ln(x^-1/2)

Is this correct?

Thanks again!
2. I would've done this?
y=e^(x-2) - 1

y+1 = e^(x-2)
ln (y+1) = (x-2) ln e
x= 2 + ln ( y+1)

therefore f-(x) = 2+ ln (x+1)

( we're taught to rearrange then swap x and y and the end)

3. No, we require a function f such that x = e^(f(x)-2) - 1.

Rearranging gives f(x) = ln(x+1) +2.
4. y = e^(x - 2) - 1

Swap x and y:

x = e^(y - 2) - 1

Move 1 over:

x + 1 = e^(y - 2)

Taking logs of both sides:

ln (x + 1) = y - 2

Move 2 over:

y = 2 + ln (x + 1)
5. Yeah, agree with all the above replies
6. (Original post by Chinesegirl)
Yeah, agree with all the above replies
Ok thanks everyone - think I need some more practise!

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Updated: March 21, 2004
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