Inverse function domain, image and rule
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Inverse function domain, image and rule

[From: ] [author: ] [Date: 12-02-11] [Hit: ]
Start with [AS]; the 12 is subtracted,Done with that, now onto [MD]; the 2 is multiplying the x-term,Done with that, now onto E; x-term is raised to the 2nd power,Then just subtract 1 to get your inverse function.......
Can someone explain how to find the rule of inverse functions such as:

y=2(x+1)^2-12

I understand the method I just dont know how to deal with the '2' before (x+1).

Thanks in advance! :)

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Just do PE[MD][AS] in reverse

y = 2*(x + 1)² - 12

Start with [AS]; the 12 is subtracted, so add it to both sides

y + 12 = 2*(x + 1)²

Done with that, now onto [MD]; the 2 is multiplying the x-term, so divide by it:

(y + 12)/2 = (x + 1)²

Done with that, now onto E; x-term is raised to the 2nd power, so raise both sides to the 1/2 power

[(y + 12)/2]^(1/2) = x + 1

Then just subtract 1 to get your inverse function.

x = [(y + 12)/2]^(1/2) - 1

or in standard notation

y = f⁻¹(x) = √[(x + 12)/2] - 1
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