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! :)
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
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