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#2
(Original post by KOH)
Givving that f(x) = X^2-8X
Find f^-1 (x)
Givving that f(x) = X^2-8X
Find f^-1 (x)
just complete the square.
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(Original post by Ralfskini)
just complete the square.
just complete the square.
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#4
(Original post by KOH)
Givving that f(x) = X^2-8X
Find f^-1 (x)
Givving that f(x) = X^2-8X
Find f^-1 (x)
x=y^2-8y
x=(y-4)^2-16
(x+16)^0.5=y-4
so y=4+-sqrt(x+16)
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#5
You need to restrict the domain of f to {x >= 4} to get a unique inverse. The range of f is then {y >= -16}. The domain of f^-1 is the range of f.
For all x >= 4 with y = x^2 - 8x we have
y = (x - 4)^2 - 16,
y + 16 = (x - 4)^2,
sqrt(y + 16) = x - 4,
x = 4 + sqrt(y + 16).
So f^-1(y) = 4 + sqrt(y + 16) for all y >= -16.
For all x >= 4 with y = x^2 - 8x we have
y = (x - 4)^2 - 16,
y + 16 = (x - 4)^2,
sqrt(y + 16) = x - 4,
x = 4 + sqrt(y + 16).
So f^-1(y) = 4 + sqrt(y + 16) for all y >= -16.
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