Given f(x)=x^2+2x-3, find f^-1(x). If u didnt know f^-1 is basically the inverse of f(x)
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replace x with y and vice versa.
thus:
x=y^2+2y-3
Add 3 to both sides:
x+3=y^2+2y
Complete the square for the right:
x+3 (+1)=y^2+2y+1=(y+1)^2
Square root both sides:
(x+4)^.5=y+1.
thus the inverse is:
-1+(x+4)^.5
QED
thus:
x=y^2+2y-3
Add 3 to both sides:
x+3=y^2+2y
Complete the square for the right:
x+3 (+1)=y^2+2y+1=(y+1)^2
Square root both sides:
(x+4)^.5=y+1.
thus the inverse is:
-1+(x+4)^.5
QED
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f^-1(x) = 1/(x^2+2x-3)
The inverse of anything is just 1/anything.
Note: You could also factorise giving 1/((x-1)(x+3)) OR (1/(x-1))*(1/(x+3))... they're all the same.
The inverse of anything is just 1/anything.
Note: You could also factorise giving 1/((x-1)(x+3)) OR (1/(x-1))*(1/(x+3))... they're all the same.