Asked by Moe
                Find the inverse of this function:
f(x) = x^2 + 2x - 1, x>0
So far I just switched the x and y's...so it looks like this now:
x = y^2 + 2y - 1
I know you're solving for y, and I want to factor (or use quadratic formula)...but x is in the way. So I'm not sure what to do.
Thank you
            
        f(x) = x^2 + 2x - 1, x>0
So far I just switched the x and y's...so it looks like this now:
x = y^2 + 2y - 1
I know you're solving for y, and I want to factor (or use quadratic formula)...but x is in the way. So I'm not sure what to do.
Thank you
Answers
                    Answered by
            Reiny
            
    You will have to treat the x as part of the c term in ax^2 + bx + c = 0
so you have
y^2 + 2y - 1 - x = 0
so a = 1 , by = 2, and c = -1-x
y = (-2 ± √(4 - 4(1)(-1-x)))/2
= (-2 ± √(8+4x))/2
= (-2 ± 2√(2+x))/2
= -1 ± √(2+x)
check with some value of x
e.g. f(5) = 25 + 10 - 1 = 34
so put that in the inverse,
y = -1 ± √(2+34)
= -1 ± √36
= -1 ± 6 = 5 or -7
notice the new inverse is clearly not a function, but we did get our 5 back.
    
so you have
y^2 + 2y - 1 - x = 0
so a = 1 , by = 2, and c = -1-x
y = (-2 ± √(4 - 4(1)(-1-x)))/2
= (-2 ± √(8+4x))/2
= (-2 ± 2√(2+x))/2
= -1 ± √(2+x)
check with some value of x
e.g. f(5) = 25 + 10 - 1 = 34
so put that in the inverse,
y = -1 ± √(2+34)
= -1 ± √36
= -1 ± 6 = 5 or -7
notice the new inverse is clearly not a function, but we did get our 5 back.
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