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January 30, 2015

January 30, 2015

Posted by **Anonymous** on Monday, February 17, 2014 at 11:10pm.

f(x) = x^2 − 6x, x ≥ 3

- pre-calc -
**Reiny**, Tuesday, February 18, 2014 at 12:00amYour given function will be the part of the parabola which lies to the right of the vertex (3, -9)

so y = x^2 - 6x

inverse is

x = y^2- 6y

y^2 - 6y = x

complete the square:

y^2 - 6y+ 9 = x+9

(y-3)^2= x+9

y - 3= ± √(x+9)

y = 3 ± √(x+9)

but we only used the part for x≥3 of the original, so

the inverse is

f^-1 (x) = 3 + √(x+9) , x ≥ -9

check:

let x = 4

f(4) = 16 - 24 = -8

f^-1 (-8) = 3 + √(-8+9)

= 3 + √1 = 4

let x = 5.67

f(5.67) = -1.8711

f^-1 (-1.8711) = 3 + √(-1.8711+9)

= 3+√7.1289

=3 + 2.67

= 5.67

I am very confident my inverse is correct!

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