2007-11-16 02:41:20 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
f(x)= 4 – 12x -9x^2 = (4+3x)(1-3x)
Zeros: x = -4/3, 1/3
When 2 ≤ x ≤10, f(x) is decreasing. Therefore, f^-1(x) exists.
Switch x and y,
x = 4 - 12y - 9y^2, 2 ≤ y ≤10
9y^2 + 12y + (x-4) = 0
y = f^-1(x) = [-2 + 2√(2-36x)]/3, use "+" because y must be positive.
2007-11-16 02:54:52 · answer #1 · answered by sahsjing 7 · 0⤊ 0⤋
1
2016-05-23 09:47:05 · answer #2 · answered by audrey 3 · 0⤊ 0⤋
3
2007-11-16 02:44:11 · answer #3 · answered by Anonymous · 0⤊ 1⤋
y = 4 - 12x - 9x^2
9 x^2 + 12x +(y-4) = 0
x = (-12 +/- sqrt(144-36(y-4))/18
x>2, then choose the + sign
x= f^(-1)(y) = (-2/3) + sqrt(144-36(y-4))/18
2007-11-16 02:52:27 · answer #4 · answered by GusBsAs 6 · 0⤊ 0⤋
A plate of chips and a pepsi? (long shot)
2007-11-16 02:49:17 · answer #5 · answered by Ghost Boy 7 · 0⤊ 1⤋
if this is your question
f(x)=4-12x-9x^2
f(x)=-9x^2-12x+4
f '(x)=-18x-12
12=-18x
x=12/-18
x=-2/3
2007-11-16 02:46:40 · answer #6 · answered by Anonymous · 0⤊ 1⤋
omg i hate math lol
2007-11-17 12:28:31 · answer #7 · answered by Anonymous · 0⤊ 0⤋
My head hurts :(
2007-11-16 02:48:18 · answer #8 · answered by Baby Boy due 12/12/09 3 · 0⤊ 1⤋
blah blah, feckidy blah
2007-11-16 02:44:08 · answer #9 · answered by Anonymous · 0⤊ 1⤋