Let f(x)=2/3x -6. Find f^-1(x)
Solve: x^2/3 - x^1/3 - 6 = 0
Solve: SQRT(x - 4) = 6 - x
Solve using quadratic formula: 2x^2= -2x -1
2007-04-26 09:57:58 · 3 answers · asked by mrsjablo 1 in Science & Mathematics ➔ Mathematics
1.Solve: x^2/3 - x^1/3 - 6 = 0
put x^(1/3)=y, then it becomes:
y^2-y-6=0 which gives y=3,-2which leads to x=27,-8.
2007-04-26 10:08:04 · answer #1 · answered by Anonymous · 0⤊ 0⤋
1)
y+6=2/3 x
x=(y+6)*3/2
f^-1 (x)= 3(x+6)/2
2)
x^2/3 - x^1/3 - 6 = 0
Let x^1/3=t
t^2-t-6=0
t1=-2
t2=3
x^1/3=-2
x1=-8
x^1/3=3
x2=27
3)
SQRT(x - 4) = 6 - x
take square
x-4=36+x^2-12x
x^2-13x+40=0
x1=5
x2=8 (does not satisfy)
answer: x=5
4)
2x^2= -2x -1
2x^2+2x +1=0
a=2
b=2
c=1
no real roots
2007-04-26 17:10:50 · answer #2 · answered by iyiogrenci 6 · 0⤊ 0⤋
wtf
2007-04-26 17:04:57 · answer #3 · answered by Anonymous · 0⤊ 0⤋