First, we'll impose the constraints of existence of the square
root:
x - 3>
=0
x>=3
Now, we'll solve the
inequality by raising to square both sides:
(x - 3) >=1/(x -
3)^2
Now, we'll subtract 1/(x - 3)^2 both
sides:
(x - 3) - 1/(x - 3)^2
>=0
We'll multiply by (x-3)^2 the
inequality:
(x - 3)^3 -
1>=0
We'll solve the difference of cubes using the
formula:
a^3 - b^3 = (a-b)(a^2 + ab +
b^2)
(x - 3)^3 - 1 = (x - 3 -1)[(x-3)^2 + x - 3 +
1]
We'll combine like terms inside
brackets:
(x - 4)[(x-3)^2 + x - 3 + 1]
>=0
A product is zero if both factors have the same sign.
We'll get 2 cases to analyze:
Case
1)
x-4>=0
x>=4
x^2
- 6x + 9 + x - 2 >=0
x^2 - 5x + 7
>=0
x1 = [5+sqrt(25 -
28)]/2
Since delta = 25-28 = -3<0, the
expression
x^2 - 5x + 7 > 0 for any
x.
The common solution is the interval [4,
+infinity).
Case
2)
x-4=<0
x=<4
x^2
- 5x + 7 <0 impossible, because x^2 - 5x + 7 >0 for any value of
x.
The solution of the inequality is the interval [4,
+infinity).
No comments:
Post a Comment