log ½
(x 2 + 2x) > 0
log ½
(x 2 + 2x) > log ½ 1
(x 2
+ 2x) > 0 => x € (-∞, 2) U (0, +∞)
(x 2
+ 2x) > 1
(x 2
+ 2x) > 0 (conform ecuatiei de gradul doi in care Δ = b 2 - 4ac )
Δ = 4 – 4 *
1 * 0
Δ = 4
x 1
= (-2 + 2) / 2 = 0
x 2
= (- 2 – 2) / 2 = -2
x € (-∞,
-2) U (0, +∞)
x
|
-∞
|
-2
|
0
|
+∞
|
f(x)
|
++++++++++++
|
+++++0---------
|
---------0+++++
|
++++++++++++
|
x 2
+ 2x > 1
x 2
+ 2x - 1 > 0
Δ = 4 – 4 *
1 (-1)
Δ = 4 + 4
Δ = 8
√Δ = √8
√Δ = √4*2
√Δ = 2√2
=>
x 1
= (-2 + 2√2) / 2 = 2 (-1 + √2)/2
/:2 => x 1 = -1 +√2
x 2
= (-2 - 2√2) / 2 = 2 (-1 - √2)/2
/:2 => x 2 = -1 -√2
x
|
-∞
|
-1 -√2
|
-1 +√2
|
+∞
|
f(x)
|
++++++++++++
|
+++++0---------
|
---------0+++++
|
++++++++++++
|