wyznacz dziedzinę funkcji f(x)=
D:
log (2x + 10) - log (6 - x) ≥ 0 i 2x + 10 > 0 i 6 - x > 0
log (2x +10) ≥ log (6 - x) i 2x > - 10 i - x > - 6
2x + 10 ≥ 6 - x x > - 5 i x < 6
2x + x ≥ 6 - 10
3x ≥ - 4
x ≥ - 4/3 i x > - 5 i x < 6
odp. D: x ∈ <-4/3 , 6)
Liczba pierwiastkowana mie może byc ujemna, zatem:
log(2x+10)-log(6-x) ≥ 0 ∧ 2x+10 > 0 i 6-x > 0
log[(2x+10)/(6-x) ≥ log1 x > -5 i x < 6
(2x+10)/6-x) -1 ≥ 0
(2x+10)/(6-x) - (6-x)/(6-x) ≥ 0
(2x+10-6+x)/(6-x) ≥ 0
(3x+4)/(6-x) ≥ 0
(3x+4)(6-x) ≥ 0
18x-3x²+24-4x ≥ 0
-3x²+14x+24 ≥ 0
Δ = 196+288 = 484
√Δ = 22
x1 = (-14-22)/(-6) = 6
x2 = (-14+22)/(-6) = -4/3
x ∈ [<-4/3; 6>] n [ (-5; 6)]
x ∈ <-4/3; 6)
Odp. x ∈ <-4/3; 6)
=============
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D:
log (2x + 10) - log (6 - x) ≥ 0 i 2x + 10 > 0 i 6 - x > 0
log (2x +10) ≥ log (6 - x) i 2x > - 10 i - x > - 6
2x + 10 ≥ 6 - x x > - 5 i x < 6
2x + x ≥ 6 - 10
3x ≥ - 4
x ≥ - 4/3 i x > - 5 i x < 6
odp. D: x ∈ <-4/3 , 6)
Liczba pierwiastkowana mie może byc ujemna, zatem:
log(2x+10)-log(6-x) ≥ 0 ∧ 2x+10 > 0 i 6-x > 0
log[(2x+10)/(6-x) ≥ log1 x > -5 i x < 6
(2x+10)/6-x) -1 ≥ 0
(2x+10)/(6-x) - (6-x)/(6-x) ≥ 0
(2x+10-6+x)/(6-x) ≥ 0
(3x+4)/(6-x) ≥ 0
(3x+4)(6-x) ≥ 0
18x-3x²+24-4x ≥ 0
-3x²+14x+24 ≥ 0
Δ = 196+288 = 484
√Δ = 22
x1 = (-14-22)/(-6) = 6
x2 = (-14+22)/(-6) = -4/3
x ∈ [<-4/3; 6>] n [ (-5; 6)]
x ∈ <-4/3; 6)
Odp. x ∈ <-4/3; 6)
=============