task/29412801 --------------------
Найдите область определения функции
1 . f(x) =√(2x²+3x -2)
2x²+3x -2 ≥0 ⇔2(x+2)(x -1/2) ≥ 0 методом интервалов
"+" "- " "+"
//////////////////// [-2] -------------- [1/2] //////////////////////
ответ: D(f) = (-∞ ; -2 ] ∪ [1 /2 ; ∞) .
2. f(x) = Log₃ (x²-14x +45) / (3 - x)
x²-14x +45) / (3 - x) >0 ⇔ (x- 5)(x-9 ) / (x-3) <0 ⇔ (x-3)(x-5)(x-9) <0 </p>
" - " "+" "- " "+"
///////////////// (3) --------------(5)/////////////// (9)--------------
ответ: D(f) = (-∞ ; 3 ) ∪ ( 5 ; 9) .
3. f(x) =√(64-x²) / Log₃₂ (x +6)
{ 64 - x² ≥0 ; x +6 >0 ; x+6 ≠1 .⇔ { (x+8)(x-8) ≤ 0 ; x > - 6, x ≠ -5 .⇔
{ x ∈[ -8 ;8] ; x ∈( -6 ,-5) ∪(-5 ;∞) . ⇒x ∈( - 6 ,-5) ∪(-5 ;8 ]
ответ: D(f) = ( - 6 ,-5) ∪(-5 ;8 ]
4. f(x) =√cosx +Log₄₈ (-x² -3x -2)
{ cosx ≥ 0 ; -x² - 3x -2 > 0.⇔{cosx ≥ 0 ; x²+3x +2< 0 .⇔{cosx ≥ 0 ; (x +2)(x +1 < 0 .
{ 2πn - π/2 ≤ x ≤ π/2+ 2πn ,n ∈ ℤ ; -2 < x < - 1 . имеет реш. только при n =0
{ - π/2 ≤ x ≤ π/2 ; -2 < x < - 1 . ⇒ x ∈ [- π/2 ; - 1) .
--------------------------- [- π/2] ///////////////////////// [ π/2 ] -------------
---------------- (-2) ////////////////////// (-1) -----------------------------------
ответ: D(f) = [- π/2 ; - 1) .