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inkognitous
@inkognitous
August 2022
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Алгебра ( 2 задания = 50 баллов)
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oganesbagoyan
Verified answer
F(x) =(x²+3x)/(x+4) . || = x -x/(x+4) . ||
f'(x) = ( (x²+3x)/(x+4) ) ' =
(
(x²+3x)'(x+4) -(x²+3x)*(x+4)'
) /(
x+4)² =
= ((2x+3)(x+4) -(x² +3x)) /
(
x+4)² =(x²+8x+12)/(x+4) =(x+2)(x+6)/(x+4)² .
* * * f(x) =(x²+3x)/(x+4) = x -x/(x+4) . * * *
f '(x) =( x -x/(x+4) )' =1 -(1*(x+4) -1*x)/(x+4)² = 1 -4/(x+4)² =((x+4)² -4)/(x+4)² =
(x+4-2)*(x+4+2)/(x+4)² =(x+2)(x+6)/(x+4)² .
f'(x) =0 ⇔(x+2)(x+6)=0 ⇒
x=2, x=4
.
------------
f(x) = -x³ -3x² +9x + 3 .
f '(x) =( -x³ -3x² +9x + 3)' = -3x² -6x +9 = -3(x² +2x -3) = -2(x+3)(x-1).
Функция возрастает(↑) , если f '(x)≥0 .
-2(x+3)(x-1)≥0⇔ (x+3)(x-1)≤0⇒
x∈[ -3;1].
2 votes
Thanks 1
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Answers & Comments
Verified answer
F(x) =(x²+3x)/(x+4) . || = x -x/(x+4) . ||f'(x) = ( (x²+3x)/(x+4) ) ' =((x²+3x)'(x+4) -(x²+3x)*(x+4)' ) /(x+4)² =
= ((2x+3)(x+4) -(x² +3x)) / (x+4)² =(x²+8x+12)/(x+4) =(x+2)(x+6)/(x+4)² .
* * * f(x) =(x²+3x)/(x+4) = x -x/(x+4) . * * *
f '(x) =( x -x/(x+4) )' =1 -(1*(x+4) -1*x)/(x+4)² = 1 -4/(x+4)² =((x+4)² -4)/(x+4)² =
(x+4-2)*(x+4+2)/(x+4)² =(x+2)(x+6)/(x+4)² .
f'(x) =0 ⇔(x+2)(x+6)=0 ⇒ x=2, x=4.
------------
f(x) = -x³ -3x² +9x + 3 .
f '(x) =( -x³ -3x² +9x + 3)' = -3x² -6x +9 = -3(x² +2x -3) = -2(x+3)(x-1).
Функция возрастает(↑) , если f '(x)≥0 .
-2(x+3)(x-1)≥0⇔ (x+3)(x-1)≤0⇒ x∈[ -3;1].