y = 7 + 12x - x³
y’ = 12 - 3x² = 3•(2 - x)(2 + x)
y’ > 0 при х ∈ [-2; 2]
Длина промежутка возрастания функции: 2 - (-2) = 4
5(3x-4)=4(2x+1)
15x-20=8x+4
7x=24
X=24/7
-x+5y=-1
-x=8
y=1
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