Точка в которой производная = 0 - экстремум функции
1) y' = ((x²+3)(x-1)+(x²+3x+12))/(x-1)²
((2x+3)(x-1)-(x²+3x+12))/(x-1)² = 0
(x-1)²<>0 x <> 1
((2x+3)(x-1)-(x²+3x+12)) = 0
2x²-2x+3x-3-x²-3x-12 = 0
x²-2x-15 = 0 - парабола ветви вверх
x1 = 5 - минимум x2 = -3 - максимум
y(5) = (25+15+12)/4 = 52/4 = 13
y(-3) = (9 - 9 + 12) /-4 = -12 / 4 = - 3
2) y' = ln(x) + 1
ln(x) + 1 = 0
ln(x) = -1
x = 1/e
y(x) = (1/e)*ln(1/e) = -1/e
3) y' = 2x*
- x²*
2x* - x²* = 0
*(2x - x²) = 0
(2x - x²) = 0
x(2-x) = 0
x1 = 0 x2 = 2
= 0 нет корней
y(0) = 0 y(2) = 4 *
= 4/e²
1) y' = ((x²+3)(x-1)+(x²+3x+12))/(x-1)²
((2x+3)(x-1)-(x²+3x+12))/(x-1)² = 0
(x-1)²<>0 x <> 1
((2x+3)(x-1)-(x²+3x+12)) = 0
2x²-2x+3x-3-x²-3x-12 = 0
x²-2x-15 = 0 - парабола ветви вверх
x1 = 5 - минимум x2 = -3 - максимум
y(5) = (25+15+12)/4 = 52/4 = 13
y(-3) = (9 - 9 + 12) /-4 = -12 / 4 = - 3
2) y' = ln(x) + 1
ln(x) + 1 = 0
ln(x) = -1
x = 1/e
y(x) = (1/e)*ln(1/e) = -1/e
3) y' = 2x*
2x*
(2x - x²) = 0
x(2-x) = 0
x1 = 0 x2 = 2
y(0) = 0 y(2) = 4 *