extreme f(x)=ln(y-x^2)+ln(2x-y)

解答

极值点

f (x) = ln (y - x^{2}) + ln (2 x - y)

极大值 (1, \frac{3}{2})

求解步骤

找到临界点:

(1, \frac{3}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{- x ^{4} + 2 x ^{2} y - 4 x ^{2} + 4 x y - 2 y ^{2}}{( y - x ^{2} ) ^{2} ( 2 x - y ) ^{2}}

D (x, y) = \frac{8 x ^{8} - 8 x ^{7} y - 32 x ^{7} + 2 x ^{6} y ^{2} + 24 x ^{6} y - 16 x ^{6} + 112 x ^{5} y - 2 x ^{4} y ^{3} - 116 x ^{4} y ^{2} + 40 x ^{3} y ^{3} - 64 x ^{3} y ^{2} - 4 x ^{2} y ^{4} + 80 x ^{2} y ^{3} + 16 x ^{2} y ^{2} - 32 x y ^{4} - 16 x y ^{3} + 4 y ^{5} + 4 y ^{4}}{( - x ^{2} + y ) ^{4} ( - y + 2 x ) ^{4}}

检查在每个临界点处的符号

检查临界点

(1, \frac{3}{2}) :

极大值

极大值 (1, \frac{3}{2})

e x t re m e f (x) = - e^{x} (x + 5)

e x t re m e f (x) = - 14 x^{3} + 42 x + 42

e x t re m e f (x) = 15 x - x^{2} - 54

e x t re m e x^{2} + 2 x - x

e x t re m e y = 4 (x - 2) (x + 1)