extreme f(x)=4(e^{-0.05x}-e^{-0.55x})

解答

极值点

f (x) = 4 (e^{- 0.05 x} - e^{- 0.55 x})

极小值 (\frac{ln ( 0.55 ) - ln ( 0.05 )}{0.5}, 2.86106 \dots)

求解步骤

f^{'} (x) = 4 (- 0.05 e^{- 0.05 x} + 0.55 e^{- 0.55 x})

Find intervals:

递增

: - \infty < x < ln (11),

递减

: ln (11) < x < \infty

将

x = \frac{ln ( 0.55 ) - ln ( 0.05 )}{0.5}

代入

4 (e^{- 0.05 x} - e^{- 0.55 x}) : 2.86106 \dots

极小值 (\frac{ln ( 0.55 ) - ln ( 0.05 )}{0.5}, 2.86106 \dots)

e x t re m e f (x) = - x^{3} + 1.5 x^{2} + 36 x + 2

e x t re m e f (x, y) = 2 x^{2} + y^{2} - 4 x + 6 y

e x t re m e y = x e^{- 2 x}

e x t re m e f (x) = x + 3 y^{2} = 25

e x t re m e f (x) = 110.2 x - 0.4 x^{2} - 20.2 - 2000