What is the extreme f(x)=4(e^{-0.05x}-e^{-0.55x}) ?

The extreme f(x)=4(e^{-0.05x}-e^{-0.55x}) is Minimum((ln(0.55)-ln(0.05))/(0.5),2.86106…)

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

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

Decreasing

: ln (11) < x < \infty

Plug

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

into

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

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

f (x) = - x^{3} + 1.5 x^{2} + 36 x + 2

f (x, y) = 2 x^{2} + y^{2} - 4 x + 6 y

y = x e^{- 2 x}

F (X, Y) = 2 X + 3 Y X^{2}

f (x) = x + 3 y^{2} = 25

What is the extreme f(x)=4(e^{-0.05x}-e^{-0.55x}) ?
The extreme f(x)=4(e^{-0.05x}-e^{-0.55x}) is Minimum((ln(0.55)-ln(0.05))/(0.5),2.86106…)