What is the extreme f(x)=(2e^{-3x})/(2-3x) ?

The extreme f(x)=(2e^{-3x})/(2-3x) is Minimum(1/3 , 2/e)

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extreme f(x)=(2e^{-3x})/(2-3x)

extreme points

f (x) = \frac{2 e ^{- 3 x}}{2 - 3 x}

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

Solution steps

Find the critical points:

x = \frac{1}{3}

Domain of

\frac{2 e ^{- 3 x}}{2 - 3 x} : x < \frac{2}{3} or x > \frac{2}{3}

Find intervals:

Decreasing

: - \infty < x < \frac{1}{3},

Increasing

: \frac{1}{3} < x < \frac{2}{3},

Increasing

: \frac{2}{3} < x < \infty

Plug the extreme point

x = \frac{1}{3}

into

\frac{2 e ^{- 3 x}}{2 - 3 x} \Rightarrow y = \frac{2}{e}

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

f (x) = \frac{x}{ln ( 9 x )}

1 x^{3} - 4 x^{2} - 3 x + 30

f (x) = - 4 x^{3} - 6 x^{2} + 144 x - 11

f (x) = 3 x^{\frac{4}{3}} - 2 x

y = 43 x + 8 cos (x), 0 \leq x \leq 2 π

What is the extreme f(x)=(2e^{-3x})/(2-3x) ?
The extreme f(x)=(2e^{-3x})/(2-3x) is Minimum(1/3 , 2/e)