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

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

Maximum (\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:

Increasing

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

Decreasing

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

Decreasing

: \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}

Maximum (\frac{1}{3}, - \frac{2}{e})

f (x) = x^{2} - 4 x, - \infty < x \leq 4

f (x) = (x - 2) (x + 2) (x + 4)

f (x) = \frac{x + 1}{x ^{2} - 6 x - 7}

f (x) = 11 + 5 x + x^{2}

f (x) = x^{2} lo g_{3} (x)

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