What is the extreme f(x)=(2x+1)e^{3x} ?

The extreme f(x)=(2x+1)e^{3x} is Minimum(-5/6 ,-2/(3e^{5/2)})

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extreme f(x)=(2x+1)e^{3x}

extreme points

f (x) = (2 x + 1) e^{3 x}

Minimum (- \frac{5}{6}, - \frac{2}{3 e ^{\frac{5}{2}}})

Solution steps

f^{'} (x) = e^{3 x} (6 x + 5)

Find intervals:

Decreasing

: - \infty < x < - \frac{5}{6},

Increasing

: - \frac{5}{6} < x < \infty

Plug

x = - \frac{5}{6}

into

(2 x + 1) e^{3 x} : - \frac{2}{3 e ^{\frac{5}{2}}}

Minimum (- \frac{5}{6}, - \frac{2}{3 e ^{\frac{5}{2}}})

2 x^{2} - 5

f (x) = 5 x - 15 x^{\frac{1}{3}}

\frac{x - 4}{x ^{2} - 16}

f (x) = \frac{- 5 x - 6}{- 15 x - 28}

\frac{1}{x - 2}

What is the extreme f(x)=(2x+1)e^{3x} ?
The extreme f(x)=(2x+1)e^{3x} is Minimum(-5/6 ,-2/(3e^{5/2)})