What is the extreme f(x)=(x^2-7/3 x+13/9)*e^{3x} ?

The extreme f(x)=(x^2-7/3 x+13/9)*e^{3x} is Maximum(2/3 ,(e^2)/3),Minimum(1,(e^3)/9)

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extreme f(x)=(x^2-7/3 x+13/9)*e^{3x}

extreme points

f (x) = (x^{2} - \frac{7}{3} x + \frac{13}{9}) \cdot e^{3 x}

Maximum (\frac{2}{3}, \frac{e ^{2}}{3}), Minimum (1, \frac{e ^{3}}{9})

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Increasing

: 1 < x < \infty

Plug

x = \frac{2}{3}

into

(x^{2} - \frac{7}{3} x + \frac{13}{9}) \cdot e^{3 x} : \frac{e ^{2}}{3}

Maximum (\frac{2}{3}, \frac{e ^{2}}{3})

Plug

x = 1

into

(x^{2} - \frac{7}{3} x + \frac{13}{9}) \cdot e^{3 x} : \frac{e ^{3}}{9}

Minimum (1, \frac{e ^{3}}{9})

Maximum (\frac{2}{3}, \frac{e ^{2}}{3}), Minimum (1, \frac{e ^{3}}{9})

f (x) = 315 x^{2} + 1260 x^{3}, 0 \leq x \leq 0.25

f (x, y) = 34 - 9 x^{2} - 4 y^{2}

f (x, y) = ln (y^{2} + x) - \frac{1}{2} x + y

(x - 5) e^{x}

What is the extreme f(x)=(x^2-7/3 x+13/9)*e^{3x} ?
The extreme f(x)=(x^2-7/3 x+13/9)*e^{3x} is Maximum(2/3 ,(e^2)/3),Minimum(1,(e^3)/9)