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

The extreme f(x)=(x^2-3x+5)e^{-x/3} is Minimum(2, 3/(e^{2/3)}),Maximum(7,(33)/(e^{7/3)})

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extreme f(x)=(x^2-3x+5)e^{-x/3}

extreme points

f (x) = (x^{2} - 3 x + 5) e^{- \frac{x}{3}}

Minimum (2, \frac{3}{e ^{\frac{2}{3}}}), Maximum (7, \frac{33}{e ^{\frac{7}{3}}})

Solution steps

f^{'} (x) = e^{- \frac{x}{3}} (\frac{- x ^{2} - 5}{3} + 3 x - 3)

Find intervals:

Decreasing

: - \infty < x < 2,

Increasing

: 2 < x < 7,

Decreasing

: 7 < x < \infty

Plug

x = 2

into

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

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

Plug

x = 7

into

(x^{2} - 3 x + 5) e^{- \frac{x}{3}} : \frac{33}{e ^{\frac{7}{3}}}

Maximum (7, \frac{33}{e ^{\frac{7}{3}}})

Minimum (2, \frac{3}{e ^{\frac{2}{3}}}), Maximum (7, \frac{33}{e ^{\frac{7}{3}}})

f (x) = x^{4} - 4 x^{3} + 16 x

f (x, y) = (6 - x) (6 - y) (x + y - 6)

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

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

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

What is the extreme f(x)=(x^2-3x+5)e^{-x/3} ?
The extreme f(x)=(x^2-3x+5)e^{-x/3} is Minimum(2, 3/(e^{2/3)}),Maximum(7,(33)/(e^{7/3)})