What is the extreme f(x)=(x^2-12)*e^{-2x} ?

The extreme f(x)=(x^2-12)*e^{-2x} is Minimum(-3,-3e^6),Maximum(4, 4/(e^8))

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extreme f(x)=(x^2-12)*e^{-2x}

extreme points

f (x) = (x^{2} - 12) \cdot e^{- 2 x}

Minimum (- 3, - 3 e^{6}), Maximum (4, \frac{4}{e ^{8}})

Solution steps

f^{'} (x) = - 2 e^{- 2 x} (x + 3) (x - 4)

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = - 3

into

(x^{2} - 12) \cdot e^{- 2 x} : - 3 e^{6}

Minimum (- 3, - 3 e^{6})

Plug

x = 4

into

(x^{2} - 12) \cdot e^{- 2 x} : \frac{4}{e ^{8}}

Maximum (4, \frac{4}{e ^{8}})

Minimum (- 3, - 3 e^{6}), Maximum (4, \frac{4}{e ^{8}})

f (x) = 49 x^{2} - 10 x + 17

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

f (x, y) = - 3 y^{3} - 3 x y + 13 y - 2 \cdot 2 + 5 x + 19

f (x) = x y + y - 16 x

y = x^{7} e^{x} + 4

What is the extreme f(x)=(x^2-12)*e^{-2x} ?
The extreme f(x)=(x^2-12)*e^{-2x} is Minimum(-3,-3e^6),Maximum(4, 4/(e^8))