What is the extreme-2e^{-2x}(x^4-2x^3) ?

The extreme-2e^{-2x}(x^4-2x^3) is Saddle(0,0),Maximum(1, 2/(e^2)),Minimum(3,-(54)/(e^6))

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extreme-2e^{-2x}(x^4-2x^3)

extreme points

- 2 e^{- 2 x} (x^{4} - 2 x^{3})

Saddle (0, 0), Maximum (1, \frac{2}{e ^{2}}), Minimum (3, - \frac{54}{e ^{6}})

Solution steps

Find the critical points:

x = 0, x = 1, x = 3

Domain of

- 2 e^{- 2 x} (x^{4} - 2 x^{3}) : - \infty < x < \infty

Find intervals:

Increasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug the extreme point

x = 0

into

- 2 e^{- 2 x} (x^{4} - 2 x^{3}) \Rightarrow y = 0

Saddle (0, 0)

Plug the extreme point

x = 1

into

- 2 e^{- 2 x} (x^{4} - 2 x^{3}) \Rightarrow y = \frac{2}{e ^{2}}

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

Plug the extreme point

x = 3

into

- 2 e^{- 2 x} (x^{4} - 2 x^{3}) \Rightarrow y = - \frac{54}{e ^{6}}

Minimum (3, - \frac{54}{e ^{6}})

Saddle (0, 0), Maximum (1, \frac{2}{e ^{2}}), Minimum (3, - \frac{54}{e ^{6}})

f (x) = x^{4} + 6 x^{3} - 5 x^{2} + 2 x + 3

f (x) = x^{2} - 8 x + 17

f (x) = 8 x^{2} - x^{4}

f (t) = r (t) e^{- a t}

What is the extreme-2e^{-2x}(x^4-2x^3) ?
The extreme-2e^{-2x}(x^4-2x^3) is Saddle(0,0),Maximum(1, 2/(e^2)),Minimum(3,-(54)/(e^6))