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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme

Minimum (1, \frac{2}{e}), Maximum (2, \frac{6}{e ^{2}})

Solution steps

Find the critical points:

x = 1, x = 2

Domain of

2 e^{- x} (x^{2} - x + 1) : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < 2,

Decreasing

: 2 < x < \infty

Plug the extreme point

x = 1

into

2 e^{- x} (x^{2} - x + 1) \Rightarrow y = \frac{2}{e}

Minimum (1, \frac{2}{e})

Plug the extreme point

x = 2

into

2 e^{- x} (x^{2} - x + 1) \Rightarrow y = \frac{6}{e ^{2}}

Maximum (2, \frac{6}{e ^{2}})

Minimum (1, \frac{2}{e}), Maximum (2, \frac{6}{e ^{2}})

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