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

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

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

extreme points

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

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

Solution steps

Find the critical points:

x = 3

Domain of

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

Find intervals:

Decreasing

: - \infty < x < 3,

Increasing

: 3 < x < \infty

Plug the extreme point

x = 3

into

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

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

f (x, y) = x^{4}

f (x) = (x - 5) (x + 2)

e^{x} - 2 e^{- x} - 3 x

f (x) = x^{3} + 3 x^{2} - 4 x - 12

f (x) = x^{3} + 3 x^{2} - 24 x + 6

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