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

The extreme f(x)=e^{-((x-3)^2)/2}(-x+3) is Maximum(2, 1/(e^{1/2)}),Minimum(4,-1/(e^{1/2)})

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

extreme points

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

Maximum (2, \frac{1}{e ^{\frac{1}{2}}}), Minimum (4, - \frac{1}{e ^{\frac{1}{2}}})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 2

into

e^{- \frac{( x - 3 ) ^{2}}{2}} (- x + 3) : \frac{1}{e ^{\frac{1}{2}}}

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

Plug

x = 4

into

e^{- \frac{( x - 3 ) ^{2}}{2}} (- x + 3) : - \frac{1}{e ^{\frac{1}{2}}}

Minimum (4, - \frac{1}{e ^{\frac{1}{2}}})

Maximum (2, \frac{1}{e ^{\frac{1}{2}}}), Minimum (4, - \frac{1}{e ^{\frac{1}{2}}})

\frac{8 x ^{3}}{3} + 6 x^{2} - 8 x, - 3 \leq x \leq 1

f (x) = \frac{x ^{4}}{4} - (\frac{x}{3})^{3} - x^{2} + 1

f (x) = cos (x) - 9 x, 0 \leq x \leq 4 π

f (x, y) = 3 x^{2} - 3 y^{2} + 8

y = x e^{- \frac{x ^{2}}{18}}

What is the extreme f(x)=e^{-((x-3)^2)/2}(-x+3) ?
The extreme f(x)=e^{-((x-3)^2)/2}(-x+3) is Maximum(2, 1/(e^{1/2)}),Minimum(4,-1/(e^{1/2)})