What is the extreme f(x)=(x-5)e^{-x+5-(x-5)^2} ?

The extreme f(x)=(x-5)e^{-x+5-(x-5)^2} is Minimum(4,-1),Maximum(11/2 , 1/(2e^{3/4)})

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extreme f(x)=(x-5)e^{-x+5-(x-5)^2}

extreme points

f (x) = (x - 5) e^{- x + 5 - (x - 5)^{2}}

Minimum (4, - 1), Maximum (\frac{11}{2}, \frac{1}{2 e ^{\frac{3}{4}}})

Solution steps

f^{'} (x) = - e^{- (x - 4) (x - 5)} (x - 4) (2 x - 11)

Find intervals:

Decreasing

: - \infty < x < 4,

Increasing

: 4 < x < \frac{11}{2},

Decreasing

: \frac{11}{2} < x < \infty

Plug

x = 4

into

(x - 5) e^{- x + 5 - (x - 5)^{2}} : - 1

Minimum (4, - 1)

Plug

x = \frac{11}{2}

into

(x - 5) e^{- x + 5 - (x - 5)^{2}} : \frac{1}{2 e ^{\frac{3}{4}}}

Maximum (\frac{11}{2}, \frac{1}{2 e ^{\frac{3}{4}}})

Minimum (4, - 1), Maximum (\frac{11}{2}, \frac{1}{2 e ^{\frac{3}{4}}})

f (x, y) = 2 x^{2} + 2 x y + 2 y^{2} - 6 x

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

f (x, y) = x y (1 + 2 x - 3 y)

f (x) = \frac{x}{2} - sin (x)

What is the extreme f(x)=(x-5)e^{-x+5-(x-5)^2} ?
The extreme f(x)=(x-5)e^{-x+5-(x-5)^2} is Minimum(4,-1),Maximum(11/2 , 1/(2e^{3/4)})