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

The extreme f(x)=(x-3)e^{-x+3-(x-3)^2} is Minimum(2,-1),Maximum(7/2 , 1/(2e^{3/4)})

Popular >

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

extreme points

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

Minimum (2, - 1), Maximum (\frac{7}{2}, \frac{1}{2 e ^{\frac{3}{4}}})

Solution steps

f^{'} (x) = - e^{- (x - 2) (x - 3)} (x - 2) (2 x - 7)

Find intervals:

Decreasing

: - \infty < x < 2,

Increasing

: 2 < x < \frac{7}{2},

Decreasing

: \frac{7}{2} < x < \infty

Plug

x = 2

into

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

Minimum (2, - 1)

Plug

x = \frac{7}{2}

into

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

Maximum (\frac{7}{2}, \frac{1}{2 e ^{\frac{3}{4}}})

Minimum (2, - 1), Maximum (\frac{7}{2}, \frac{1}{2 e ^{\frac{3}{4}}})

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 20

f (x) = 4 x - x, 2 \leq x \leq 8

f (x) = (x - 9)^{3} (x + 5)^{2}

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

sin (4 x), 0 \leq x \leq \frac{π}{2}

What is the extreme f(x)=(x-3)e^{-x+3-(x-3)^2} ?
The extreme f(x)=(x-3)e^{-x+3-(x-3)^2} is Minimum(2,-1),Maximum(7/2 , 1/(2e^{3/4)})