What is the extreme f(x)=1+x^2e^{5-3x} ?

The extreme f(x)=1+x^2e^{5-3x} is Minimum(0,1),Maximum(2/3 ,1+(4e^3)/9)

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extreme f(x)=1+x^2e^{5-3x}

extreme points

f (x) = 1 + x^{2} e^{5 - 3 x}

Minimum (0, 1), Maximum (\frac{2}{3}, 1 + \frac{4 e ^{3}}{9})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{2}{3},

Decreasing

: \frac{2}{3} < x < \infty

Plug

x = 0

into

1 + x^{2} e^{5 - 3 x} : 1

Minimum (0, 1)

Plug

x = \frac{2}{3}

into

1 + x^{2} e^{5 - 3 x} : 1 + \frac{4 e ^{3}}{9}

Maximum (\frac{2}{3}, 1 + \frac{4 e ^{3}}{9})

Minimum (0, 1), Maximum (\frac{2}{3}, 1 + \frac{4 e ^{3}}{9})

y = - 4 e^{- 4 t} + 5 e^{- 5 t}

x^{3} + y^{3} + 9 x^{2} - 3 y^{2} + 15 x - 9 y

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

\frac{x ^{2} - 2 x + 4}{x - 2}

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

What is the extreme f(x)=1+x^2e^{5-3x} ?
The extreme f(x)=1+x^2e^{5-3x} is Minimum(0,1),Maximum(2/3 ,1+(4e^3)/9)