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

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})

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)