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

The extreme f(x)=e^{x^2-3x-1} is Minimum(3/2 , 1/(e^{13/4)})

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

extreme points

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

Minimum (\frac{3}{2}, \frac{1}{e ^{\frac{13}{4}}})

Solution steps

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

Find intervals:

Decreasing

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

Increasing

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

Plug

x = \frac{3}{2}

into

e^{x^{2} - 3 x - 1} : \frac{1}{e ^{\frac{13}{4}}}

Minimum (\frac{3}{2}, \frac{1}{e ^{\frac{13}{4}}})

f (x) = 3 x^{4} + 16 x^{3}

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

f (x) = (x - 1) e^{x}

\frac{x ^{2} + 1}{x ^{2} - 4}

f (x, y) = - x^{2} - y^{2} + 8 x + 6 y

What is the extreme f(x)=e^{x^2-3x-1} ?
The extreme f(x)=e^{x^2-3x-1} is Minimum(3/2 , 1/(e^{13/4)})