What is the extreme f(x)=e^{(x^2-7x-1)} ?

The extreme f(x)=e^{(x^2-7x-1)} is Minimum(7/2 , 1/(e^{53/4)})

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extreme f(x)=e^{(x^2-7x-1)}

extreme points

f (x) = e^{(x^{2} - 7 x - 1)}

Minimum (\frac{7}{2}, \frac{1}{e ^{\frac{53}{4}}})

Solution steps

f^{'} (x) = e^{x^{2} - 7 x - 1} (2 x - 7)

Find intervals:

Decreasing

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

Increasing

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

Plug

x = \frac{7}{2}

into

e^{x^{2} - 7 x - 1} : \frac{1}{e ^{\frac{53}{4}}}

Minimum (\frac{7}{2}, \frac{1}{e ^{\frac{53}{4}}})

x^{\frac{2}{3}} - 6 x^{- \frac{1}{3}}

f (x, y) = \frac{4}{x + y}

f (x) = 3 x^{3} - 36 x + 3

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

What is the extreme f(x)=e^{(x^2-7x-1)} ?
The extreme f(x)=e^{(x^2-7x-1)} is Minimum(7/2 , 1/(e^{53/4)})