What is the extreme f(x)=e^{x^2-9x-1} ?

The extreme f(x)=e^{x^2-9x-1} is Minimum(9/2 , 1/(e^{85/4)})

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extreme f(x)=e^{x^2-9x-1}

extreme points

f (x) = e^{x^{2} - 9 x - 1}

Minimum (\frac{9}{2}, \frac{1}{e ^{\frac{85}{4}}})

Solution steps

f^{'} (x) = e^{x^{2} - 9 x - 1} (2 x - 9)

Find intervals:

Decreasing

: - \infty < x < \frac{9}{2},

Increasing

: \frac{9}{2} < x < \infty

Plug

x = \frac{9}{2}

into

e^{x^{2} - 9 x - 1} : \frac{1}{e ^{\frac{85}{4}}}

Minimum (\frac{9}{2}, \frac{1}{e ^{\frac{85}{4}}})

f (x, y) = 5 + 2 x y + \frac{20}{x} + \frac{25}{y}

f (x) = x 32 - x^{2}

x \cdot ln (x)

f (x, y) = 3 + x y - x - 2 y

f (x, y) = x^{3} + 6 x y + y^{3} + 3

What is the extreme f(x)=e^{x^2-9x-1} ?
The extreme f(x)=e^{x^2-9x-1} is Minimum(9/2 , 1/(e^{85/4)})