What is the extreme f(x)=e^{x^2-4x-1} ?

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

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extreme f(x)=e^{x^2-4x-1}

extreme points

f (x) = e^{x^{2} - 4 x - 1}

Minimum (2, \frac{1}{e ^{5}})

Solution steps

f^{'} (x) = e^{x^{2} - 4 x - 1} (2 x - 4)

Find intervals:

Decreasing

: - \infty < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 2

into

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

Minimum (2, \frac{1}{e ^{5}})

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