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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 1

into

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

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

f (x) = 137

f (x) = ∣ x ∣ - 3 ∣ x + 1 ∣

f (x) = 2 cos (x) + cos^{2} (x)

f (x) = - x^{3} - 4.5 x^{2} - 6 x + 2

f (x, y) = x^{4} + y^{4} + 4 x y

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