What is the extreme f(x)=e^{x^2-3x-1}[-3.3] ?

The extreme f(x)=e^{x^2-3x-1}[-3.3] is Maximum(3/2 ,-0.12795…)

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extreme f(x)=e^{x^2-3x-1}[-3.3]

extreme points

f (x) = e^{x^{2} - 3 x - 1} [- 3.3]

Maximum (\frac{3}{2}, - 0.12795 \dots)

Solution steps

Find the critical points:

x = \frac{3}{2}

Domain of

e^{x^{2} - 3 x - 1} [- 3.3] : - \infty < x < \infty

Find intervals:

Increasing

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

Decreasing

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

Plug the extreme point

x = \frac{3}{2}

into

e^{x^{2} - 3 x - 1} [- 3.3] \Rightarrow y = - 0.12795 \dots

Maximum (\frac{3}{2}, - 0.12795 \dots)

f (x, y) = x^{2} y - 2 x y + 3 y^{3} - 3 y

f (x) = \frac{4 x ^{2}}{x - 8}

f (x) = - 2 x^{2} + 12 x + 3

f (x, y) = - y^{2} + x y - x^{3} + y + 3

f (x, y) = 2 x^{2} + 3 y^{2} + 2 x y + 10 x - 20 y

What is the extreme f(x)=e^{x^2-3x-1}[-3.3] ?
The extreme f(x)=e^{x^2-3x-1}[-3.3] is Maximum(3/2 ,-0.12795…)