What is the extreme f(x)=e^{x^2-8x-1}[-8.8] ?

The extreme f(x)=e^{x^2-8x-1}[-8.8] is Maximum(4,-3.64315E-7)

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extreme f(x)=e^{x^2-8x-1}[-8.8]

extreme points

f (x) = e^{x^{2} - 8 x - 1} [- 8.8]

Maximum (4, - 3.64315 E - 7)

Solution steps

Find the critical points:

x = 4

Domain of

e^{x^{2} - 8 x - 1} [- 8.8] : - \infty < x < \infty

Find intervals:

Increasing

: - \infty < x < 4,

Decreasing

: 4 < x < \infty

Plug the extreme point

x = 4

into

e^{x^{2} - 8 x - 1} [- 8.8] \Rightarrow y = - 3.64315 E - 7

Maximum (4, - 3.64315 E - 7)

f (x) = x^{4} + 6 x^{3} + 12 x^{2} + 8 x + 10

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

3 x^{4} - x^{6}

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

f (x) = \frac{( x ^{2} + x + 4 )}{( 3 + x ^{2} )}

What is the extreme f(x)=e^{x^2-8x-1}[-8.8] ?
The extreme f(x)=e^{x^2-8x-1}[-8.8] is Maximum(4,-3.64315E-7)