What is the extreme+sqrt(x^2-2x-8) ?

The extreme+sqrt(x^2-2x-8) is Minimum(-2,0),Minimum(4,0)

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extreme+sqrt(x^2-2x-8)

extreme points

+ x^{2} - 2 x - 8

Minimum (- 2, 0), Minimum (4, 0)

Solution steps

f^{'} (x) = \frac{x - 1}{x ^{2} - 2 x - 8}

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: 4 < x < \infty

Plug

x = - 2

into

+ x^{2} - 2 x - 8 : 0

Minimum (- 2, 0)

Plug

x = 4

into

+ x^{2} - 2 x - 8 : 0

Minimum (4, 0)

Minimum (- 2, 0), Minimum (4, 0)

f (x) = 7 x y - λ (6 x + 11 y - 2)

\frac{x + 3}{x ^{2} - 4 x - 21}

x^{2} - 20 x + 150

f (x) = 2 + \frac{7}{x} + \frac{7}{x ^{2}}

What is the extreme+sqrt(x^2-2x-8) ?
The extreme+sqrt(x^2-2x-8) is Minimum(-2,0),Minimum(4,0)