What is the extreme f(x)=4x+8/x ?

The extreme f(x)=4x+8/x is Maximum(-sqrt(2),-8sqrt(2)),Minimum(sqrt(2),8sqrt(2))

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extreme f(x)=4x+8/x

extreme points

f (x) = 4 x + \frac{8}{x}

Maximum (- 2, - 82), Minimum (2, 82)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

4 x + \frac{8}{x} : - 82

Maximum (- 2, - 82)

Plug

x = 2

into

4 x + \frac{8}{x} : 82

Minimum (2, 82)

Maximum (- 2, - 82), Minimum (2, 82)

f (x) = - x^{2} + 229 x

f (x) = 6 x - 0.2 x^{2}, 0 \leq x \leq 30

5 x^{5} - 3 x^{4} + 2

f (x) = 4 - x, - 3 \leq x \leq 3

f (x, y) = (x^{2} - 9)^{2} + (y^{2} - 36)^{2}

What is the extreme f(x)=4x+8/x ?
The extreme f(x)=4x+8/x is Maximum(-sqrt(2),-8sqrt(2)),Minimum(sqrt(2),8sqrt(2))