What is the extreme f(x)=(2x^2+8)/x ?

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

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extreme f(x)=(2x^2+8)/x

extreme points

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

Maximum (- 2, - 8), Minimum (2, 8)

Solution steps

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

\frac{2 x ^{2} + 8}{x} : - 8

Maximum (- 2, - 8)

Plug

x = 2

into

\frac{2 x ^{2} + 8}{x} : 8

Minimum (2, 8)

Maximum (- 2, - 8), Minimum (2, 8)

f (x) = 9 x^{3} - 2 7^{2} + 5 x + 42

f (x) = x^{3} - 2 x^{2} - 15 x + 5 (2.2)

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

b - sin (x)

f (x) = x^{3} + 15 x - 2

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