What is the extreme f(x)=2x+(32)/x ,x\ne 0 ?

The extreme f(x)=2x+(32)/x ,x\ne 0 is Maximum(-4,-16),Minimum(4,16)

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extreme f(x)=2x+(32)/x ,x\ne 0

extreme points

f (x) = 2 x + \frac{32}{x}, x \neq = 0

Maximum (- 4, - 16), Minimum (4, 16)

Solution steps

Find the critical points:

x = - 4, x = 4

Domain of

2 x + \frac{32}{x}, x \neq = 0 : x < 0 or x > 0

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug the extreme point

x = - 4

into

2 x + \frac{32}{x} \Rightarrow y = - 16

Maximum (- 4, - 16)

Plug the extreme point

x = 4

into

2 x + \frac{32}{x} \Rightarrow y = 16

Minimum (4, 16)

Maximum (- 4, - 16), Minimum (4, 16)

f (\frac{24}{13}, \frac{88}{39}) = 3 x y - 4 x^{2} - 3 y^{2} + 8 x + 8 y - 4

- x^{3} + 75 x + 80

f (x) = x^{3} + 6 x^{2} + 5

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

f (x, y) = (x - x^{2}) (y^{2} - 4)

What is the extreme f(x)=2x+(32)/x ,x\ne 0 ?
The extreme f(x)=2x+(32)/x ,x\ne 0 is Maximum(-4,-16),Minimum(4,16)