What is the extreme f(x)=x^3+(48)/x ?

The extreme f(x)=x^3+(48)/x is Maximum(-2,-32),Minimum(2,32)

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extreme f(x)=x^3+(48)/x

extreme points

f (x) = x^{3} + \frac{48}{x}

Maximum (- 2, - 32), Minimum (2, 32)

Solution steps

f^{'} (x) = 3 x^{2} - \frac{48}{x ^{2}}

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

x^{3} + \frac{48}{x} : - 32

Maximum (- 2, - 32)

Plug

x = 2

into

x^{3} + \frac{48}{x} : 32

Minimum (2, 32)

Maximum (- 2, - 32), Minimum (2, 32)

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

f (x) = \frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8

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

f (x, y) = 9 - 2 x + 4 y - x^{2} - 4 y^{2}

f (x) = x^{2} e^{- x}, - 1 \leq x \leq 1

What is the extreme f(x)=x^3+(48)/x ?
The extreme f(x)=x^3+(48)/x is Maximum(-2,-32),Minimum(2,32)