What is the extreme f(x)=(x^3)/3-6x^2+32x ?

The extreme f(x)=(x^3)/3-6x^2+32x is Maximum(4, 160/3),Minimum(8, 128/3)

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extreme f(x)=(x^3)/3-6x^2+32x

extreme points

f (x) = \frac{x ^{3}}{3} - 6 x^{2} + 32 x

Maximum (4, \frac{160}{3}), Minimum (8, \frac{128}{3})

Solution steps

f^{'} (x) = x^{2} - 12 x + 32

Find intervals:

Increasing

: - \infty < x < 4,

Decreasing

: 4 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 4

into

\frac{x ^{3}}{3} - 6 x^{2} + 32 x : \frac{160}{3}

Maximum (4, \frac{160}{3})

Plug

x = 8

into

\frac{x ^{3}}{3} - 6 x^{2} + 32 x : \frac{128}{3}

Minimum (8, \frac{128}{3})

Maximum (4, \frac{160}{3}), Minimum (8, \frac{128}{3})

y^{2} - x^{2}

f (x) = 4 x - 5 x^{\frac{4}{5}}

f (x) = 3 x^{4} + 3 x^{3} + 3

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

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

What is the extreme f(x)=(x^3)/3-6x^2+32x ?
The extreme f(x)=(x^3)/3-6x^2+32x is Maximum(4, 160/3),Minimum(8, 128/3)