What is the extreme f(x)= 1/3 x^3-3x^2+8x ?

The extreme f(x)= 1/3 x^3-3x^2+8x is Maximum(2, 20/3),Minimum(4, 16/3)

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extreme f(x)= 1/3 x^3-3x^2+8x

extreme points

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

Maximum (2, \frac{20}{3}), Minimum (4, \frac{16}{3})

Solution steps

f^{'} (x) = x^{2} - 6 x + 8

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 2

into

\frac{1}{3} x^{3} - 3 x^{2} + 8 x : \frac{20}{3}

Maximum (2, \frac{20}{3})

Plug

x = 4

into

\frac{1}{3} x^{3} - 3 x^{2} + 8 x : \frac{16}{3}

Minimum (4, \frac{16}{3})

Maximum (2, \frac{20}{3}), Minimum (4, \frac{16}{3})

\frac{1}{x - 6}

2 x + 2

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

f (x) = - 3 - \frac{36}{x - 2}

f (x) = - x^{4} + 3 x^{3} - 2

What is the extreme f(x)= 1/3 x^3-3x^2+8x ?
The extreme f(x)= 1/3 x^3-3x^2+8x is Maximum(2, 20/3),Minimum(4, 16/3)