What is the extreme f(x)= 2/3 x^3-6x^2+16x ?

The extreme f(x)= 2/3 x^3-6x^2+16x is Maximum(2, 40/3),Minimum(4, 32/3)

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extreme f(x)= 2/3 x^3-6x^2+16x

extreme points

f (x) = \frac{2}{3} x^{3} - 6 x^{2} + 16 x

Maximum (2, \frac{40}{3}), Minimum (4, \frac{32}{3})

Solution steps

f^{'} (x) = 2 x^{2} - 12 x + 16

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 2

into

\frac{2}{3} x^{3} - 6 x^{2} + 16 x : \frac{40}{3}

Maximum (2, \frac{40}{3})

Plug

x = 4

into

\frac{2}{3} x^{3} - 6 x^{2} + 16 x : \frac{32}{3}

Minimum (4, \frac{32}{3})

Maximum (2, \frac{40}{3}), Minimum (4, \frac{32}{3})

What is the extreme f(x)= 2/3 x^3-6x^2+16x ?
The extreme f(x)= 2/3 x^3-6x^2+16x is Maximum(2, 40/3),Minimum(4, 32/3)