What is the extreme f(x)= 1/6 (x^3-6x^2+9x+6) ?

The extreme f(x)= 1/6 (x^3-6x^2+9x+6) is Maximum(1, 5/3),Minimum(3,1)

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extreme f(x)= 1/6 (x^3-6x^2+9x+6)

extreme points

f (x) = \frac{1}{6} (x^{3} - 6 x^{2} + 9 x + 6)

Maximum (1, \frac{5}{3}), Minimum (3, 1)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 1

into

\frac{1}{6} (x^{3} - 6 x^{2} + 9 x + 6) : \frac{5}{3}

Maximum (1, \frac{5}{3})

Plug

x = 3

into

\frac{1}{6} (x^{3} - 6 x^{2} + 9 x + 6) : 1

Minimum (3, 1)

Maximum (1, \frac{5}{3}), Minimum (3, 1)

What is the extreme f(x)= 1/6 (x^3-6x^2+9x+6) ?
The extreme f(x)= 1/6 (x^3-6x^2+9x+6) is Maximum(1, 5/3),Minimum(3,1)