What is the extreme f(1)= 2/3 x^3-4x^2+6x+2 ?

The extreme f(1)= 2/3 x^3-4x^2+6x+2 is Maximum(1, 14/3),Minimum(3,2)

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extreme f(1)= 2/3 x^3-4x^2+6x+2

extreme points

f (1) = \frac{2}{3} x^{3} - 4 x^{2} + 6 x + 2

Maximum (1, \frac{14}{3}), Minimum (3, 2)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 1

into

\frac{2}{3} x^{3} - 4 x^{2} + 6 x + 2 : \frac{14}{3}

Maximum (1, \frac{14}{3})

Plug

x = 3

into

\frac{2}{3} x^{3} - 4 x^{2} + 6 x + 2 : 2

Minimum (3, 2)

Maximum (1, \frac{14}{3}), Minimum (3, 2)

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

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} + 1 [- 2.3]

f (x) = \frac{1}{3} x^{3} - x^{2} + x + 1, 0 \leq x \leq 5

f (x) = 8 x^{3} - 3 x y + y^{3}

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

What is the extreme f(1)= 2/3 x^3-4x^2+6x+2 ?
The extreme f(1)= 2/3 x^3-4x^2+6x+2 is Maximum(1, 14/3),Minimum(3,2)