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

The extreme f(x)=(x^3)/3-3x^2+8x-2 is Maximum(2, 14/3),Minimum(4, 10/3)

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

extreme points

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

Maximum (2, \frac{14}{3}), Minimum (4, \frac{10}{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{x ^{3}}{3} - 3 x^{2} + 8 x - 2 : \frac{14}{3}

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

Plug

x = 4

into

\frac{x ^{3}}{3} - 3 x^{2} + 8 x - 2 : \frac{10}{3}

Minimum (4, \frac{10}{3})

Maximum (2, \frac{14}{3}), Minimum (4, \frac{10}{3})

f (x) = \frac{x ^{3}}{3} - 3 x^{2} + 8 x - 4

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

f (x) = 3 x^{2} - 18 x + 5

f (x) = \frac{7}{4} x^{2} - \frac{21}{2} x - \frac{43}{3}

x \cdot e^{- x^{2} + 1}

What is the extreme f(x)=(x^3)/3-3x^2+8x-2 ?
The extreme f(x)=(x^3)/3-3x^2+8x-2 is Maximum(2, 14/3),Minimum(4, 10/3)