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

The extreme f(x)=(x^3)/3-2x^2+3x is Maximum(1, 4/3),Minimum(3,0)

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

extreme points

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

Maximum (1, \frac{4}{3}), Minimum (3, 0)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 1

into

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

Maximum (1, \frac{4}{3})

Plug

x = 3

into

\frac{x ^{3}}{3} - 2 x^{2} + 3 x : 0

Minimum (3, 0)

Maximum (1, \frac{4}{3}), Minimum (3, 0)

f (x) = x + \frac{64}{x}

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

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

f (x, y) = y^{3} + 9 y^{2} - 8 x y + 5 x^{2} - 2

\frac{e ^{- x}}{x + 1}

What is the extreme f(x)=(x^3)/3-2x^2+3x ?
The extreme f(x)=(x^3)/3-2x^2+3x is Maximum(1, 4/3),Minimum(3,0)