What is the extreme f(x)=3x^3-5x^2+x+3 ?

The extreme f(x)=3x^3-5x^2+x+3 is Maximum(1/9 , 742/243),Minimum(1,2)

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extreme f(x)=3x^3-5x^2+x+3

extreme points

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

Maximum (\frac{1}{9}, \frac{742}{243}), Minimum (1, 2)

Solution steps

f^{'} (x) = 9 x^{2} - 10 x + 1

Find intervals:

Increasing

: - \infty < x < \frac{1}{9},

Decreasing

: \frac{1}{9} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = \frac{1}{9}

into

3 x^{3} - 5 x^{2} + x + 3 : \frac{742}{243}

Maximum (\frac{1}{9}, \frac{742}{243})

Plug

x = 1

into

3 x^{3} - 5 x^{2} + x + 3 : 2

Minimum (1, 2)

Maximum (\frac{1}{9}, \frac{742}{243}), Minimum (1, 2)

u (x, y) = y^{2} e^{x} + 3 x^{2}

f (x) = - \frac{2 x}{5 x ^{2} + 8}

f (x) = 2 x^{3} + 6 x - 90 x + 5, - 5 < x < 4

y = 4 x^{2} - 24 x + 1

f (x) = \frac{x ^{4}}{4} - \frac{8 x ^{3}}{3} - 32 x^{2} + 512 x

What is the extreme f(x)=3x^3-5x^2+x+3 ?
The extreme f(x)=3x^3-5x^2+x+3 is Maximum(1/9 , 742/243),Minimum(1,2)