What is the extreme f(x)=x^3-5x^2+3x+12 ?

The extreme f(x)=x^3-5x^2+3x+12 is Maximum(1/3 , 337/27),Minimum(3,3)

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extreme f(x)=x^3-5x^2+3x+12

extreme points

f (x) = x^{3} - 5 x^{2} + 3 x + 12

Maximum (\frac{1}{3}, \frac{337}{27}), Minimum (3, 3)

Solution steps

f^{'} (x) = 3 x^{2} - 10 x + 3

Find intervals:

Increasing

: - \infty < x < \frac{1}{3},

Decreasing

: \frac{1}{3} < x < 3,

Increasing

: 3 < x < \infty

Plug

x = \frac{1}{3}

into

x^{3} - 5 x^{2} + 3 x + 12 : \frac{337}{27}

Maximum (\frac{1}{3}, \frac{337}{27})

Plug

x = 3

into

x^{3} - 5 x^{2} + 3 x + 12 : 3

Minimum (3, 3)

Maximum (\frac{1}{3}, \frac{337}{27}), Minimum (3, 3)

f (x) = (x - 8) e^{- 9 x}

3 x^{4} - 8 x^{3} + 6 x^{2}

\frac{1}{x ^{2} + 1}

f (x, y) = x^{2} + y^{2} + 1

f (x, y) = 4 x^{2} + 2 y^{2} - 2 x y

What is the extreme f(x)=x^3-5x^2+3x+12 ?
The extreme f(x)=x^3-5x^2+3x+12 is Maximum(1/3 , 337/27),Minimum(3,3)