What is the extreme f(x)=3x^3-3x^2-3x+10 ?

The extreme f(x)=3x^3-3x^2-3x+10 is Maximum(-1/3 , 95/9),Minimum(1,7)

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extreme f(x)=3x^3-3x^2-3x+10

extreme points

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

Maximum (- \frac{1}{3}, \frac{95}{9}), Minimum (1, 7)

Solution steps

f^{'} (x) = 9 x^{2} - 6 x - 3

Find intervals:

Increasing

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

Decreasing

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

Increasing

: 1 < x < \infty

Plug

x = - \frac{1}{3}

into

3 x^{3} - 3 x^{2} - 3 x + 10 : \frac{95}{9}

Maximum (- \frac{1}{3}, \frac{95}{9})

Plug

x = 1

into

3 x^{3} - 3 x^{2} - 3 x + 10 : 7

Minimum (1, 7)

Maximum (- \frac{1}{3}, \frac{95}{9}), Minimum (1, 7)

0.5 x + 15 + \frac{5000}{x}

f (x) = 5 x^{3} - 6

f (x) = x^{2} + \frac{10}{x}

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

y = \frac{x \cdot ( 1 + x ) \cdot ( 1 - x )}{1 + 2 x}

What is the extreme f(x)=3x^3-3x^2-3x+10 ?
The extreme f(x)=3x^3-3x^2-3x+10 is Maximum(-1/3 , 95/9),Minimum(1,7)