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

The extreme f(x)=3x^3-3x^2-3x+5 is Maximum(-1/3 , 50/9),Minimum(1,2)

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

extreme points

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

Maximum (- \frac{1}{3}, \frac{50}{9}), Minimum (1, 2)

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 + 5 : \frac{50}{9}

Maximum (- \frac{1}{3}, \frac{50}{9})

Plug

x = 1

into

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

Minimum (1, 2)

Maximum (- \frac{1}{3}, \frac{50}{9}), Minimum (1, 2)

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

y = 2 x^{3} + 3 x^{2} - 12 x + 5

f (x) = 4 x + 4 sin (x), 0 \leq x \leq 2 π

f (x, y) = x^{2} + 7 x y + y^{2}

f (x) = 2 x^{2} - 16 ln (x)

What is the extreme f(x)=3x^3-3x^2-3x+5 ?
The extreme f(x)=3x^3-3x^2-3x+5 is Maximum(-1/3 , 50/9),Minimum(1,2)