What is the extreme f(x)=-3x^3+3x^2-x-1 ?

The extreme f(x)=-3x^3+3x^2-x-1 is Saddle(1/3 ,-10/9)

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extreme f(x)=-3x^3+3x^2-x-1

extreme points

f (x) = - 3 x^{3} + 3 x^{2} - x - 1

Saddle (\frac{1}{3}, - \frac{10}{9})

Solution steps

f^{'} (x) = - 9 x^{2} + 6 x - 1

Find intervals:

Decreasing

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

Decreasing

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

Plug

x = \frac{1}{3}

into

- 3 x^{3} + 3 x^{2} - x - 1 : - \frac{10}{9}

Saddle (\frac{1}{3}, - \frac{10}{9})

- x^{3} - 6 x^{2}

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

f (x) = \frac{x ^{3}}{4 - x ^{2}}

f (x) = sin^{2} (x), 0 \leq x \leq \frac{2 π}{3}

3 x^{2} e^{x}

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