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

The extreme f(x)=-3x^3+3x^2-x-5 is Saddle(1/3 ,-46/9)

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

extreme points

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

Saddle (\frac{1}{3}, - \frac{46}{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 - 5 : - \frac{46}{9}

Saddle (\frac{1}{3}, - \frac{46}{9})

5 x^{\frac{2}{3}} - x^{\frac{1}{3}}

2 sin^{2} (x)

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

y = 5 - 2 x^{2}, - 4 \leq x \leq 2

f (x) = x^{3} - \frac{7}{2} x^{2} + 2 x - 5

What is the extreme f(x)=-3x^3+3x^2-x-5 ?
The extreme f(x)=-3x^3+3x^2-x-5 is Saddle(1/3 ,-46/9)