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

The extreme f(x)=(2x^3)/3-x^2+x/2+1/2 is Saddle(1/2 , 7/12)

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

extreme points

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

Saddle (\frac{1}{2}, \frac{7}{12})

Solution steps

f^{'} (x) = 2 x^{2} - 2 x + \frac{1}{2}

Find intervals:

Increasing

: - \infty < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = \frac{1}{2}

into

\frac{2 x ^{3}}{3} - x^{2} + \frac{x}{2} + \frac{1}{2} : \frac{7}{12}

Saddle (\frac{1}{2}, \frac{7}{12})

y = x^{5} - 15 x^{3} + 100

f (x) = (ln (x^{2} + 1))

f (x) = 3 x + 5 x^{- 1}

f (x) = (x - 1)^{5}

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

What is the extreme f(x)=(2x^3)/3-x^2+x/2+1/2 ?
The extreme f(x)=(2x^3)/3-x^2+x/2+1/2 is Saddle(1/2 , 7/12)