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

The extreme f(x)=-(x^2)/2+x/3 is Maximum(1/3 , 1/18)

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

extreme points

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

Maximum (\frac{1}{3}, \frac{1}{18})

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Plug

x = \frac{1}{3}

into

- \frac{x ^{2}}{2} + \frac{x}{3} : \frac{1}{18}

Maximum (\frac{1}{3}, \frac{1}{18})

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

f (x) = 14 x + 19 x^{\frac{14}{19}}

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

f (x) = 3 cos (5 x) + sin (5 x)

f (x) = 11 x_{1} + 6 x_{1} x_{2} + 3 x_{2}

What is the extreme f(x)=-(x^2)/2+x/3 ?
The extreme f(x)=-(x^2)/2+x/3 is Maximum(1/3 , 1/18)