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

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

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

extreme points

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

Minimum (- \frac{1}{3}, - \frac{1}{18})

Solution steps

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

Find intervals:

Decreasing

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

Increasing

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

Plug

x = - \frac{1}{3}

into

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

Minimum (- \frac{1}{3}, - \frac{1}{18})

y = x^{4} - 4 x^{3}

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

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

f (x, y) = ∣ x y ∣

f (x) = \frac{x ^{2} - 8}{x - 3}

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