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

The extreme f(x)=(3x^2)/2-2x-3 is Minimum(2/3 ,-11/3)

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

extreme points

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

Minimum (\frac{2}{3}, - \frac{11}{3})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < \frac{2}{3},

Increasing

: \frac{2}{3} < x < \infty

Plug

x = \frac{2}{3}

into

\frac{3 x ^{2}}{2} - 2 x - 3 : - \frac{11}{3}

Minimum (\frac{2}{3}, - \frac{11}{3})

y = x^{2} e^{- z}

f (x) = x^{12} - 12 x

y = \frac{1}{3} x^{2} + 2 x + 5

f (x) = 2 x^{2} + 8 y^{2}

f (x) = 3 x^{5} - 25 x^{3} + 60 x + 1

What is the extreme f(x)=(3x^2)/2-2x-3 ?
The extreme f(x)=(3x^2)/2-2x-3 is Minimum(2/3 ,-11/3)