What is the extreme f(x)=(x^4)/2+3x^2-2x ?

The extreme f(x)=(x^4)/2+3x^2-2x is Minimum(0.32218…,-0.32757…)

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extreme f(x)=(x^4)/2+3x^2-2x

extreme points

f (x) = \frac{x ^{4}}{2} + 3 x^{2} - 2 x

Minimum (0.32218 \dots, - 0.32757 \dots)

Solution steps

Find the critical points:

x \approx 0.32218 \dots

Domain of

\frac{x ^{4}}{2} + 3 x^{2} - 2 x : - \infty < x < \infty

Find intervals:

Decreasing

: - \infty < x < 0.32218 \dots,

Increasing

: 0.32218 \dots < x < \infty

Plug the extreme point

x = 0.32218 \dots

into

\frac{x ^{4}}{2} + 3 x^{2} - 2 x \Rightarrow y = - 0.32757 \dots

Minimum (0.32218 \dots, - 0.32757 \dots)

y = \frac{x ^{3}}{x ^{2} - 7}

g (x) = \frac{x ^{2} + 5}{x + 2}

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

x^{2} + 6 x - 2

What is the extreme f(x)=(x^4)/2+3x^2-2x ?
The extreme f(x)=(x^4)/2+3x^2-2x is Minimum(0.32218…,-0.32757…)