Question

discuss the function f(x) = x^4/4 - 3x^2/2 for where mimimum and maximum

Answer 1

Solution.

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

Find $f'(x)=4\frac{x^3}{4}-3\frac{2x}{2}=x^3-3x.$

Make and solve equation $f'(x)=0.$

$$x^3-3x=0,\newline x(x^2-3)=0,\newline x_1=0, \text{ or }x_2=\sqrt{3},\text{ or }x_3=-\sqrt{3}.$$

Investigate the sign of the derivative at each interval:

From here $\sqrt{3}$ and $-\sqrt{3}$ are minimum points, is maximum point.

$f_{min}(\sqrt{3})=\frac{\sqrt{3}^4}{4}-\frac{3\sqrt{3}^2}{2}=\frac{9}{4}-\frac{9}{2}= -\frac{9}{4}=-2\frac{1}{4}.$

$f_{min}(-\sqrt{3})=\frac{(-\sqrt{3})^4}{4}-\frac{3(-\sqrt{3})^2}{2}=\frac{9}{4}-\frac{9}{2}= -\frac{9}{4}=-2\frac{1}{4}.$

$f_{max}(0)=0.$

Answer. $f_{min}=-2\frac{1}{4}, f_{max}=0.$