Question

Draw the graph of f(x)= 3x4-6X3+3x2 by using the extreme points from differentiation.

Answer 1

By using differentiation and equating to zero, the extreme points can be found

$f(x)=3x^4−6x^3+3x^2 \newline \frac{\partial f}{\partial x} =12x^3−18x^2+6x=0$

which results in x=0 , x=0.5, x=1 and the corresponding points are (0,0) ,(0,1) ,(0.5,0.1875) .

The second derivative is

$\frac{\partial^2 f}{\partial x^2} =6(6x^2-6x+1)$ .

For $x \in (-\infin, \frac{3-\sqrt{3}}{6}) \cup (\frac{3+\sqrt{3}}{6},\infin)$ the second derivative has a positive sign, it means that the graph of $f(x)$ is concave up there.

For $x \in (\frac{3-\sqrt{3}}{6},\frac{3+\sqrt{3}}{6})$ the second derivative has a negative sign, it means that the graph of $f(x)$ is concave down there.

Substituting (0,0) and (1,0) the second derivative has a positive sign, it means both points are relative minimum values.

Substituting (0.5,0.1875) the second derivative has a negative value, it means it is a relative maximum value.