Critical Points — Question 5

Question 5 - Solution

We are given $$f(x) = x^4 - 4x^3 + 6x^2.$$

(a) Find the critical points.

Differentiate: $$f'(x) = 4x^3 - 12x^2 + 12x.$$

Factor: $$f'(x) = 4x(x^2 - 3x + 3).$$

Set $f'(x)=0$: $$4x(x^2 - 3x + 3)=0.$$

This gives $x=0$, or $x^2 - 3x + 3=0$. The quadratic has discriminant $$(-3)^2 - 4(1)(3) = 9 - 12 = -3 < 0,$$ so it has no real solutions. Therefore,

$$\text{Only critical point: } x=0.$$

(b) Classify the critical point.

Compute the second derivative: $$f''(x)=12x^2 - 24x + 12.$$

Evaluate at $x=0$: $$f''(0)=12>0,$$ so $f(x)$ has a local minimum at $x=0$. Also $f(0)=0$, so the minimum point is $(0,0)$.

Conclusion $$\boxed{\text{Local minimum at } x=0 \text{ (point } (0,0)\text{)}}$$

Graph of $f(x)$ with the critical point marked