Question

Find (and classify) the critical points of the following function and determine if they are local max, local min, or neither: f(x) =x^4 - 2x^2

Answer 1

Solution:

Given:

$f(x)=x^4-2x^2$

To get the local maximum and minimum point, we differentiate the function twice.

$\begin{gathered} f^{\prime}(x)=4x^3-4x \\ \\ To\text{ get the critical points, f'(x)=0} \\ 4x^3-4x=0 \\ \text{Factorizing,} \\ 4x(x^2-1)=0 \\ \text{Thus,} \\ (4x)(x-1)(x+1)=0 \\ \text{Hence, the critical points are;} \\ x=0,x=1,x=-1 \end{gathered}$

To get the maximum and minimum points, we differentiate further,

$\begin{gathered} f^{\prime}(x)=4x^3-4x \\ f^{\doubleprime}(x)=12x^2-4 \\ \\ \text{Inputting the critical points into f''(x),} \\ \text{when x = -1,} \\ f^{\doubleprime}(-1)=12(-1)^2-4=12-4=8 \\ S\text{ ince f''(x)>0, then this is a minimum point.} \\ x=-1\text{ is a minimum} \\ \\ \\ \text{when x = 1,} \\ f^{\doubleprime}(1)=12(1^2)-4=12-4=8 \\ S\text{ ince f''(x)>0, then this is a minimum point.} \\ x=1\text{ is a minimum} \\ \\ \text{when x = 0,} \\ f^{\doubleprime}(0)=12(0^2)-4=0-4=-4 \\ S\text{ ince f''(x)<0, then this is a maximum point.} \\ x=0\text{ is a maximum} \end{gathered}$

Thus,

$\begin{gathered} At\text{ x = 0,} \\ f(x)=x^4-2x^2 \\ f(0)=0^4-2(0^2)=0-0=0 \\ \\ \text{Hence, (0,0) is a local maximum} \\ \\ \\ At\text{ x = -1,} \\ f(x)=x^4-2x^2 \\ f(-1)=(-1)^4-2(-1)^2=1-2=-1 \\ \\ \text{Hence, (-1,-1) is a local minimum} \\ \\ \\ At\text{ x = 1,} \\ f(x)=x^4-2x^2 \\ f(1)=1^4-2(1^2) \\ f(1)=1-2=-1 \\ \\ \text{Hence, (1,-1) is a local minimum} \end{gathered}$

The graph is as shown below;

Therefore,

The local minimum occurs at (-1,-1) and (1,-1)

The local maximum occurs at (0,0)