Question

Consider the cubic function \(f(x)=a x^{3}+b x^{2}+c x+d\) where \(a \neq 0 .\) Show that \(f\) can have zero, one, or two critical numbers and give an example of each case.

Short answer

A cubic function \(f(x)=ax^{3}+bx^{2}+cx+d\) can have zero, one or two critical numbers. The number of critical numbers is determined by the discriminant \(b^{2}-4ac\). However, a zero critical numbers scenario is existentially elusive in nature. Meanwhile, cubic functions, such as \(f(x)=x^{3}\) and \(f(x)=x^{3}-3x\), can serve as practical examples of one and two critical numbers scenarios respectively.

Step-by-step solution

Compute the derivative of the function

The derivative of the function \(f(x)=ax^{3}+bx^{2}+cx+d\) is \(f'(x)=3ax^{2}+2bx+c\).

Solving the derivative equation

Critical numbers are obtained by solving the equation \(f'(x)=0\), which reduces to \(3ax^{2}+2bx+c=0\). The solutions to this quadratic equation are given by \(x_{1,2}=[-b \pm \sqrt{b^{2}-4ac}]/(2a)\). To have real roots, the condition is \(b^{2}-4ac \geq 0\). If \(b^{2}-4ac < 0\), then the function has no real critical numbers (zero critical numbers); if \(b^{2}-4ac = 0\), the function has one real critical number; and if \(b^{2}-4ac > 0\), the function has two real critical numbers. These are the three possible situations for critical numbers of the cubic function.

Zero Critical Numbers Example

Consider \(f(x)=x^{3}-3x+2\) where \(a=1, b=0, c=-3\) from our standard form. Here, \(b^{2}-4ac = {-3}^{2}-4*1*2 = 9-8 =1 > 0\), it indicates there are 2 critical numbers, which is not desired. A cubic function without a real critical number is not easily producible due to the inherent nature of cubic functions.

One Critical Number Example

Let us consider \(f(x)=x^{3}\). Here, \(a=1, b=0, c=0\). The derivative is \(f'(x)=3x^{2}\) and solving \(f'(x)=0\) gives only one root \(x=0\). So, it only has one critical point.

Two Critical Numbers Example

Consider \(f(x)=x^{3}-3x\). Here, \(a=1, b=0, c=-3\). The derivative is \(f'(x)=3x^{2}-3\) and solving \(f'(x)=0\) gives two roots \(x=\sqrt{1},-\sqrt{1}\). So, there are two critical points, namely \(x=1,-1\).