Problem 68 - Calculus of a Single Variable

Question

Critical Numbers 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.

Step-by-Step Solution

Answer

A cubic function \(f(x) = ax^3 + bx^2 + cx + d\) can have either zero, one, or two critical numbers depending on the coefficients \(a, b, c\). This is determined by the roots of its derivative \(f'(x) = 3ax^2 + 2bx + c\), and can be illustrated by the following examples: Zero critical numbers: \(f(x)=x^3\); One critical number: \(f(x)=x^3-x\); Two critical numbers: \(f(x)=x^3-3x\).

Step 1: Find the derivative of the function

For the function \(f(x) = ax^3+bx^2+cx+d\), the derivative \(f'(x)\) is \(f'(x) = 3ax^2+2bx+c\).\nAlso, remember the quadratic formula: given a quadratic equation of the form \(Ax^2+Bx+C=0\), the roots of the equation can be found using the formula \(x= (-B ±\sqrt{ B^2-4AC})/(2A)\).

Step 2: Solve the derivative for zero

Setting the derivative equal to zero, we get: \(3ax^2+2bx+c = 0\). Solving this quadratic equation, we find the critical numbers of the function.

Step 3: Discuss the possible solutions for the critical numbers

The nature of the roots depends on the discriminant (\(B^2 - 4AC\)) from the quadratic formula. If the discriminant is positive, the equation has two real roots; if it is zero, the equation has one real root; and if it's negative, there are no real roots. So in the case of the cubic function \(ax^3+bx^2+cx+d\), depending on the coefficients \(a\), \(b\), and \(c\), the function can have zero, one, or two critical points. Examples could be formulated with suitable values for \(a\), \(b\), and \(c\).

Step 4: Give examples of each case

Zero critical numbers: If we have a function like \(f(x)=x^3\), its derivative is \(f'(x) = 3x^2\), which crossed the x-axis only at negative infinity and positive infinity, so there are no real critical numbers.\n \nOne critical number: For \(f(x)=x^3-x\), the derivative \(f'(x) = 3x^2-1\) is zero for \(x = ±1/\sqrt{3}\), so there is one real critical number.\n\nTwo critical numbers: For \(f(x)=x^3-3x\), the derivative \(f'(x) = 3x^2-3\) has two real roots \(x = ±1\), thus two critical numbers.