Problem 100 - Essential Calculus

Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. An \(n\) th-degree polynomial has at most \((n-1)\) critical numbers.

Step-by-Step Solution

Answer

The statement is false. An nth-degree polynomial can have at most n critical numbers, not \(n-1\). A counter example is the polynomial \(P(x) = x^2\), which has one critical number, not zero, as the statement would suggest.

Step 1: Understanding the terms

Firstly, understand what an nth-degree polynomial and critical numbers are. An nth-degree polynomial is a polynomial where the highest power of the variable is n. Critical numbers of a function are points in the function where the derivative of the function is either zero or undefined.

Step 2: Determining the validity of the statement

To determine the accuracy of the statement, consider a polynomial of nth-degree. According to the fundamental theorem of calculus, such a polynomial has at most n roots. Now, recall the definition of a critical number - it is a value in the domain of the function where the derivative is either zero or non-existent.

Step 3: Contradicting the Statement

Since the derivative of an nth-degree polynomial (which is an n-minues-1 degree polynomial) also has at most n roots, it implies that there can exist at most n critical points for an nth-degree polynomial. Therefore, the original statement is incorrect.

Step 4: Providing counter example

Consider the polynomial \(P(x) = x^2\). This is a 2nd degree polynomial. Its derivative is \(P'(x) = 2x\), which is zero at \(x = 0\). Thus the polynomial \(x^2\) has one critical point, and not at most \(2 - 1 = 1\), proving that the statement is false.