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Chapter 3: Problem 100

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.

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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.
04

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.

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